(a) We would like to rewrite the expression $color{#4257b2}sin acos 2a+cos asin 2a$ as a single trigonometric ratio.

First, we note that our expression is on the form of the addition formula of the sine function $color{#4257b2}sin left(x+yright)=sin xcos y+cos xsin y$ where $color{#4257b2}x$ and $color{#4257b2}y$ are considered to be $color{#4257b2}a$ and $color{#4257b2}2a$ in our expression, so we can use the addition formula for the sine function to rewrite our expression as a single trigonometric ratio.

$$
begin{align*}
sin acos 2a+cos asin 2a&=sinleft(a+2aright)
\ \
&=boxed{ sin 3a }
end{align*}
$$

(b) We would like to rewrite the expression $color{#4257b2}cos 4xcos 3x-sin 4xsin 3x$ as a single trigonometric ratio.

First, we note that our expression is on the form of the addition formula of the cosine function $color{#4257b2}cos left(a+bright)=cos acos b-sin asin b$ where $color{#4257b2}a$ and $color{#4257b2}b$ are considered to be $color{#4257b2}4x$ and $color{#4257b2}3x$ in our expression, so we can use the addition formula for the cosine function to rewrite our expression as a single trigonometric ratio.

$$
begin{align*}
cos 4xcos 3x-sin 4xsin 3x&=cosleft(4x+3xright)
\ \
&=boxed{ cos 7x }
end{align*}
$$

$$
text{color{#c34632}$(a) sin 3a$ $(b) cos 7x$}
$$

(a) We would like to rewrite the expression $color{#4257b2}dfrac{tan 170text{textdegree}-tan 110text{textdegree}}{1+tan 170text{textdegree}tan 110text{textdegree}}$ as a single trigonometric ratio.

First, we note that our expression is on the form of the subtraction formula of the tangentt function $color{#4257b2}tan left(a-bright)=dfrac{tan a-tan b}{1+tan atan b}$ where $color{#4257b2}a$ and $color{#4257b2}b$ are considered to be $color{#4257b2}170text{textdegree}$ and $color{#4257b2}110text{textdegree}$ in our expression, so we can use the subtraction formula for the tangent function to rewrite our expression as a single trigonometric ratio.

$$
begin{align*}
dfrac{tan 170text{textdegree}-tan 110text{textdegree}}{1+tan 170text{textdegree}tan 110text{textdegree}}&=tanleft(170text{textdegree}-110text{textdegree}right)
\ \
&=tan 60text{textdegree}
\ \
&=boxed{ sqrt{3} }
end{align*}
$$

Note that $color{#4257b2}60text{textdegree}$ is a special angle and the value of the tangent function for it is $color{#4257b2}sqrt{3}$

(b) We would like to rewrite the expression $color{#4257b2}cos dfrac{5pi}{12}cos dfrac{pi}{12}+sin dfrac{5pi}{12}sin dfrac{pi}{12}$ as a single trigonometric ratio.

First, we note that our expression is on the form of the subtraction formula of the cosine function $color{#4257b2}cos left(a-bright)=cos acos b+sin asin b$ where $color{#4257b2}a$ and $color{#4257b2}b$ are considered to be $color{#4257b2}dfrac{5pi}{12}$ and $color{#4257b2}dfrac{pi}{12}$ in our expression, so we can use the subtraction formula for the cosine function to rewrite our expression as a single trigonometric ratio.

$$
begin{align*}
cos dfrac{5pi}{12}cos dfrac{pi}{12}+sin dfrac{5pi}{12}sin dfrac{pi}{12}&=cosleft(dfrac{5pi}{12}-dfrac{pi}{12}right)
\ \
&=cos dfrac{4pi}{12}
\ \
&=cos dfrac{pi}{3}
\ \
&=boxed{ dfrac{1}{2} }
end{align*}
$$

Note that $color{#4257b2}dfrac{pi}{3}$ is a special angle and the value of the cosine function for it is $color{#4257b2}dfrac{1}{2}$

$$
text{color{#c34632}$(a) sqrt{3}$ $(b) dfrac{1}{2}$}
$$

(a) We would like to express the angle $color{#4257b2}75text{textdegree}$ as a compound angle using pair of angles from the special triangles. First, we know that $color{#4257b2}45text{textdegree}+30text{textdegree}=75text{textdegree}$ and the angles $color{#4257b2}45text{textdegree}$ and $color{#4257b2}30text{textdegree}$ both of them are special angles, so we can express $color{#4257b2}75text{textdegree}$ as $color{#4257b2}45text{textdegree}+30text{textdegree}$.

$$
boxed{ 75text{textdegree}=45text{textdegree}+30text{textdegree} }
$$

(b) We would like to express the angle $color{#4257b2}-15text{textdegree}$ as a compound angle using pair of angles from the special triangles. First, we know that $color{#4257b2}30text{textdegree}-45text{textdegree}=-15text{textdegree}$ and the angles $color{#4257b2}30text{textdegree}$ and $color{#4257b2}45text{textdegree}$ both of them are special angles, so we can express $color{#4257b2}-15text{textdegree}$ as $color{#4257b2}30text{textdegree}-45text{textdegree}$.

$$
boxed{ -15text{textdegree}=30text{textdegree}-45text{textdegree} }
$$

(c) We would like to express the angle $color{#4257b2}-dfrac{pi}{6}$ as a compound angle using pair of angles from the special triangles. First, we know that $color{#4257b2}dfrac{pi}{3}-dfrac{pi}{2}=-dfrac{pi}{6}$ and the angles $color{#4257b2}dfrac{pi}{3}$ and $color{#4257b2}dfrac{pi}{2}$ both of them are special angles, so we can express $color{#4257b2}-dfrac{pi}{6}$ as $color{#4257b2}dfrac{pi}{3}-dfrac{pi}{2}$.

$$
boxed{ -dfrac{pi}{6}=dfrac{pi}{3}-dfrac{pi}{2} }
$$

(d) We would like to express the angle $color{#4257b2}dfrac{pi}{12}$ as a compound angle using pair of angles from the special triangles. First, we know that $color{#4257b2}dfrac{pi}{4}-dfrac{pi}{6}=dfrac{pi}{12}$ and the angles $color{#4257b2}dfrac{pi}{4}$ and $color{#4257b2}dfrac{pi}{6}$ both of them are special angles, so we can express $color{#4257b2}dfrac{pi}{12}$ as $color{#4257b2}dfrac{pi}{4}-dfrac{pi}{6}$.

$$
boxed{ dfrac{pi}{12}=dfrac{pi}{4}-dfrac{pi}{6} }
$$

(e) We would like to express the angle $color{#4257b2}105text{textdegree}$ as a compound angle using pair of angles from the special triangles. First, we know that $color{#4257b2}45text{textdegree}+60text{textdegree}=105text{textdegree}$ and the angles $color{#4257b2}45text{textdegree}$ and $color{#4257b2}60text{textdegree}$ both of them are special angles, so we can express $color{#4257b2}105text{textdegree}$ as $color{#4257b2}45text{textdegree}+60text{textdegree}$.

$$
boxed{ 105text{textdegree}=45text{textdegree}+60text{textdegree} }
$$

(f) We would like to express the angle $color{#4257b2}dfrac{5pi}{6}$ as a compound angle using pair of angles from the special triangles. First, we know that $color{#4257b2}dfrac{pi}{3}+dfrac{pi}{2}=dfrac{5pi}{6}$ and the angles $color{#4257b2}dfrac{pi}{3}$ and $color{#4257b2}dfrac{pi}{2}$ both of them are special angles, so we can express $color{#4257b2}dfrac{5pi}{6}$ as $color{#4257b2}dfrac{pi}{3}+dfrac{pi}{2}$.

$$
boxed{ dfrac{5pi}{6}=dfrac{pi}{3}+dfrac{pi}{2} }
$$

$$
text{color{#c34632}$(a) 45text{textdegree}+30text{textdegree}$ $(b) 30text{textdegree}-45text{textdegree}$ $(c) dfrac{pi}{3}-dfrac{pi}{2}$
\
\
Large{color{#c34632}$(d) dfrac{pi}{4}-dfrac{pi}{6}$ $(e) 45text{textdegree}+60text{textdegree}$ $(f) dfrac{pi}{3}+dfrac{pi}{2}$}}
$$

(a) We would like to determine the exact value of the trigonometric ratio $color{#4257b2}sin 75text{textdegree}$. First, we know that $color{#4257b2}75text{textdegree}=45text{textdegree}+30text{textdegree}$, so we can rewrite $color{#4257b2}sin 75text{textdegree}$ as $color{#4257b2}sin left(45text{textdegree}+30text{textdegree}right)$. Now we have a compound angle, so we can use the addition formula for the sine function to find the exact value where
$color{#4257b2}sinleft(a+bright)=sin acos b+cos asin b$.

$$
begin{align*}
sin 75text{textdegree}&=sin left(45text{textdegree}+30text{textdegree}right)
\ \
&=sin 45text{textdegree}cos 30text{textdegree}+cos 45text{textdegree}sin 30text{textdegree}
\ \
&=dfrac{sqrt{2}}{2}cdot dfrac{sqrt{3}}{2}+dfrac{sqrt{2}}{2}cdot dfrac{1}{2}
\ \
&=dfrac{sqrt{6}}{4}+dfrac{sqrt{2}}{4}=dfrac{sqrt{6}+sqrt{2}}{4}
end{align*}
$$

Note that $color{#4257b2}45text{textdegree}$ and $color{#4257b2}30text{textdegree}$ are special angles and we know the sine and cosine values of them.

So the exact value of $color{#4257b2}sin 75text{textdegree}$ is $boxed{ dfrac{sqrt{6}+sqrt{2}}{4} }$

(b) We would like to determine the exact value of the trigonometric ratio $color{#4257b2}cos 15text{textdegree}$. First, we know that $color{#4257b2}15text{textdegree}=45text{textdegree}-30text{textdegree}$, so we can rewrite $color{#4257b2}cos 15text{textdegree}$ as $color{#4257b2}cos left(45text{textdegree}-30text{textdegree}right)$. Now we have a compound angle, so we can use the subtraction formula for the cosine function to find the exact value where
$color{#4257b2}cosleft(a-bright)=cos acos b+sin asin b$.

$$
begin{align*}
cos 15text{textdegree}&=cos left(45text{textdegree}-30text{textdegree}right)
\ \
&=cos 45text{textdegree}cos 30text{textdegree}+sin 45text{textdegree}sin 30text{textdegree}
\ \
&=dfrac{sqrt{2}}{2}cdot dfrac{sqrt{3}}{2}+dfrac{sqrt{2}}{2}cdot dfrac{1}{2}
\ \
&=dfrac{sqrt{6}}{4}+dfrac{sqrt{2}}{4}=dfrac{sqrt{6}+sqrt{2}}{4}
end{align*}
$$

Note that $color{#4257b2}45text{textdegree}$ and $color{#4257b2}30text{textdegree}$ are special angles and we know the sine and cosine values of them.

So the exact value of $color{#4257b2}cos 15text{textdegree}$ is $boxed{ dfrac{sqrt{6}+sqrt{2}}{4} }$

(c) We would like to determine the exact value of the trigonometric ratio $color{#4257b2}tan dfrac{5pi}{12}$. First, we know that $color{#4257b2}dfrac{5pi}{12}=dfrac{3pi}{12}+dfrac{2pi}{12}$, so we can rewrite $color{#4257b2}tan dfrac{5pi}{12}$ as $color{#4257b2}tan left(dfrac{3pi}{12}+dfrac{2pi}{12}right)$. Now we have a compound angle, so we can use the addition formula for the tangent function to find the exact value where $color{#4257b2}tanleft(a+bright)=dfrac{tan a+tan b}{1-tan atan b}$.

$$
begin{align*}
tan dfrac{5pi}{12}&=tan left(dfrac{3pi}{12}+dfrac{2pi}{12}right)
\ \
&=tan left(dfrac{pi}{4}+dfrac{pi}{6}right)
\ \
&=dfrac{tan dfrac{pi}{4}+tan dfrac{pi}{6}}{1-tan dfrac{pi}{4}tan dfrac{pi}{6}}
\ \
&=dfrac{1+dfrac{1}{sqrt{3}}}{1-1cdot dfrac{1}{sqrt{3}}}
\ \
&=dfrac{sqrt{3}left(1+dfrac{1}{sqrt{3}}right)}{sqrt{3}left(1-dfrac{1}{sqrt{3}}right)} text{multiply and divide by $sqrt{3}$}
\ \
&=dfrac{sqrt{3}+1}{sqrt{3}-1}
\ \
&=dfrac{left(sqrt{3}+1right)left(sqrt{3}+1right)}{left(sqrt{3}-1right)left(sqrt{3}+1right)}=dfrac{left(sqrt{3}+1right)^{2}}{left(sqrt{3}right)^{2}-1^{2}} text{multiply and divide by $sqrt{3}+1$}
\ \
&=dfrac{3+2sqrt{3}+1}{3-1}=dfrac{4+2sqrt{3}}{2}=2+sqrt{3}
end{align*}
$$

Note that $color{#4257b2}dfrac{pi}{4}$ and $color{#4257b2}dfrac{pi}{6}$ are special angles and we know the tangent values of them.

So the exact value of $color{#4257b2}tan dfrac{5pi}{12}$ is $boxed{ 2+sqrt{3} }$

(d) We would like to determine the exact value of the trigonometric ratio $color{#4257b2}sin left(-dfrac{pi}{12}right)$. First, we know that $color{#4257b2}-dfrac{pi}{12}=dfrac{2pi}{12}-dfrac{3pi}{12}$, so we can rewrite $color{#4257b2}sin left(-dfrac{pi}{12}right)$ as $color{#4257b2}sin left(dfrac{2pi}{12}-dfrac{3pi}{12}right)$. Now we have a compound angle, so we can use the subtraction formula for the sine function to find the exact value where
$color{#4257b2}sinleft(a-bright)=sin acos b-cos asin b$.

$$
begin{align*}
sin left(-dfrac{pi}{12}right)&=sin left(dfrac{2pi}{12}-dfrac{3pi}{12}right)
\ \
&=sin left(dfrac{pi}{6}-dfrac{pi}{4}right)
\ \
&=sin dfrac{pi}{6}cos dfrac{pi}{4}-cos dfrac{pi}{6}sin dfrac{pi}{4}
\ \
&=dfrac{1}{2}cdot dfrac{sqrt{2}}{2}-dfrac{sqrt{3}}{2}cdot dfrac{sqrt{2}}{2}
\ \
&=dfrac{sqrt{2}}{4}-dfrac{sqrt{6}}{4}=dfrac{sqrt{2}-sqrt{6}}{4}
end{align*}
$$

Note that $color{#4257b2}dfrac{pi}{6}$ and $color{#4257b2}dfrac{pi}{4}$ are special angles and we know the sine and cosine values of them.

So the exact value of $color{#4257b2}sin left(-dfrac{pi}{12}right)$ is $boxed{ dfrac{sqrt{2}-sqrt{6}}{4} }$

(e) We would like to determine the exact value of the trigonometric ratio $color{#4257b2}cos 105text{textdegree}$. First, we know that $color{#4257b2}105text{textdegree}=45text{textdegree}+60text{textdegree}$, so we can rewrite $color{#4257b2}cos 105text{textdegree}$ as $color{#4257b2}cos left(45text{textdegree}+60text{textdegree}right)$. Now we have a compound angle, so we can use the addition formula for the cosine function to find the exact value where
$color{#4257b2}cosleft(a+bright)=cos acos b-sin asin b$.

$$
begin{align*}
cos 105text{textdegree}&=cos left(45text{textdegree}+60text{textdegree}right)
\ \
&=cos 45text{textdegree}cos 60text{textdegree}-sin 45text{textdegree}sin 60text{textdegree}
\ \
&=dfrac{sqrt{2}}{2}cdot dfrac{1}{2}-dfrac{sqrt{2}}{2}cdot dfrac{sqrt{3}}{2}
\ \
&=dfrac{sqrt{2}}{4}-dfrac{sqrt{6}}{4}=dfrac{sqrt{2}-sqrt{6}}{4}
end{align*}
$$

Note that $color{#4257b2}45text{textdegree}$ and $color{#4257b2}60text{textdegree}$ are special angles and we know the sine and cosine values of them.

So the exact value of $color{#4257b2}cos 105text{textdegree}$ is $boxed{ dfrac{sqrt{2}-sqrt{6}}{4} }$

(f) We would like to determine the exact value of the trigonometric ratio $color{#4257b2}tan dfrac{23pi}{12}$. First, we know that $color{#4257b2}dfrac{23pi}{12}=2pi-dfrac{pi}{12}$, so we can rewrite $color{#4257b2}tan dfrac{23pi}{12}$ as $color{#4257b2}tan left(2pi-dfrac{pi}{12}right)$. But we know from the transformations that
$color{#4257b2}tan left(2pi-thetaright)=-tan theta$, so we can rewrite $color{#4257b2}tan left(2pi-dfrac{pi}{12}right)$ as $color{#4257b2}-tan dfrac{pi}{12}$ which equals $color{#4257b2}-tan left(dfrac{3pi}{12}-dfrac{2pi}{12}right)$. Now we have a compound angle, so we can use the subtraction formula for the tangent function to find the exact value where $color{#4257b2}tanleft(a-bright)=dfrac{tan a-tan b}{1+tan atan b}$.

$$
begin{align*}
tan dfrac{23pi}{12}&=-tan left(dfrac{3pi}{12}-dfrac{2pi}{12}right)
\ \
&=-tan left(dfrac{pi}{4}-dfrac{pi}{6}right)
\ \
&=-left(dfrac{tan dfrac{pi}{4}-tan dfrac{pi}{6}}{1+tan dfrac{pi}{4}tan dfrac{pi}{6}}right)
\ \
&=-left(dfrac{1-dfrac{1}{sqrt{3}}}{1+1cdot dfrac{1}{sqrt{3}}}right)
\ \
&=-left[dfrac{sqrt{3}left(1-dfrac{1}{sqrt{3}}right)}{sqrt{3}left(1+dfrac{1}{sqrt{3}}right)}right] text{multiply and divide by $sqrt{3}$}
\ \
&=-left(dfrac{sqrt{3}-1}{sqrt{3}+1}right)
\ \
&=-left[dfrac{left(sqrt{3}-1right)left(sqrt{3}-1right)}{left(sqrt{3}+1right)left(sqrt{3}-1right)}right]=-left[dfrac{left(sqrt{3}-1right)^{2}}{left(sqrt{3}right)^{2}-1^{2}}right] text{multiply and divide by $sqrt{3}-1$}
\ \
&=-left(dfrac{3-2sqrt{3}+1}{3-1}right)=-left(dfrac{4-2sqrt{3}}{2}right)=-2+sqrt{3}
end{align*}
$$

Note that $color{#4257b2}dfrac{pi}{4}$ and $color{#4257b2}dfrac{pi}{6}$ are special angles and we know the tangent values of them.

So the exact value of $color{#4257b2}tan dfrac{23pi}{12}$ is $boxed{ -2+sqrt{3} }$

$$
text{color{#c34632}$(a) dfrac{sqrt{6}+sqrt{2}}{4}$ $(b) dfrac{sqrt{6}+sqrt{2}}{4}$ $(c) 2+sqrt{3}$
\
\
Large{color{#c34632}$(d) dfrac{sqrt{2}-sqrt{6}}{4}$ $(e) dfrac{sqrt{2}-sqrt{6}}{4}$ $(f) -2+sqrt{3}$}}
$$

(a) We would like to determine the exact value of the expression $color{#4257b2}sinleft(pi+dfrac{pi}{6}right)$ using an appropriate compound angle formula. First, we note that our expression is a sine function which consists of two sum angles, so we can use the addition formula for the sine function where
$color{#4257b2}sinleft(a+bright)=sin acos b+cos asin b$.

$$
begin{align*}
sinleft(pi+dfrac{pi}{6}right)&=sin picos dfrac{pi}{6}+cos pisin dfrac{pi}{6}
\ \
&=0cdot dfrac{sqrt{3}}{2}+(-1)cdot dfrac{1}{2}
\ \
&=0-dfrac{1}{2}=-dfrac{1}{2}
end{align*}
$$

Note that $color{#4257b2}pi$ and $color{#4257b2}dfrac{pi}{6}$ are special angles and we know the sine and cosine values of them.

So the exact value of $color{#4257b2}sinleft(pi+dfrac{pi}{6}right)$ is $boxed{ -dfrac{1}{2} }$

(b) We would like to determine the exact value of the expression $color{#4257b2}cosleft(pi-dfrac{pi}{4}right)$ using an appropriate compound angle formula. First, we note that our expression is a cosine function which consists of two subtraction angles, so we can use the subtraction formula for the cosine function where
$color{#4257b2}cosleft(a-bright)=cos acos b+sin asin b$.

$$
begin{align*}
cosleft(pi-dfrac{pi}{4}right)&=cos picos dfrac{pi}{4}+sin pisin dfrac{pi}{4}
\ \
&=(-1)cdot dfrac{sqrt{2}}{2}+0cdot dfrac{sqrt{2}}{2}
\ \
&=-dfrac{sqrt{2}}{2}+0=-dfrac{sqrt{2}}{2}
end{align*}
$$

Note that $color{#4257b2}pi$ and $color{#4257b2}dfrac{pi}{4}$ are special angles and we know the sine and cosine values of them.

So the exact value of $color{#4257b2}cosleft(pi-dfrac{pi}{4}right)$ is $boxed{ -dfrac{sqrt{2}}{2} }$

(c) We would like to determine the exact value of the expression $color{#4257b2}tanleft(dfrac{pi}{4}+piright)$ using an appropriate compound angle formula. First, we note that our expression is a tangent function which consists of two sum angles, so we can use the addition formula for the tangent function where
$color{#4257b2}tanleft(a+bright)=dfrac{tan a+tan b}{1-tan atan b}$.

$$
begin{align*}
tanleft(dfrac{pi}{4}+piright)&=dfrac{tan dfrac{pi}{4}+tan pi}{1-tan dfrac{pi}{4}tan pi}
\ \
&=dfrac{1+0}{1-1cdot 0}
\ \
&=dfrac{1}{1-0}
\ \
&=dfrac{1}{1}=1
end{align*}
$$

Note that $color{#4257b2}pi$ and $color{#4257b2}dfrac{pi}{4}$ are special angles and we know the tangent values of them.

So the exact value of $color{#4257b2}tanleft(dfrac{pi}{4}+piright)$ is $boxed{ 1 }$

(d) We would like to determine the exact value of the expression $color{#4257b2}sinleft(-dfrac{pi}{2}+dfrac{pi}{3}right)$ using an appropriate compound angle formula. First, we note that our expression is a sine function which consists of two sum angles, so we can use the addition formula for the sine function where
$color{#4257b2}sinleft(a+bright)=sin acos b+cos asin b$.

$$
begin{align*}
sinleft(-dfrac{pi}{2}+dfrac{pi}{3}right)&=sin left(-dfrac{pi}{2}right)cos dfrac{pi}{3}+cos left(-dfrac{pi}{2}right)sin dfrac{pi}{3}
\ \
&=(-1)cdot dfrac{1}{2}+0cdot dfrac{sqrt{3}}{2}
\ \
&=-dfrac{1}{2}+0=-dfrac{1}{2}
end{align*}
$$

Note that $color{#4257b2}-dfrac{pi}{2}$ and $color{#4257b2}dfrac{pi}{3}$ are special angles and we know the sine and cosine values of them.

So the exact value of $color{#4257b2}sinleft(-dfrac{pi}{2}+dfrac{pi}{3}right)$ is $boxed{ -dfrac{1}{2} }$

(e) We would like to determine the exact value of the expression $color{#4257b2}tanleft(dfrac{pi}{3}-dfrac{pi}{6}right)$ using an appropriate compound angle formula. First, we note that our expression is a tangent function which consists of two subtraction angles, so we can use the subtraction formula for the tangent function where
$color{#4257b2}tanleft(a-bright)=dfrac{tan a-tan b}{1+tan atan b}$.

$$
begin{align*}
tanleft(dfrac{pi}{3}-dfrac{pi}{6}right)&=dfrac{tan dfrac{pi}{3}-tan dfrac{pi}{6}}{1+tan dfrac{pi}{3}tan dfrac{pi}{6}}
\ \
&=dfrac{sqrt{3}-dfrac{1}{sqrt{3}}}{1+sqrt{3}cdot dfrac{1}{sqrt{3}}}
\ \
&=dfrac{sqrt{3}-dfrac{1}{sqrt{3}}}{1+1}
\ \
&=dfrac{sqrt{3}-dfrac{1}{sqrt{3}}}{2}
\ \
&=dfrac{sqrt{3}left(sqrt{3}-dfrac{1}{sqrt{3}}right)}{2sqrt{3}} text{multiply and divide by $sqrt{3}$}
\ \
&=dfrac{3-1}{2sqrt{3}}=dfrac{2}{2sqrt{3}}=dfrac{1}{sqrt{3}}
end{align*}
$$

Note that $color{#4257b2}dfrac{pi}{3}$ and $color{#4257b2}dfrac{pi}{6}$ are special angles and we know the tangent values of them.

So the exact value of $color{#4257b2}tanleft(dfrac{pi}{3}-dfrac{pi}{6}right)$ is $boxed{ dfrac{1}{sqrt{3}} }$

(f) We would like to determine the exact value of the expression $color{#4257b2}cosleft(dfrac{pi}{2}+dfrac{pi}{3}right)$ using an appropriate compound angle formula. First, we note that our expression is a cosine function which consists of two sum angles, so we can use the addition formula for the cosine function where
$color{#4257b2}cosleft(a+bright)=cos acos b-sin asin b$.

$$
begin{align*}
cosleft(dfrac{pi}{2}+dfrac{pi}{3}right)&=cos dfrac{pi}{2}cos dfrac{pi}{3}-sin dfrac{pi}{2}sin dfrac{pi}{3}
\ \
&=0cdot dfrac{1}{2}-1cdot dfrac{sqrt{3}}{2}
\ \
&=0-dfrac{sqrt{3}}{2}=-dfrac{sqrt{3}}{2}
end{align*}
$$

Note that $color{#4257b2}dfrac{pi}{2}$ and $color{#4257b2}dfrac{pi}{3}$ are special angles and we know the sine and cosine values of them.

So the exact value of $color{#4257b2}cosleft(dfrac{pi}{2}+dfrac{pi}{3}right)$ is $boxed{ -dfrac{sqrt{3}}{2} }$

$$
text{color{#c34632}$(a) -dfrac{1}{2}$ $(b) -dfrac{sqrt{2}}{2}$ $(c) 1$ $(d) -dfrac{1}{2}$ $(e) dfrac{1}{sqrt{3}}$ $(f) -dfrac{sqrt{3}}{2}$}
$$

(a) We would like to create an equivalent expression for the expression $color{#4257b2}sinleft(pi+xright)$ using an appropriate compound angle formula. First, we note that our expression is a sine function which consists of two sum angles, so we can use the addition formula for the sine function where
$color{#4257b2}sinleft(a+bright)=sin acos b+cos asin b$.

$$
begin{align*}
sinleft(pi+xright)&=sin picos x+cos pisin x
\ \
&=0cdot cos x+(-1)cdot sin x
\ \
&=0-sin x
\ \
&=-sin x
end{align*}
$$

Note that $color{#4257b2}pi$ is a special angle and we know the sine and cosine values of it where $color{#4257b2}sin pi=0$ and $color{#4257b2}cos pi=-1$.

So the equivalent expression of $color{#4257b2}sinleft(pi+xright)$ is $boxed{ -sin x }$

(b) We would like to create an equivalent expression for the expression $color{#4257b2}cosleft(x+dfrac{3pi}{2}right)$ using an appropriate compound angle formula. First, we note that our expression is a cosine function which consists of two sum angles, so we can use the addition formula for the cosine function where
$color{#4257b2}cosleft(a+bright)=cos acos b-sin asin b$.

$$
begin{align*}
cosleft(x+dfrac{3pi}{2}right)&=cos xcos dfrac{3pi}{2}-sin xsin dfrac{3pi}{2}
\ \
&=cos xcdot (0)-sin xcdot (-1)
\ \
&=0-left(-sin xright)
\ \
&=sin x
end{align*}
$$

Note that $color{#4257b2}dfrac{3pi}{2}$ is a special angle and we know the sine and cosine values of it where $color{#4257b2}sin dfrac{3pi}{2}=-1$ and $color{#4257b2}cos dfrac{3pi}{2}=0$.

So the equivalent expression of $color{#4257b2}cosleft(x+dfrac{3pi}{2}right)$ is $boxed{ sin x }$

(c) We would like to create an equivalent expression for the expression $color{#4257b2}cosleft(x+dfrac{pi}{2}right)$ using an appropriate compound angle formula. First, we note that our expression is a cosine function which consists of two sum angles, so we can use the addition formula for the cosine function where
$color{#4257b2}cosleft(a+bright)=cos acos b-sin asin b$.

$$
begin{align*}
cosleft(x+dfrac{pi}{2}right)&=cos xcos dfrac{pi}{2}-sin xsin dfrac{pi}{2}
\ \
&=cos xcdot (0)-sin xcdot (1)
\ \
&=0-sin x
\ \
&=-sin x
end{align*}
$$

Note that $color{#4257b2}dfrac{pi}{2}$ is a special angle and we know the sine and cosine values of it where $color{#4257b2}sin dfrac{pi}{2}=1$ and $color{#4257b2}cos dfrac{pi}{2}=0$.

So the equivalent expression of $color{#4257b2}cosleft(x+dfrac{pi}{2}right)$ is $boxed{ -sin x }$

(d) We would like to create an equivalent expression for the expression $color{#4257b2}tanleft(x+piright)$ using an appropriate compound angle formula. First, we note that our expression is a tangent function which consists of two sum angles, so we can use the addition formula for the tangent function where
$color{#4257b2}tanleft(a+bright)=dfrac{tan a+tan b}{1-tan atan b}$.

$$
begin{align*}
tanleft(x+piright)&=dfrac{tan x+tan pi}{1-tan xtan pi}
\ \
&=dfrac{tan x+0}{1-tan xcdot (0)}
\ \
&=dfrac{tan x}{1-0}
\ \
&=dfrac{tan x}{1}
\ \
&=tan x
end{align*}
$$

Note that $color{#4257b2}pi$ is a special angle and we know the tangent value of it where $color{#4257b2}tan pi=0$.

So the equivalent expression of $color{#4257b2}tanleft(x+piright)$ is $boxed{ tan x }$

(e) We would like to create an equivalent expression for the expression $color{#4257b2}sinleft(x-piright)$ using an appropriate compound angle formula. First, we note that our expression is a sine function which consists of two subtraction angles, so we can use the subtraction formula for the sine function where
$color{#4257b2}sinleft(a-bright)=sin acos b-cos asin b$.

$$
begin{align*}
sinleft(x-piright)&=sin xcos pi-cos xsin pi
\ \
&=sin xcdot (-1)-cos xcdot 0
\ \
&=-sin x-0
\ \
&=-sin x
end{align*}
$$

Note that $color{#4257b2}pi$ is a special angle and we know the sine and cosine values of it where $color{#4257b2}sin pi=0$ and $color{#4257b2}cos pi=-1$.

So the equivalent expression of $color{#4257b2}sinleft(x-piright)$ is $boxed{ -sin x }$

(f) We would like to create an equivalent expression for the expression $color{#4257b2}tanleft(2pi-xright)$ using an appropriate compound angle formula. First, we note that our expression is a tangent function which consists of two subtraction angles, so we can use the subtraction formula for the tangent function where
$color{#4257b2}tanleft(a-bright)=dfrac{tan a-tan b}{1+tan atan b}$.

$$
begin{align*}
tanleft(2pi-xright)&=dfrac{tan 2pi-tan x}{1+tan 2pitan x}
\ \
&=dfrac{0-tan x}{1+0cdot tan x}
\ \
&=dfrac{-tan x}{1+0}
\ \
&=dfrac{-tan x}{1}
\ \
&=-tan x
end{align*}
$$

Note that $color{#4257b2}2pi$ is a special angle and we know the tangent value of it where $color{#4257b2}tan 2pi=0$.

So the equivalent expression of $color{#4257b2}tanleft(2pi-xright)$ is $boxed{ -tan x }$

$$
text{color{#c34632}$(a) -sin x$ $(b) sin x$ $(c) -sin x$ $(d) tan x$ $(e) -sin x$ $(f) -tan x$}
$$

(a) We would like to determine the exact value of the trigonometric ratio $color{#4257b2}cos 75text{textdegree}$. First, we know that $color{#4257b2}75text{textdegree}=45text{textdegree}+30text{textdegree}$, so we can rewrite $color{#4257b2}cos 75text{textdegree}$ as $color{#4257b2}cos left(45text{textdegree}+30text{textdegree}right)$. Now we have a compound angle, so we can use the addition formula for the cosine function to find the exact value where
$color{#4257b2}cosleft(a+bright)=cos acos b-sin asin b$.

$$
begin{align*}
cos 75text{textdegree}&=cos left(45text{textdegree}+30text{textdegree}right)
\ \
&=cos 45text{textdegree}cos 30text{textdegree}-sin 45text{textdegree}sin 30text{textdegree}
\ \
&=dfrac{sqrt{2}}{2}cdot dfrac{sqrt{3}}{2}-dfrac{sqrt{2}}{2}cdot dfrac{1}{2}
\ \
&=dfrac{sqrt{6}}{4}-dfrac{sqrt{2}}{4}
\ \
&=dfrac{sqrt{6}-sqrt{2}}{4}
end{align*}
$$

Note that $color{#4257b2}45text{textdegree}$ and $color{#4257b2}30text{textdegree}$ are special angles and we know the sine and cosine values of them.

So the exact value of $color{#4257b2}cos 75text{textdegree}$ is $boxed{ dfrac{sqrt{6}-sqrt{2}}{4} }$

(b) We would like to determine the exact value of the trigonometric ratio $color{#4257b2}tan (-15text{textdegree})$. First, we know that $color{#4257b2}-15text{textdegree}=30text{textdegree}-45text{textdegree}$, so we can rewrite $color{#4257b2}tan (-15text{textdegree})$ as $color{#4257b2}tan left(30text{textdegree}-45text{textdegree}right)$. Now we have a compound angle, so we can use the subtraction formula for the tangent function to find the exact value where $color{#4257b2}tanleft(a-bright)=dfrac{tan a-tan b}{1+tan atan b}$.

$$
begin{align*}
tan (-15text{textdegree})&=tan left(30text{textdegree}-45text{textdegree}right)
\ \
&=dfrac{tan 30text{textdegree}-tan 45text{textdegree}}{1+tan 30text{textdegree}tan 45text{textdegree}}
\ \
&=dfrac{dfrac{1}{sqrt{3}}-1}{1+dfrac{1}{sqrt{3}}cdot 1}
\ \
&=dfrac{sqrt{3}left(dfrac{1}{sqrt{3}}-1right)}{sqrt{3}left(1+dfrac{1}{sqrt{3}}right)} text{multiply and divide by $sqrt{3}$}
\ \
&=dfrac{1-sqrt{3}}{sqrt{3}+1}
\ \
&=dfrac{left(1-sqrt{3}right)left(1-sqrt{3}right)}{left(1+sqrt{3}right)left(1-sqrt{3}right)}=dfrac{left(1-sqrt{3}right)^{2}}{1^{2}-left(sqrt{3}right)^{2}} text{multiply and divide by $1-sqrt{3}$}
\ \
&=dfrac{1-2sqrt{3}+3}{1-3}=dfrac{4-2sqrt{3}}{-2}=-2+sqrt{3}
end{align*}
$$

Note that $color{#4257b2}30text{textdegree}$ and $color{#4257b2}45text{textdegree}$ are special angles and we know the tangent values of them.

So the exact value of $color{#4257b2}tan (-15text{textdegree})$ is $boxed{ -2+sqrt{3} }$

(c) We would like to determine the exact value of the trigonometric ratio $color{#4257b2}cos dfrac{11pi}{12}$. First, we know that $color{#4257b2}dfrac{11pi}{12}=pi-dfrac{pi}{12}$, so we can rewrite $color{#4257b2}cos dfrac{11pi}{12}$ as $color{#4257b2}cos left(pi-dfrac{pi}{12}right)$. But we know from the transformations that
$color{#4257b2}cos left(pi-thetaright)=-cos theta$, so we can rewrite $color{#4257b2}cos left(pi-dfrac{pi}{12}right)$ as $color{#4257b2}-cos dfrac{pi}{12}$ which equals $color{#4257b2}-cos left(dfrac{3pi}{12}-dfrac{2pi}{12}right)$. Now we have a compound angle, so we can use the subtraction formula for the cosine function to find the exact value where $color{#4257b2}cosleft(a-bright)=cos acos b+sin asin b$.

$$
begin{align*}
cos dfrac{11pi}{12}&=cos left(pi-dfrac{pi}{12}right)
\ \
&=-cos dfrac{pi}{12}
\ \
&=-cos left(dfrac{3pi}{12}-dfrac{2pi}{12}right)
\ \
&=-cos left(dfrac{pi}{4}-dfrac{pi}{6}right)
\ \
&=-left(cos dfrac{pi}{4}cos dfrac{pi}{6}+sin dfrac{pi}{4}sin dfrac{pi}{6}right)
\ \
&=-left(dfrac{sqrt{2}}{2}cdot dfrac{sqrt{3}}{2}+dfrac{sqrt{2}}{2}cdot dfrac{1}{2}right)
\ \
&=-left(dfrac{sqrt{6}}{4}+dfrac{sqrt{2}}{4}right)=-dfrac{sqrt{6}+sqrt{2}}{4}
end{align*}
$$

Note that $color{#4257b2}dfrac{pi}{4}$ and $color{#4257b2}dfrac{pi}{6}$ are special angles and we know the sine and cosine values of them.

So the exact value of $color{#4257b2}cos dfrac{11pi}{12}$ is $boxed{ -dfrac{sqrt{6}+sqrt{2}}{4} }$

(d) We would like to determine the exact value of the trigonometric ratio $color{#4257b2}sin dfrac{13pi}{12}$. First, we know that $color{#4257b2}dfrac{13pi}{12}=pi+dfrac{pi}{12}$, so we can rewrite $color{#4257b2}sin dfrac{13pi}{12}$ as $color{#4257b2}sin left(pi+dfrac{pi}{12}right)$. But we know from the transformations that
$color{#4257b2}sin left(pi+thetaright)=-sin theta$, so we can rewrite $color{#4257b2}sin left(pi+dfrac{pi}{12}right)$ as $color{#4257b2}-sin dfrac{pi}{12}$ which equals $color{#4257b2}-sin left(dfrac{3pi}{12}-dfrac{2pi}{12}right)$. Now we have a compound angle, so we can use the subtraction formula for the sine function to find the exact value where $color{#4257b2}sinleft(a-bright)=sin acos b-cos asin b$.

$$
begin{align*}
sin dfrac{13pi}{12}&=sin left(pi+dfrac{pi}{12}right)
\ \
&=-sin dfrac{pi}{12}
\ \
&=-sin left(dfrac{3pi}{12}-dfrac{2pi}{12}right)
\ \
&=-sin left(dfrac{pi}{4}-dfrac{pi}{6}right)
\ \
&=-left(sin dfrac{pi}{4}cos dfrac{pi}{6}-cos dfrac{pi}{4}sin dfrac{pi}{6}right)
\ \
&=-left(dfrac{sqrt{2}}{2}cdot dfrac{sqrt{3}}{2}-dfrac{sqrt{2}}{2}cdot dfrac{1}{2}right)
\ \
&=-left(dfrac{sqrt{6}}{4}-dfrac{sqrt{2}}{4}right)=dfrac{-sqrt{6}+sqrt{2}}{4}
end{align*}
$$

Note that $color{#4257b2}dfrac{pi}{4}$ and $color{#4257b2}dfrac{pi}{6}$ are special angles and we know the sine and cosine values of them.

So the exact value of $color{#4257b2}sin dfrac{13pi}{12}$ is $boxed{ dfrac{-sqrt{6}+sqrt{2}}{4} }$

(e) We would like to determine the exact value of the trigonometric ratio $color{#4257b2}tan dfrac{7pi}{12}$. First, we know that $color{#4257b2}dfrac{7pi}{12}=dfrac{3pi}{12}+dfrac{4pi}{12}$, so we can rewrite $color{#4257b2}tan dfrac{7pi}{12}$ as $color{#4257b2}tan left(dfrac{3pi}{12}+dfrac{4pi}{12}right)$. Now we have a compound angle, so we can use the addition formula for the tangent function to find the exact value where $color{#4257b2}tanleft(a+bright)=dfrac{tan a+tan b}{1-tan atan b}$.

$$
begin{align*}
tan dfrac{7pi}{12}&=tan left(dfrac{3pi}{12}+dfrac{4pi}{12}right)
\ \
&=tan left(dfrac{pi}{4}+dfrac{pi}{3}right)
\ \
&=dfrac{tan dfrac{pi}{4}+tan dfrac{pi}{3}}{1-tan dfrac{pi}{4}tan dfrac{pi}{3}}
\ \
&=dfrac{1+sqrt{3}}{1-1cdot sqrt{3}}
\ \
&=dfrac{1+sqrt{3}}{1-sqrt{3}}
\ \
&=dfrac{left(1+sqrt{3}right)left(1+sqrt{3}right)}{left(1-sqrt{3}right)left(1+sqrt{3}right)}=dfrac{left(1+sqrt{3}right)^{2}}{1^{2}-left(sqrt{3}right)^{2}} text{multiply and divide by $1+sqrt{3}$}
\ \
&=dfrac{1+2sqrt{3}+3}{1-3}=dfrac{4+2sqrt{3}}{-2}=-2-sqrt{3}
end{align*}
$$

Note that $color{#4257b2}dfrac{pi}{4}$ and $color{#4257b2}dfrac{pi}{3}$ are special angles and we know the tangent values of them.

So the exact value of $color{#4257b2}tan dfrac{7pi}{12}$ is $boxed{ -2-sqrt{3} }$

(f) We would like to determine the exact value of the trigonometric ratio $color{#4257b2}tan dfrac{-5pi}{12}$. First, we know that $tan (-theta)=-tan theta$, so we can say that $color{#4257b2}tan dfrac{-5pi}{12}=-tan dfrac{5pi}{12}$. We know also that $color{#4257b2}dfrac{5pi}{12}=dfrac{3pi}{12}+dfrac{2pi}{12}$, so we can rewrite $color{#4257b2}tan dfrac{5pi}{12}$ as $color{#4257b2}tan left(dfrac{3pi}{12}+dfrac{2pi}{12}right)$. Now we have a compound angle, so we can use the addition formula for the tangent function to find the exact value where $color{#4257b2}tanleft(a+bright)=dfrac{tan a+tan b}{1-tan atan b}$.

$$
begin{align*}
tan dfrac{-5pi}{12}&=-tan dfrac{5pi}{12}
\ \
&=-tan left(dfrac{3pi}{12}+dfrac{2pi}{12}right)
\ \
&=-tan left(dfrac{pi}{4}+dfrac{pi}{6}right)
\ \
&=-left(dfrac{tan dfrac{pi}{4}+tan dfrac{pi}{6}}{1-tan dfrac{pi}{4}tan dfrac{pi}{6}}right)
\ \
&=-left(dfrac{1+dfrac{1}{sqrt{3}}}{1-1cdot dfrac{1}{sqrt{3}}}right)
\ \
&=-left[dfrac{sqrt{3}left(1+dfrac{1}{sqrt{3}}right)}{sqrt{3}left(1-dfrac{1}{sqrt{3}}right)}right]=-left(dfrac{sqrt{3}+1}{sqrt{3}-1}right)
\ \
&=-left[dfrac{left(sqrt{3}+1right)left(sqrt{3}+1right)}{left(sqrt{3}-1right)left(sqrt{3}+1right)}right]=-left[dfrac{left(sqrt{3}+1right)^{2}}{left(sqrt{3}right)^{2}-1^{2}}right] text{multiply and divide by $sqrt{3}+1$}
\ \
&=-left(dfrac{3+2sqrt{3}+1}{3-1}right)=-left(dfrac{4+2sqrt{3}}{2}right)=-2-sqrt{3}
end{align*}
$$

Note that $color{#4257b2}dfrac{pi}{4}$ and $color{#4257b2}dfrac{pi}{6}$ are special angles and we know the tangent values of them.

So the exact value of $color{#4257b2}tan dfrac{-5pi}{12}$ is $boxed{ -2-sqrt{3} }$

$$
text{color{#c34632}$(a) dfrac{sqrt{6}-sqrt{2}}{4}$ $(b) -2+sqrt{3}$ $(c) -dfrac{sqrt{6}+sqrt{2}}{4}$
\
\
Large{color{#c34632}$(d) dfrac{-sqrt{6}+sqrt{2}}{4}$ $(e) -2-sqrt{3}$ $(f) -2-sqrt{3}$}}
$$

In this problem we would like to evaluate the value of some trigonometric functions which consists of two angles $color{#4257b2}x$ and $color{#4257b2}y$, so the first step is to find all trigonometric functions for each angle and then substitute in the addition and subtraction formulas for these functions.

First, we note that $color{#4257b2}sin x=dfrac{4}{5}$ where $color{#4257b2}0 < x < dfrac{pi}{2}$, so we can use the Pythagorean identity $color{#4257b2}sin^{2}theta+cos^{2}theta=1$ to find the value of $color{#4257b2}cos x$.

$$
sin^{2}x+cos^{2}x=1
$$

$$
left(dfrac{4}{5}right)^{2}+cos^{2}x=1
$$

$$
dfrac{16}{25}+cos^{2}x=1
$$

$$
cos^{2}x=1-dfrac{16}{25}=dfrac{9}{25}
$$

Now we can take the square root for each side to find the value of $color{#4257b2}cos x$

$$
cos x=pm sqrt{dfrac{9}{25}}=pm dfrac{3}{5}
$$

But we know that $color{#4257b2}0 < x < dfrac{pi}{2}$ which means that $color{#4257b2}x$ is in quadrant $1$, so the value of $color{#4257b2}cos x$ is positive and the negative solution is refused.

$$
boxed{ cos x=dfrac{3}{5} }
$$

Now we have the values of $color{#4257b2}sin x$ and $color{#4257b2}cos x$, so we can use the identity $color{#4257b2}tan theta=dfrac{sin theta}{cos theta}$ to find $color{#4257b2}tan x$

$$
tan x=dfrac{sin x}{cos x}
$$

$$
tan x=dfrac{dfrac{4}{cancel{5}}}{dfrac{3}{cancel{5}}}
$$

$$
boxed{ tan x=dfrac{4}{3} }
$$

Now we found the values of the trigonometric functions for the angle $color{#4257b2}x$. The next step is to find the trigonometric functions for the angle $color{#4257b2}y$ by the same way.

We note that $color{#4257b2}sin y=-dfrac{12}{13}$ where $color{#4257b2}dfrac{3pi}{2} < y < 2pi$, so we can use the Pythagorean identity $color{#4257b2}sin^{2}theta+cos^{2}theta=1$ to find the value of $color{#4257b2}cos y$.

$$
sin^{2}y+cos^{2}y=1
$$

$$
left(-dfrac{12}{13}right)^{2}+cos^{2}y=1
$$

$$
dfrac{144}{169}+cos^{2}y=1
$$

$$
cos^{2}y=1-dfrac{144}{169}=dfrac{25}{169}
$$

Now we can take the square root for each side to find the value of $color{#4257b2}cos y$

$$
cos y=pm sqrt{dfrac{25}{169}}=pm dfrac{5}{13}
$$

But we know that $color{#4257b2}dfrac{3pi}{2} < y < 2pi$ which means that $color{#4257b2}y$ is in quadrant $4$, so the value of $color{#4257b2}cos y$ is positive and the negative solution is refused.

$$
boxed{ cos y=dfrac{5}{13} }
$$

Now we have the values of $color{#4257b2}sin y$ and $color{#4257b2}cos y$, so we can use the identity $color{#4257b2}tan theta=dfrac{sin theta}{cos theta}$ to find $color{#4257b2}tan y$

$$
tan y=dfrac{sin y}{cos y}=dfrac{-dfrac{12}{cancel{13}}}{dfrac{5}{cancel{13}}}
$$

$$
boxed{ tan y=-dfrac{12}{5} }
$$

Now we have the values of all trigonometric functions for the two angles $color{#4257b2}x$ and $color{#4257b2}y$, so we can evaluate each requirement as follows:

(a) We would like to evaluate the value of $color{#4257b2}cos left(x+yright)$, so we can use the addition formula for the cosine function $color{#4257b2}cos left(a+bright)=cos acos b-sin asin b$.

$$
begin{align*}
cos left(x+yright)&=cos xcos y-sin xsin y
\ \
&=dfrac{3}{5}cdot dfrac{5}{13}-dfrac{4}{5}cdot dfrac{-12}{13}
\ \
&=dfrac{15}{65}+dfrac{48}{65}=boxed{ dfrac{63}{65} }
end{align*}
$$

(b) We would like to evaluate the value of $color{#4257b2}sin left(x+yright)$, so we can use the addition formula for the sine function $color{#4257b2}sin left(a+bright)=sin acos b+cos asin b$.

$$
begin{align*}
sin left(x+yright)&=sin xcos y+cos xsin y
\ \
&=dfrac{4}{5}cdot dfrac{5}{13}+dfrac{3}{5}cdot dfrac{-12}{13}
\ \
&=dfrac{20}{65}-dfrac{36}{65}=boxed{ -dfrac{16}{65} }
end{align*}
$$

(c) We would like to evaluate the value of $color{#4257b2}cos left(x-yright)$, so we can use the subtraction formula for the cosine function $color{#4257b2}cos left(a-bright)=cos acos b+sin asin b$.

$$
begin{align*}
cos left(x-yright)&=cos xcos y+sin xsin y
\ \
&=dfrac{3}{5}cdot dfrac{5}{13}+dfrac{4}{5}cdot dfrac{-12}{13}
\ \
&=dfrac{15}{65}-dfrac{48}{65}=boxed{ -dfrac{33}{65} }
end{align*}
$$

(d) We would like to evaluate the value of $color{#4257b2}sin left(x-yright)$, so we can use the subtraction formula for the sine function $color{#4257b2}sin left(a+bright)=sin acos b-cos asin b$.

$$
begin{align*}
sin left(x-yright)&=sin xcos y-cos xsin y
\ \
&=dfrac{4}{5}cdot dfrac{5}{13}-dfrac{3}{5}cdot dfrac{-12}{13}
\ \
&=dfrac{20}{65}+dfrac{36}{65}=boxed{ dfrac{56}{65} }
end{align*}
$$

(e) We would like to evaluate the value of $color{#4257b2}tan left(x+yright)$, so we can use the addition formula for the tangent function $color{#4257b2}tan left(a+bright)=dfrac{tan a+tan b}{1-tan atan b}$.

$$
begin{align*}
tan left(x+yright)&=dfrac{tan x+tan y}{1-tan xtan y}
\ \
&=dfrac{dfrac{4}{3}+dfrac{-12}{5}}{1-dfrac{4}{3}cdot dfrac{-12}{5}}
\ \
&=boxed{ -dfrac{16}{63} }
end{align*}
$$

(f) We would like to evaluate the value of $color{#4257b2}tan left(x-yright)$, so we can use the subtraction formula for the tangent function $color{#4257b2}tan left(a-bright)=dfrac{tan a-tan b}{1+tan atan b}$.

$$
begin{align*}
tan left(x-yright)&=dfrac{tan x-tan y}{1+tan xtan y}
\ \
&=dfrac{dfrac{4}{3}-dfrac{-12}{5}}{1+dfrac{4}{3}cdot dfrac{-12}{5}}
\ \
&=boxed{ -dfrac{56}{33} }
end{align*}
$$

$$
text{color{#c34632}$(a) dfrac{63}{65}$ $(b) -dfrac{16}{65}$ $(c) -dfrac{33}{65}$ $(d) dfrac{56}{65}$ $(e) -dfrac{16}{63}$ $(f) -dfrac{56}{33}$}
$$

In this problem we would like to evaluate the trigonometric functions $color{#4257b2}sin left(alpha+betaright)$ and $color{#4257b2}tan left(alpha+betaright)$, so the first step is to find all trigonometric functions for the two angles $color{#4257b2}alpha$ and $color{#4257b2}beta$ and then substitute in the addition formulas for these functions.

First, we note that $color{#4257b2}sin alpha=dfrac{7}{25}$, so we can use the Pythagorean identity $color{#4257b2}sin^{2}theta+cos^{2}theta=1$ to find the value of $color{#4257b2}cos alpha$.

$$
sin^{2}alpha+cos^{2}alpha=1
$$

$$
left(dfrac{7}{25}right)^{2}+cos^{2}alpha=1
$$

$$
dfrac{49}{625}+cos^{2}alpha=1
$$

$$
cos^{2}alpha=1-dfrac{49}{625}=dfrac{576}{625}
$$

Now we can take the square root for each side to find the value of $color{#4257b2}cos alpha$

$$
cos alpha=pm sqrt{dfrac{576}{625}}=pm dfrac{24}{25}
$$

But we know that $color{#4257b2}alpha$ is in quadrant $1$, so the value of $color{#4257b2}cos alpha$ is positive and the negative solution is refused.

$$
boxed{ cos alpha=dfrac{24}{25} }
$$

Now we have the values of $color{#4257b2}sin alpha$ and $color{#4257b2}cos alpha$, so we can use the identity $color{#4257b2}tan theta=dfrac{sin theta}{cos theta}$ to find $color{#4257b2}tan alpha$

$$
tan alpha=dfrac{sin alpha}{cos alpha}
$$

$$
tan alpha=dfrac{dfrac{7}{cancel{25}}}{dfrac{24}{cancel{25}}}
$$

$$
boxed{ tan alpha=dfrac{7}{24} }
$$

Now we found the values of the trigonometric functions for the angle $color{#4257b2}alpha$. The next step is to find the trigonometric functions for the angle $color{#4257b2}beta$ by the same way.

We note that $color{#4257b2}cos beta=dfrac{5}{13}$, so we can use the Pythagorean identity
$color{#4257b2}sin^{2}theta+cos^{2}theta=1$ to find the value of $color{#4257b2}sin beta$.

$$
sin^{2}beta+cos^{2}beta=1
$$

$$
sin^{2}beta+left(dfrac{5}{13}right)^{2}=1
$$

$$
sin^{2}beta+dfrac{25}{169}=1
$$

$$
sin^{2}beta=1-dfrac{25}{169}=dfrac{144}{169}
$$

Now we can take the square root for each side to find the value of $color{#4257b2}sin beta$

$$
sin beta=pm sqrt{dfrac{144}{169}}=pm dfrac{12}{13}
$$

But we know that $color{#4257b2}beta$ is in quadrant $1$, so the value of $color{#4257b2}sin beta$ is positive and the negative solution is refused.

$$
boxed{ sin beta=dfrac{12}{13} }
$$

Now we have the values of $color{#4257b2}sin beta$ and $color{#4257b2}cos beta$, so we can use the identity $color{#4257b2}tan theta=dfrac{sin theta}{cos theta}$ to find $color{#4257b2}tan beta$

$$
tan beta=dfrac{sin beta}{cos beta}=dfrac{dfrac{12}{cancel{13}}}{dfrac{5}{cancel{13}}}
$$

$$
boxed{ tan beta=dfrac{12}{5} }
$$

Now we have the values of all trigonometric functions for the two angles $color{#4257b2}alpha$ and $color{#4257b2}beta$, so we can evaluate each requirement as follows:

(a) We would like to evaluate the value of $color{#4257b2}sin left(alpha+betaright)$, so we can use the addition formula for the sine function $color{#4257b2}sin left(a+bright)=sin acos b+cos asin b$.

$$
begin{align*}
sin left(alpha+betaright)&=sin alphacos beta+cos alphasin beta
\ \
&=dfrac{7}{25}cdot dfrac{5}{13}+dfrac{24}{25}cdot dfrac{12}{13}
\ \
&=dfrac{35}{325}+dfrac{288}{325}=boxed{ dfrac{323}{325} }
end{align*}
$$

(b) We would like to evaluate the value of $color{#4257b2}tan left(alpha+betaright)$, so we can use the addition formula for the tangent function $color{#4257b2}tan left(a+bright)=dfrac{tan a+tan b}{1-tan atan b}$.

$$
begin{align*}
tan left(alpha+betaright)&=dfrac{tan alpha+tan beta}{1-tan alphatan beta}
\ \
&=dfrac{dfrac{7}{24}+dfrac{12}{5}}{1-dfrac{7}{24}cdot dfrac{12}{5}}
\ \
&=boxed{ dfrac{323}{36} }
end{align*}
$$

$$
text{color{#c34632}$(a) sinleft(alpha+betaright)=dfrac{323}{325}$ $(b) tanleft(alpha+betaright)=dfrac{323}{36}$}
$$

(a) We would like to use the compound angle formulas to verify the identity
$$
color{#4257b2}sin x=cos left(dfrac{pi}{2}-xright)
$$

First, we note that the right side is a cosine function which consists of two subtraction angles, so we can use the subtraction formula for the cosine function $color{#4257b2}cos left(a-bright)=cos acos b+sin asin b$

$$
begin{align*}
cosleft(dfrac{pi}{2}-xright)&=cos dfrac{pi}{2}cos x+sin dfrac{pi}{2}sin x
\ \
&=0cdot cos x+1cdot sin x
\ \
&=0+sin x=sin x
end{align*}
$$

Note that the angle $color{#4257b2}dfrac{pi}{2}$ is a special angle which we know the sine and cosine values of it $color{#4257b2}sin dfrac{pi}{2}=1$ and $color{#4257b2}cos dfrac{pi}{2}=0$

We proved that the right side is $color{#4257b2}sin x$, so the left side=the right side.

(b) We would like to use the compound angle formulas to verify the identity
$$
color{#4257b2}cos x=sin left(dfrac{pi}{2}-xright)
$$

First, we note that the right side is a sine function which consists of two subtraction angles, so we can use the subtraction formula for the sine function $color{#4257b2}sin left(a-bright)=sin acos b-cos asin b$

$$
begin{align*}
sinleft(dfrac{pi}{2}-xright)&=sin dfrac{pi}{2}cos x-cos dfrac{pi}{2}sin x
\ \
&=1cdot cos x-0cdot sin x
\ \
&=cos x-0=cos x
end{align*}
$$

We proved that the right side is $color{#4257b2}cos x$, so the left side=the right side.

$$
text{color{#c34632}$(a) sin x=cos left(dfrac{pi}{2}-xright)$ $(b) cos x=sin left(dfrac{pi}{2}-xright)$}
$$

(a) We would like to simplify the expression $color{#4257b2}sinleft(pi+xright)+sin left(pi-xright)$. First, we note that our expression is a sum of two sine functions each of them consists of two compound angles, so we can use the addition and subtraction formulas for the sine function where $color{#4257b2}sinleft(a+bright)=sin acos b+cos asin b$ and $color{#4257b2}sinleft(a-bright)=sin acos b-cos asin b$.

$$
begin{align*}
sinleft(pi+xright)+sin left(pi-xright)&=left(sin picos x+cos pisin xright)+left(sin picos x-cos pisin xright)
\ \
&=left(sin picos x+cancel{cos pisin x}right)+left(sin picos xcancel{-cos pisin x}right)
\ \
&=sin picos x+sin picos x
\ \
&=2sin picos x
\ \
&=2cdot 0cdot cos x
\ \
&=0
end{align*}
$$

Note that $color{#4257b2}pi$ is a special angle and we know the sine value of it where $color{#4257b2}sin pi=0$.

So the expression of $color{#4257b2}sinleft(pi+xright)+sin left(pi-xright)$ is $boxed{ 0 }$

(b) We would like to simplify the expression $color{#4257b2}cosleft(x+dfrac{pi}{3}right)-sin left(x+dfrac{pi}{6}right)$. First, we note that our expression is a subtraction between a cosine and sine functions each of them consists of two sum angles, so we can use the addition formulas for the sine and the cosine functions where
$color{#4257b2}sinleft(a+bright)=sin acos b+cos asin b$ and $color{#4257b2}cosleft(a+bright)=cos acos b-sin asin b$.

$$
begin{align*}
cosleft(x+dfrac{pi}{3}right)-sin left(x+dfrac{pi}{6}right)&=left(cos xcos dfrac{pi}{3}-sin xsin dfrac{pi}{3}right)-left(sin xcos dfrac{pi}{6}+cos xsin dfrac{pi}{6}right)
\ \
&=left(dfrac{1}{2}cos x-dfrac{sqrt{3}}{2} sin xright)-left(dfrac{sqrt{3}}{2} sin x+dfrac{1}{2} cos xright)
\ \
&=dfrac{1}{2}cos x-dfrac{sqrt{3}}{2} sin x-dfrac{sqrt{3}}{2} sin x-dfrac{1}{2} cos x
\ \
&=cancel{dfrac{1}{2}cos x}-dfrac{sqrt{3}}{2} sin x-dfrac{sqrt{3}}{2} sin xcancel{-dfrac{1}{2} cos x}
\ \
&=-dfrac{sqrt{3}}{2} sin x-dfrac{sqrt{3}}{2} sin x
\ \
&=-2cdot dfrac{sqrt{3}}{2} sin x
\ \
&=-sqrt{3} sin x
end{align*}
$$

Note that $color{#4257b2}dfrac{pi}{3}$ and $color{#4257b2}dfrac{pi}{6}$ are special angle and we know the sine value of them.

So the expression of $color{#4257b2}cosleft(x+dfrac{pi}{3}right)-sin left(x+dfrac{pi}{6}right)$ is $boxed{ -sqrt{3} sin x }$

$$
text{color{#c34632}$(a) 0$ $(b) -sqrt{3} sin x$}
$$

(a) We would like to simplify the expression $color{#4257b2}dfrac{sinleft(f+gright)+sin left(f-gright)}{cosleft(f+gright)+cos left(f-gright)}$. First, we note that our expression is consists of two sine functions and two cosine functions each of them consists of two compound angles, so we can use the addition and subtraction formulas for the sine and cosine function where $color{#4257b2}sinleft(a+bright)=sin acos b+cos asin b, sinleft(a-bright)=sin acos b-cos asin b, \ cosleft(a+bright)=cos acos b-sin asin b$ and $color{#4257b2}cosleft(a-bright)=cos acos b+sin asin b$.

$$
begin{align*}
dfrac{sinleft(f+gright)+sin left(f-gright)}{cosleft(f+gright)+cos left(f-gright)}&=dfrac{sin fcos g+cos fsin g+sin fcos g-cos fsin g}{cos fcos g-sin fsin g+cos fcos g+sin fsin g}
\ \
&=dfrac{sin fcos g+cancel{cos fsin g}+sin fcos gcancel{-cos fsin g}}{cos fcos gcancel{-sin fsin g}+cos fcos g+cancel{sin fsin g}}
\ \
&=dfrac{sin fcos g+sin fcos g}{cos fcos g+cos fcos g}
\ \
&=dfrac{2sin fcos g}{2cos fcos g}
\ \
&=dfrac{cancel{2}sin fcancel{cos g}}{cancel{2}cos fcancel{cos g}}
\ \
&=dfrac{sin f}{cos f}
\ \
&=tan f
end{align*}
$$

So the expression of $color{#4257b2}dfrac{sinleft(f+gright)+sin left(f-gright)}{cosleft(f+gright)+cos left(f-gright)}$ can be simplified to $boxed{ tan f }$

$$
color{#c34632}tan f
$$

We would like to list the compound angle formulas we used in this lesson. We know that each trigonometric function has an addition formula and subtraction formula, so we can list the addition and subtraction formulas for each trigonometric function as follows:

For the sine function:

The addition formula: $color{#4257b2}sin left(a+bright)=sin acos b+cos asin b$

The subtraction formula: $color{#4257b2}sin left(a-bright)=sin acos b-cos asin b$

For the sine function we note that the two formulas are similar in the terms but difference in sign between the two terms. Also in the addition and subtraction formulas for the sine function we note that the sign between the two terms is the same as between the two angles in the function, so we can remember the two formulas by knowing these similarities and difference.

For the cosine function:

The addition formula: $color{#4257b2}cos left(a+bright)=cos acos b-sin asin b$

The subtraction formula: $color{#4257b2}cos left(a-bright)=cos acos b+sin asin b$

For the cosine function we note that the two formulas are similar in the terms but difference in sign between the two terms. Also in the addition and subtraction formulas for the cosine function we note that the sign between the two terms is the opposite sign as between the two angles in the function, so we can remember the two formulas by knowing these similarities and difference.

For the tangent function:

The addition formula: $color{#4257b2}tan left(a+bright)=dfrac{tan a+tan b}{1-tan atan b}$

The subtraction formula: $color{#4257b2}tan left(a-bright)=dfrac{tan a-tan b}{1+tan atan b}$

For the tangent function we note that the two formulas are similar in the numerator and denominator shape but difference in sign between the two terms in the numerator and denominator. Also in the addition and subtraction formulas for the tangent function we note that the sign between the two terms in the numerator is the same between the two angles in the function and the sign between the two terms in the denominator is the opposite sign as between the two angles in the function, so we can remember the two formulas by knowing these similarities and difference.

$color{#c34632}{color{Black}For the sine function:}$
$$
begin{align*} sin (a+b)&=sin acos b+cos asin b \ \ sin (a-b)&=sin acos b-cos asin b end{align*}
$$

$color{#c34632}{color{Black}For the cosine function:}$
$$
begin{align*} cos (a+b)&=cos acos b-sin asin b \ \ cos (a-b)&=cos acos b+sin asin b end{align*}
$$

$color{#c34632}{color{Black}For the tangent function:}$
$$
begin{align*} tan (a+b)&=dfrac{tan a+tan b}{1-tan atan b} \ \ tan (a-b)&=dfrac{tan a-tan b}{1+tan atan b} end{align*}
$$

We would like to prove the identity
$$
color{#4257b2}sin C+sin D=2sin dfrac{C+D}{2}cos dfrac{C-D}{2}
$$

First, we note that the left side is a sum of two sine functions which consists of single angle, so we can write them by another way to be compound angle.

We know that $color{#4257b2}dfrac{C+D}{2}+dfrac{C-D}{2}=dfrac{C+D+C-D}{2}=dfrac{2C}{2}=C$ and also we know that $color{#4257b2}dfrac{C+D}{2}-dfrac{C-D}{2}=dfrac{C+D-C+D}{2}=dfrac{2D}{2}=D$, so we can rewrite the left side as follows:

$$
sin C+sin D=sin left(dfrac{C+D}{2}+dfrac{C-D}{2}right)+sin left(dfrac{C+D}{2}-dfrac{C-D}{2}right)
$$

Now the left side consists of two sum sine functions each of them is compound angle, so we can use the addition and subtraction formulas for the sine function $color{#4257b2}sin (a+b)=sin acos b+cos asin b$ and $color{#4257b2}sin (a-b)=sin acos b-cos asin b$

$$
begin{align*}
sin C+sin D&=sin left(dfrac{C+D}{2}+dfrac{C-D}{2}right)+sin left(dfrac{C+D}{2}-dfrac{C-D}{2}right)
\ \
&=sin dfrac{C+D}{2}cos dfrac{C-D}{2}+cos dfrac{C+D}{2}sin dfrac{C-D}{2}+sin dfrac{C+D}{2}cos dfrac{C-D}{2}-cos dfrac{C+D}{2}sin dfrac{C-D}{2}
\ \
&=sin dfrac{C+D}{2}cos dfrac{C-D}{2}+cancel{cos dfrac{C+D}{2}sin dfrac{C-D}{2}}+sin dfrac{C+D}{2}cos dfrac{C-D}{2}cancel{-cos dfrac{C+D}{2}sin dfrac{C-D}{2}}
\ \
&=sin dfrac{C+D}{2}cos dfrac{C-D}{2}+sin dfrac{C+D}{2}cos dfrac{C-D}{2}
\ \
&=2sin dfrac{C+D}{2}cos dfrac{C-D}{2}
end{align*}
$$

We proved that the left side is $color{#4257b2}2sin dfrac{C+D}{2}cos dfrac{C-D}{2}$, so the left side=the right side.

$$
color{#c34632}sin C+sin D=2sin dfrac{C+D}{2}cos dfrac{C-D}{2}
$$

We would like to determine $color{#4257b2}cot (x+y)$ in terms of $color{#4257b2}cot x$ and $color{#4257b2}cot y$. First, we know that $color{#4257b2}cot theta=dfrac{1}{tan theta}$, so we can rewrite $color{#4257b2}cot (x+y)$ as $color{#4257b2}dfrac{1}{tan (x+y)}$. Now we can use the addition formula for the tangent function
$color{#4257b2}tan (a+b)=dfrac{tan a+tan b}{1-tan atan b}$

$$
begin{align*}
cot (x+y)&=dfrac{1}{tan (x+y)}
\ \
&=dfrac{1}{dfrac{tan x+tan y}{1-tan xtan y}}
\ \
&=dfrac{1-tan xtan y}{tan x+tan y}
end{align*}
$$

Now we can divide the numerator and denominator by $color{#4257b2}tan xtan y$ to simplify.

$$
begin{align*}
dfrac{1-tan xtan y}{tan x+tan y}&=dfrac{dfrac{1}{tan xtan y}-dfrac{tan xtan y}{tan xtan y}}{dfrac{tan x}{tan xtan y}+dfrac{tan y}{tan xtan y}}
\ \
&=dfrac{dfrac{1}{tan x}cdot dfrac{1}{tan y}-dfrac{cancel{tan xtan y}}{cancel{tan xtan y}}}{dfrac{cancel{tan x}}{cancel{tan x}tan y}+dfrac{cancel{tan y}}{tan xcancel{tan y}}}
\ \
&=dfrac{dfrac{1}{tan x}cdot dfrac{1}{tan y}-1}{dfrac{1}{tan y}+dfrac{1}{tan x}}
\ \
&=dfrac{cot xcot y-1}{cot y+cot x}
end{align*}
$$

Note that in the final step we replaced $color{#4257b2}dfrac{1}{tan x}$ and $color{#4257b2}dfrac{1}{tan y}$ by $color{#4257b2}cot x$ and $color{#4257b2}tan y$.

So $color{#4257b2}cot (x+y)=dfrac{cot xcot y-1}{cot y+cot x}$

$$
color{#c34632}dfrac{cot xcot y-1}{cot y+cot x}
$$

We would like to prove the identity
$$
color{#4257b2}cos C+cos D=2cos dfrac{C+D}{2}cos dfrac{C-D}{2}
$$

First, we note that the left side is a sum of two cosine functions which consists of single angle, so we can write them by another way to be compound angle.

We know that $color{#4257b2}dfrac{C+D}{2}+dfrac{C-D}{2}=dfrac{C+D+C-D}{2}=dfrac{2C}{2}=C$ and also we know that $color{#4257b2}dfrac{C+D}{2}-dfrac{C-D}{2}=dfrac{C+D-C+D}{2}=dfrac{2D}{2}=D$, so we can rewrite the left side as follows:

$$
cos C+cos D=cos left(dfrac{C+D}{2}+dfrac{C-D}{2}right)+cos left(dfrac{C+D}{2}-dfrac{C-D}{2}right)
$$

Now the left side consists of two sum cosine functions each of them is compound angle, so we can use the addition and subtraction formulas for the cosine function $color{#4257b2}cos (a+b)=cos acos b-sin asin b$ and $color{#4257b2}cos (a-b)=cos acos b+sin asin b$

$$
begin{align*}
cos C+cos D&=cos left(dfrac{C+D}{2}+dfrac{C-D}{2}right)+cos left(dfrac{C+D}{2}-dfrac{C-D}{2}right)
\ \
&=cos dfrac{C+D}{2}cos dfrac{C-D}{2}-sin dfrac{C+D}{2}sin dfrac{C-D}{2}+cos dfrac{C+D}{2}cos dfrac{C-D}{2}+sin dfrac{C+D}{2}sin dfrac{C-D}{2}
\ \
&=cos dfrac{C+D}{2}cos dfrac{C-D}{2}cancel{-sin dfrac{C+D}{2}sin dfrac{C-D}{2}}+cos dfrac{C+D}{2}cos dfrac{C-D}{2}+cancel{sin dfrac{C+D}{2}sin dfrac{C-D}{2}}
\ \
&=cos dfrac{C+D}{2}cos dfrac{C-D}{2}+cos dfrac{C+D}{2}cos dfrac{C-D}{2}
\ \
&=2cos dfrac{C+D}{2}cos dfrac{C-D}{2}
end{align*}
$$

We proved that the left side is $color{#4257b2}2cos dfrac{C+D}{2}cos dfrac{C-D}{2}$, so the left side=the right side.

$$
color{#c34632}cos C+cos D=2cos dfrac{C+D}{2}cos dfrac{C-D}{2}
$$

We would like to prove the identity
$$
color{#4257b2}cos C-cos D=-2sin dfrac{C+D}{2}sin dfrac{C-D}{2}
$$

First, we note that the left side is a subtraction of two cosine functions which consists of single angle, so we can write them by another way to be compound angle.

We know that $color{#4257b2}dfrac{C+D}{2}+dfrac{C-D}{2}=dfrac{C+D+C-D}{2}=dfrac{2C}{2}=C$ and also we know that $color{#4257b2}dfrac{C+D}{2}-dfrac{C-D}{2}=dfrac{C+D-C+D}{2}=dfrac{2D}{2}=D$, so we can rewrite the left side as follows:

$$
cos C-cos D=cos left(dfrac{C+D}{2}+dfrac{C-D}{2}right)-cos left(dfrac{C+D}{2}-dfrac{C-D}{2}right)
$$

Now the left side consists of two subtraction cosine functions each of them is compound angle, so we can use the addition and subtraction formulas for the cosine function $color{#4257b2}cos (a+b)=cos acos b-sin asin b$ and $color{#4257b2}cos (a-b)=cos acos b+sin asin b$

$$
begin{align*}
cos C-cos D&=cos left(dfrac{C+D}{2}+dfrac{C-D}{2}right)-cos left(dfrac{C+D}{2}-dfrac{C-D}{2}right)
\ \
&=cos dfrac{C+D}{2}cos dfrac{C-D}{2}-sin dfrac{C+D}{2}sin dfrac{C-D}{2}-left(cos dfrac{C+D}{2}cos dfrac{C-D}{2}+sin dfrac{C+D}{2}sin dfrac{C-D}{2}right)
\ \
&=cos dfrac{C+D}{2}cos dfrac{C-D}{2}-sin dfrac{C+D}{2}sin dfrac{C-D}{2}-cos dfrac{C+D}{2}cos dfrac{C-D}{2}-sin dfrac{C+D}{2}sin dfrac{C-D}{2}
\ \
&=cancel{cos dfrac{C+D}{2}cos dfrac{C-D}{2}}-sin dfrac{C+D}{2}sin dfrac{C-D}{2}cancel{-cos dfrac{C+D}{2}cos dfrac{C-D}{2}}-sin dfrac{C+D}{2}sin dfrac{C-D}{2}
\ \
&=-sin dfrac{C+D}{2}sin dfrac{C-D}{2}-sin dfrac{C+D}{2}sin dfrac{C-D}{2}=-2sin dfrac{C+D}{2}sin dfrac{C-D}{2}
end{align*}
$$

We proved that the left side is $color{#4257b2}-2sin dfrac{C+D}{2}sin dfrac{C-D}{2}$, so the left side=the right side.

$$
color{#c34632}cos C-cos D=-2sin dfrac{C+D}{2}sin dfrac{C-D}{2}
$$