We can identify equivalent trigonometric expressions by comparing principal angles drawn in standard position in quadrants II, III, and IV with their related acute angle, $theta$, in quadrant I.

**Principal Angle in Quadrant II, $frac{pi}{2}lethetale pi$**
$$begin{align*}
sin(pi-theta)&=sin theta\
cos(pi-theta)&=-cos theta\
tan(pi-theta)&=-tan theta\
end{align*}tag{1}$$

**Principal Angle in Quadrant III, $pilethetalefrac{3pi}{2}$**
$$begin{align*}
sin(pi+theta)&=-sin theta\
cos(pi+theta)&=-cos theta\
tan(pi+theta)&=tan theta\
end{align*}tag{2}$$

**Principal Angle in Quadrant IV, $frac{3pi}{2}le thetale 2pi$**
$$begin{align*}
sin(2pi-theta)&=-sin theta\
cos(2pi-theta)&=cos theta\
tan(2pi-theta)&=-tan theta\
end{align*}tag{3}$$
We need to determine to which quadrant the given angle belongs, and then apply the suitable equation from set $(1)$, $(2)$, or $(3)$.

*(a)* The given angle is $frac{pi}{16}$. Note that this angle falls in the interval $0le thetalefrac{pi}{2}$. Hence, it belongs to the I quadrant, but we can rewrite the cosine function of the same sign using a set of Eq. $(3)$. The given angle we need to rewrite in the form $2pi-theta$ and compute $theta$. It yields:
$$
dfrac{pi}{16}=2pi-thetaimplies theta=2pi-dfrac{pi}{16}=dfrac{31pi}{16}
$$
Using the equation for the sine function and plugging in $theta=frac{31pi}{16}$, we obtain:
$$
cosfrac{pi}{16}=cosleft(2pi-dfrac{31pi}{16} right)=cos dfrac{31pi}{16}$$

*(b)* The given angle is $frac{7pi}{9}$. Note that this angle falls in the interval $frac{pi}{2}le thetalepi$. Hence, we need to refer to quadrant II and set of Eq. $(1)$. The given angle we need to rewrite in the form $pi-theta$ and compute $theta$. It yields:
$$
dfrac{7pi}{9}=pi-thetaimplies theta=pi-dfrac{7pi}{9}=dfrac{2pi}{9}
$$
Using the equation for the sine function and plugging in $theta=frac{2pi}{9}$, we obtain:
$$
sindfrac{7pi}{9}=sinleft(pi-dfrac{2pi}{9} right)=sindfrac{2pi}{9}$$

*(c)* The given angle is $frac{9pi}{10}$. Note that this angle falls in the interval $0le thetalefrac{pi}{2}$. Hence, it belongs to the I quadrant, but we can rewrite the tangent function of the same sign in quadrant III using a set of Eq. $(2)$ and setting $theta=frac{9pi}{10}$. It yields:

$$
tandfrac{9pi}{10}=tan left(pi+dfrac{9pi}{10} right)=tandfrac{19pi}{10}$$

*(d)* The given angle is $frac{2pi}{5}$. Note that this angle falls in the interval $0le thetalefrac{pi}{2}$. Hence, it belongs to the I quadrant, but we can rewrite the cosine function of the opposite sign using a set of Eq. $(2)$.

Using the equation for the cosine function and plugging in $theta=frac{2pi}{5}$, we obtain:
$$
-cosdfrac{2pi}{5}= cosleft(pi+dfrac{2pi}{5} right)=cos dfrac{7pi}{5}$$

*(e)* The given angle is $frac{9pi}{7}$. Note that this angle falls in the interval $pile thetalefrac{3pi}{2}$. Hence, we need to refer to quadrant III and set of Eq. $(2)$. The given angle we need to rewrite in the form $pi+theta$ and compute $theta$. It yields:
$$
dfrac{9pi}{7}=pi+thetaimplies theta=dfrac{9pi}{7}-pi=dfrac{2pi}{7}
$$
Using the equation for the sine function and plugging in $theta=frac{2pi}{9}$, we obtain:
$$
-sindfrac{9pi}{7}=sinleft(pi+dfrac{2pi}{7} right)=sindfrac{2pi}{7}$$

*(f)* The given angle is $frac{3pi}{4}$. Note that this angle falls in the interval $frac{pi}{2}le thetalepi$. Hence, it belongs to the II quadrant, but we can rewrite the tangent function of the same sign in quadrant III using a set of Eq. $(2)$ and setting $theta=frac{9pi}{10}$. It yields:

$$
tandfrac{3pi}{4}=tan left(pi+dfrac{3pi}{4} right)=tandfrac{7pi}{4}$$

a) $cos frac{31pi}{16}$
b) $sinfrac{2pi}{9}$
c) $tanfrac{19pi}{10}$
d) $cos frac{7pi}{5}$
e) $sinfrac{2pi}{7}$
f) $tanfrac{7pi}{4}$

The cofunction identities describe trigonometric relationships between the complementary angles $x$ and $frac{pi}{2}-x$ in a right triangle. For the cosine function, we have that
$$
cos left(dfrac{pi}{2}-x right)=sin xtag{1}$$
We can transform the angle and apply this formula to write the equation equivalent to the given one.

The given angle we can rewrite as
$$
dfrac{pi}{2}+x=dfrac{pi}{2}-(-x)$$
which now looks similar to the angle in the Eq. $(1)$. Setting $-x$ instead of $x$ in this equation, we obtain
$$cos left(dfrac{pi}{2}-(-x )right)=sin (-x)=-sin xtag{2}$$
Note that the sine function is an odd function and $sin (-x)=-sin x$.

Now we can rewrite the cosine function as a sine function using Eq. $(2)$ into the given equation and obtain
$$
y=-6(-sin x)+4=6sin x+4$$

$y=6sin x+4$

(a) We would like to determine a trigonometric expression that is equivalent to the expression $color{#4257b2}cos left(x+dfrac{5pi}{3}right)$. First, we note that our expression is a cosine function with two sum angles, so we can use the addition formula for the cosine function $color{#4257b2}cos(a+b)=cos acos b-sin asin b$.

$$
begin{align*}
cos left(x+dfrac{5pi}{3}right)&=cos xcos dfrac{5pi}{3}-sin xsin dfrac{5pi}{3}
\ \
&=cos xcdot left(dfrac{1}{2}right)-sin xcdot left(-dfrac{sqrt{3}}{2}right)
\ \
&=dfrac{1}{2} cos x+dfrac{sqrt{3}}{2} sin x
\ \
&=dfrac{cos x+sqrt{3}sin x}{2}
end{align*}
$$

Note that the angle $color{#4257b2}dfrac{5pi}{3}$ is a special angle which we know the values of the sine and cosine functions for it where $color{#4257b2}sin dfrac{5pi}{3}=sin left(2pi-dfrac{pi}{3}right)=-sin dfrac{pi}{3}=-dfrac{sqrt{3}}{2}$ and $color{#4257b2}cos dfrac{5pi}{3}=cos left(2pi-dfrac{pi}{3}right)=cos dfrac{pi}{3}=dfrac{1}{2}$.

So the expression $color{#4257b2}cos left(x+dfrac{5pi}{3}right)$ can be written as the equivalent expression $boxed{ dfrac{cos x+sqrt{3}sin x}{2} }$

(b) We would like to determine a trigonometric expression that is equivalent to the expression $color{#4257b2}sin left(x+dfrac{5pi}{6}right)$. First, we note that our expression is a sine function with two sum angles, so we can use the addition formula for the sine function $color{#4257b2}sin(a+b)=sin acos b+cos asin b$.

$$
begin{align*}
sin left(x+dfrac{5pi}{6}right)&=sin xcos dfrac{5pi}{6}+cos xsin dfrac{5pi}{6}
\ \
&=sin xcdot left(-dfrac{sqrt{3}}{2}right)+cos xcdot left(dfrac{1}{2}right)
\ \
&=-dfrac{sqrt{3}}{2} sin x+dfrac{1}{2} cos x
\ \
&=dfrac{-sqrt{3}sin x+cos x}{2}
end{align*}
$$

Note that the angle $color{#4257b2}dfrac{5pi}{6}$ is a special angle which we know the values of the sine and cosine functions for it where $color{#4257b2}sin dfrac{5pi}{6}=sin left(pi-dfrac{pi}{6}right)=sin dfrac{pi}{6}=dfrac{1}{2}$ and $color{#4257b2}cos dfrac{5pi}{6}=cos left(pi-dfrac{pi}{6}right)=-cos dfrac{pi}{6}=-dfrac{sqrt{3}}{2}$.

So the expression $color{#4257b2}sin left(x+dfrac{5pi}{6}right)$ can be written as the equivalent expression $boxed{ dfrac{-sqrt{3}sin x+cos x}{2} }$

(c) We would like to determine a trigonometric expression that is equivalent to the expression $color{#4257b2}tan left(x+dfrac{5pi}{4}right)$. First, we note that our expression is a tangent function with two sum angles, so we can use the addition formula for the tangent function $color{#4257b2}tan(a+b)=dfrac{tan a+tan b}{1-tan atan b}$.

$$
begin{align*}
tan left(x+dfrac{5pi}{4}right)&=dfrac{tan x+tan dfrac{5pi}{4}}{1-tan xtan dfrac{5pi}{4}}
\ \
&=dfrac{tan x+1}{1-tan xcdot (1)}
\ \
&=dfrac{tan x+1}{1-tan x}
\ \
&=dfrac{1+tan x}{1-tan x}
end{align*}
$$

Note that the angle $color{#4257b2}dfrac{5pi}{4}$ is a special angle which we know the value of the tangent function for it where $color{#4257b2}tan dfrac{5pi}{4}=tan left(pi+dfrac{pi}{4}right)=tan dfrac{pi}{4}=1$.

So the expression $color{#4257b2}tan left(x+dfrac{5pi}{4}right)$ can be written as the equivalent expression $boxed{ dfrac{1+tan x}{1-tan x} }$

(d) We would like to determine a trigonometric expression that is equivalent to the expression $color{#4257b2}cos left(x+dfrac{4pi}{3}right)$. First, we note that our expression is a cosine function with two sum angles, so we can use the addition formula for the cosine function $color{#4257b2}cos(a+b)=cos acos b-sin asin b$.

$$
begin{align*}
cos left(x+dfrac{4pi}{3}right)&=cos xcos dfrac{4pi}{3}-sin xsin dfrac{4pi}{3}
\ \
&=cos xcdot left(-dfrac{1}{2}right)-sin xcdot left(-dfrac{sqrt{3}}{2}right)
\ \
&=-dfrac{1}{2} cos x+dfrac{sqrt{3}}{2} sin x
\ \
&=dfrac{-cos x+sqrt{3}sin x}{2}
end{align*}
$$

Note that the angle $color{#4257b2}dfrac{4pi}{3}$ is a special angle which we know the values of the sine and cosine functions for it where $color{#4257b2}sin dfrac{4pi}{3}=sin left(pi+dfrac{pi}{3}right)=-sin dfrac{pi}{3}=-dfrac{sqrt{3}}{2}$ and $color{#4257b2}cos dfrac{4pi}{3}=cos left(pi+dfrac{pi}{3}right)=-cos dfrac{pi}{3}=-dfrac{1}{2}$.

So the expression $color{#4257b2}cos left(x+dfrac{pi}{3}right)$ can be written as the equivalent expression $boxed{ dfrac{-cos x+sqrt{3}sin x}{2} }$

$$
text{color{#c34632}$(a) dfrac{cos x+sqrt{3}sin x}{2}$ $(b) dfrac{-sqrt{3}sin x+cos x}{2}$
\
\
$(c) dfrac{1+tan x}{1-tan x}$ $(d) dfrac{-cos x+sqrt{3}sin x}{2}$}
$$

(a) We would like to determine a trigonometric expression that is equivalent to the expression $color{#4257b2}sin left(x-dfrac{11pi}{6}right)$. First, we note that our expression is a sine function with two subtraction angles, so we can use the subtraction formula for the sine function $color{#4257b2}sin(a-b)=sin acos b-cos asin b$.

$$
begin{align*}
sin left(x-dfrac{11pi}{6}right)&=sin xcos dfrac{11pi}{6}-cos xsin dfrac{11pi}{6}
\ \
&=sin xcdot left(dfrac{sqrt{3}}{2}right)-cos xcdot left(-dfrac{1}{2}right)
\ \
&=dfrac{sqrt{3}}{2} sin x+dfrac{1}{2} cos x
\ \
&=dfrac{sqrt{3}sin x+cos x}{2}
end{align*}
$$

Note that the angle $color{#4257b2}dfrac{11pi}{6}$ is a special angle which we know the values of the sine and cosine functions for it where
$color{#4257b2}sin dfrac{11pi}{6}=sin left(2pi-dfrac{pi}{6}right)=-sin dfrac{pi}{6}=-dfrac{1}{2}$ and $color{#4257b2}cos dfrac{11pi}{6}=cos left(2pi-dfrac{pi}{6}right)=cos dfrac{pi}{6}=dfrac{sqrt{3}}{2}$.

So the expression $color{#4257b2}sin left(x-dfrac{11pi}{6}right)$ can be written as the equivalent expression $boxed{ dfrac{sqrt{3}sin x+cos x}{2} }$

(b) We would like to determine a trigonometric expression that is equivalent to the expression $color{#4257b2}tan left(x-dfrac{pi}{3}right)$. First, we note that our expression is a tangent function with two subtraction angles, so we can use the subtraction formula for the tangent function $color{#4257b2}tan(a-b)=dfrac{tan a-tan b}{1+tan atan b}$.

$$
begin{align*}
tan left(x-dfrac{pi}{3}right)&=dfrac{tan x-tan dfrac{pi}{3}}{1+tan xtan dfrac{pi}{3}}
\ \
&=dfrac{tan x-sqrt{3}}{1+tan xcdot left(sqrt{3}right)}
\ \
&=dfrac{tan x-sqrt{3}}{1+sqrt{3} tan x}
end{align*}
$$

Note that the angle $color{#4257b2}dfrac{pi}{3}$ is a special angle which we know the value of the tangent function for it where $color{#4257b2}tan dfrac{pi}{3}=sqrt{3}$.

So the expression $color{#4257b2}tan left(x-dfrac{pi}{3}right)$ can be written as the equivalent expression $boxed{ dfrac{tan x-sqrt{3}}{1+sqrt{3} tan x} }$

(c) We would like to determine a trigonometric expression that is equivalent to the expression $color{#4257b2}cos left(x-dfrac{7pi}{4}right)$. First, we note that our expression is a cosine function with two subtraction angles, so we can use the subtraction formula for the cosine function $color{#4257b2}cos(a-b)=cos acos b+sin asin b$.

$$
begin{align*}
cos left(x-dfrac{7pi}{4}right)&=cos xcos dfrac{7pi}{4}+sin xsin dfrac{7pi}{4}
\ \
&=cos xcdot left(dfrac{sqrt{2}}{2}right)+sin xcdot left(-dfrac{sqrt{2}}{2}right)
\ \
&=dfrac{sqrt{2}}{2} cos x-dfrac{sqrt{2}}{2} sin x
\ \
&=dfrac{sqrt{2}}{2} left(cos x-sin xright)
end{align*}
$$

Note that the angle $color{#4257b2}dfrac{7pi}{4}$ is a special angle which we know the values of the sine and cosine functions for it where $color{#4257b2}sin dfrac{7pi}{4}=sin left(2pi-dfrac{pi}{4}right)=-sin dfrac{pi}{4}=-dfrac{sqrt{2}}{2}$ and $color{#4257b2}cos dfrac{7pi}{4}=cos left(2pi-dfrac{pi}{4}right)=cos dfrac{pi}{4}=dfrac{sqrt{2}}{2}$.

So the expression $color{#4257b2}cos left(x-dfrac{7pi}{4}right)$ can be written as the equivalent expression $boxed{ dfrac{sqrt{2}}{2} left(cos x-sin xright) }$

(d) We would like to determine a trigonometric expression that is equivalent to the expression $color{#4257b2}sin left(x-dfrac{2pi}{3}right)$. First, we note that our expression is a sine function with two subtraction angles, so we can use the subtraction formula for the sine function $color{#4257b2}sin(a-b)=sin acos b-cos asin b$.

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

Note that the angle $color{#4257b2}dfrac{2pi}{3}$ is a special angle which we know the values of the sine and cosine functions for it where $color{#4257b2}sin dfrac{2pi}{3}=sin left(pi-dfrac{pi}{3}right)=sin dfrac{pi}{3}=dfrac{sqrt{3}}{2}$ and $color{#4257b2}cos dfrac{2pi}{3}=cos left(pi-dfrac{pi}{3}right)=-cos dfrac{pi}{3}=-dfrac{1}{2}$.

So the expression $color{#4257b2}sin left(x-dfrac{11pi}{6}right)$ can be written as the equivalent expression $boxed{ dfrac{-sin x-sqrt{3} cos x}{2} }$

$$
text{color{#c34632}$(a) dfrac{sqrt{3}sin x+cos x}{2}$ $(b) dfrac{tan x-sqrt{3}}{1+sqrt{3} tan x}$
\
\
\
$(c) dfrac{sqrt{2}}{2} left(cos x-sin xright)$ $(d) dfrac{-sin x-sqrt{3}cos x}{2}$}
$$

(a) We would like to evaluate the expression $color{#4257b2}dfrac{tan dfrac{8pi}{9}-tan dfrac{5pi}{9}}{1+tan dfrac{8pi}{9}tan dfrac{5pi}{9}}$. First, we note that our expression is on the form of the subtraction formula for the tangent function $color{#4257b2}tan(a-b)=dfrac{tan a-tan b}{1+tan atan b}$ where $color{#4257b2}a$ and $color{#4257b2}b$ are considered to be $color{#4257b2}dfrac{8pi}{9}$ and $color{#4257b2}dfrac{5pi}{9}$ in our expression, so we can use this formula to simplify and evaluate our expression.

$$
begin{align*}
dfrac{tan dfrac{8pi}{9}-tan dfrac{5pi}{9}}{1+tan dfrac{8pi}{9}tan dfrac{5pi}{9}}&=tan left(dfrac{8pi}{9}-dfrac{5pi}{9}right)
\ \
&=tan left(dfrac{8pi-5pi}{9}right)
\ \
&=tan dfrac{3pi}{9}
\ \
&=tan dfrac{pi}{3}
\ \
&=sqrt{3}
end{align*}
$$

Note that the angle $color{#4257b2}dfrac{pi}{3}$ is a special angle which we know the value of the tangent function for it where $color{#4257b2}tan dfrac{pi}{3}=sqrt{3}$.

So the expression $color{#4257b2}dfrac{tan dfrac{8pi}{9}-tan dfrac{5pi}{9}}{1+tan dfrac{8pi}{9}tan dfrac{5pi}{9}}$ equals $boxed{ sqrt{3} }$

(b) We would like to evaluate the expression $color{#4257b2}sin dfrac{299pi}{298}cos dfrac{pi}{298}-cos dfrac{299pi}{298}sin dfrac{pi}{298}$. First, we note that our expression is on the form of the subtraction formula for the sine function $color{#4257b2}sin(a-b)=sin acos b-cos asin b$ where $color{#4257b2}a$ and $color{#4257b2}b$ are considered to be $color{#4257b2}dfrac{299pi}{298}$ and $color{#4257b2}dfrac{pi}{298}$ in our expression, so we can use this formula to simplify and evaluate our expression.

$$
begin{align*}
sin dfrac{299pi}{298}cos dfrac{pi}{298}-cos dfrac{299pi}{298}sin dfrac{pi}{298}&=sin left(dfrac{299pi}{298}-dfrac{pi}{298}right)
\ \
&=sin left(dfrac{299pi-pi}{298}right)
\
&=sin dfrac{298pi}{298}
\ \
&=sin pi
\ \
&=0
end{align*}
$$

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

So the expression $color{#4257b2}sin dfrac{299pi}{298}cos dfrac{pi}{298}-cos dfrac{299pi}{298}sin dfrac{pi}{298}$ equals $boxed{ 0 }$

(c) We would like to evaluate the expression $color{#4257b2}sin 50text{textdegree}cos 20text{textdegree}-cos 50text{textdegree}sin 20text{textdegree}$. First, we note that our expression is on the form of the subtraction formula for the sine function $color{#4257b2}sin(a-b)=sin acos b-cos asin b$ where $color{#4257b2}a$ and $color{#4257b2}b$ are considered to be $color{#4257b2}50text{textdegree}$ and $color{#4257b2}20text{textdegree}$ in our expression, so we can use this formula to simplify and evaluate our expression.

$$
begin{align*}
sin 50text{textdegree}cos 20text{textdegree}-cos 50text{textdegree}sin 20text{textdegree}&=sin left(50text{textdegree}-20text{textdegree}right)
\
&=sin 30text{textdegree}
\ \
&=dfrac{1}{2}
end{align*}
$$

Note that the angle $color{#4257b2}30text{textdegree}$ is a special angle which we know the value of the sine function for it where $color{#4257b2}sin 30text{textdegree}=dfrac{1}{2}$.

So the expression $color{#4257b2}sin 50text{textdegree}cos 20text{textdegree}-cos 50text{textdegree}sin 20text{textdegree}$ equals $boxed{ dfrac{1}{2} }$

(d) We would like to evaluate the expression $color{#4257b2}sin dfrac{3pi}{8}cos dfrac{pi}{8}+cos dfrac{3pi}{8}sin dfrac{pi}{8}$. First, we note that our expression is on the form of the addition formula for the sine function $color{#4257b2}sin(a+b)=sin acos b+cos asin b$ where $color{#4257b2}a$ and $color{#4257b2}b$ are considered to be $color{#4257b2}dfrac{3pi}{8}$ and $color{#4257b2}dfrac{pi}{8}$ in our expression, so we can use this formula to simplify and evaluate our expression.

$$
begin{align*}
sin dfrac{3pi}{8}cos dfrac{pi}{8}+cos dfrac{3pi}{8}sin dfrac{pi}{8}&=sin left(dfrac{3pi}{8}+dfrac{pi}{8}right)
\ \
&=sin left(dfrac{3pi+pi}{8}right)
\
&=sin dfrac{4pi}{8}
\ \
&=sin dfrac{pi}{2}
\ \
&=1
end{align*}
$$

Note that the angle $color{#4257b2}dfrac{pi}{2}$ is a special angle which we know the value of the sine function for it where $color{#4257b2}sin dfrac{pi}{2}=1$.

So the expression $color{#4257b2}sin dfrac{3pi}{8}cos dfrac{pi}{8}+cos dfrac{3pi}{8}sin dfrac{pi}{8}$ equals $boxed{ 1 }$

$$
text{color{#c34632}$(a) sqrt{3}$ $(b) 0$ $(c) dfrac{1}{2}$ $(d) 1$}
$$

(a) We would like to simplify the expression $color{#4257b2}dfrac{2tan x}{1-tan^{2}x}$. First, we know that $color{#4257b2}2tan x=tan x+tan x$ and $color{#4257b2}tan^{2}x=tan xtan x$, so we can rewrite our expression as follows:

$$
dfrac{2tan x}{1-tan^{2}x}=dfrac{tan x+tan x}{1-tan xtan x}
$$

Now we note that our expression is on the form of the addition formula for the tangent function $color{#4257b2}tan(a+b)=dfrac{tan a+tan b}{1-tan atan b}$ where $color{#4257b2}a$ and $color{#4257b2}b$ equal $color{#4257b2}x$ in our expression, so we can use this formula to simplify and simplify our expression.

$$
begin{align*}
dfrac{2tan x}{1-tan^{2}x}&=dfrac{tan x+tan x}{1-tan xtan x}
\ \
&=tan left(x+xright)
\ \
&=tan 2x
end{align*}
$$

So the expression $color{#4257b2}dfrac{2tan x}{1-tan^{2}x}$ can be simplified to $boxed{ tan 2x }$

(b) We would like to simplify the expression $color{#4257b2}sin dfrac{x}{5}cos dfrac{4x}{5}+cos dfrac{x}{5}sin dfrac{4x}{5}$. First, we note that our expression is on the form of the addition formula for the sine function $color{#4257b2}sin(a+b)=sin acos b+cos asin b$ where $color{#4257b2}a$ and $color{#4257b2}b$ are considered to be $color{#4257b2}dfrac{x}{5}$ and $color{#4257b2}dfrac{4x}{5}$ in our expression, so we can use this formula to simplify and simplify our expression.

$$
begin{align*}
sin dfrac{x}{5}cos dfrac{4x}{5}+cos dfrac{x}{5}sin dfrac{4x}{5}&=sin left(dfrac{x}{5}+dfrac{4x}{5}right)
\ \
&=sin dfrac{5x}{5}
\ \
&=sin x
end{align*}
$$

So the expression $color{#4257b2}sin dfrac{x}{5}cos dfrac{4x}{5}+cos dfrac{x}{5}sin dfrac{4x}{5}$ can be simplified to $boxed{ sin x }$

(c) We would like to simplify the expression $color{#4257b2}cos left(dfrac{pi}{2}-xright)$. First, we note that our expression is a cosine function with two subtraction angles, so we can use the subtraction formula for the cosine function where
$color{#4257b2}cos(a-b)=cos acos b+sin asin b$.

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

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

So the expression $color{#4257b2}cos left(dfrac{pi}{2}-xright)$ can be simplified to $boxed{ sin x }$

(d) We would like to simplify the expression $color{#4257b2}sin left(dfrac{pi}{2}+xright)$. First, we note that our expression is a sine function with two sum angles, so we can use the addition formula for the sine function where
$color{#4257b2}sin(a+b)=sin acos b+cos asin b$.

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

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

So the expression $color{#4257b2}sin left(dfrac{pi}{2}+xright)$ can be simplified to $boxed{ cos x }$

(e) We would like to simplify the expression $color{#4257b2}cos left(dfrac{pi}{4}+xright)+cos left(dfrac{pi}{4}+xright)$. First, we note that our expression is a sum for two cosine functions with the same angle, so we can add them to simplify as follows:

$$
cos left(dfrac{pi}{4}+xright)+cos left(dfrac{pi}{4}+xright)=2cos left(dfrac{pi}{4}+xright)
$$

Now we note that our expression is a cosine function with two sum angles, so we can use the addition formula for the cosine function where
$color{#4257b2}cos(a+b)=cos acos b-sin asin b$.

$$
begin{align*}
cos left(dfrac{pi}{4}+xright)+cos left(dfrac{pi}{4}+xright)&=2cos left(dfrac{pi}{4}+xright)
\ \
&=2left(cos dfrac{pi}{4}cos x-sin dfrac{pi}{4}sin xright)
\ \
&=2left(dfrac{sqrt{2}}{2}cdot cos x-dfrac{sqrt{2}}{2}cdot sin xright)
\ \
&=sqrt{2} cos x-sqrt{2} sin x
\ \
&=sqrt{2}left(cos x-sin xright)
end{align*}
$$

Note that the angle $color{#4257b2}dfrac{pi}{4}$ is a special angle which we know the values of the sine and cosine functions for it where $color{#4257b2}sin dfrac{pi}{4}=cos dfrac{pi}{4}=dfrac{sqrt{2}}{2}$.

So the expression $color{#4257b2}cos left(dfrac{pi}{4}+xright)+cos left(dfrac{pi}{4}+xright)$ can be simplified to $boxed{ sqrt{2}left(cos x-sin xright) }$

(f) We would like to simplify the expression $color{#4257b2}tan left(x-dfrac{pi}{4}right)$. First, we note that our expression is a tangent function with two subtraction angles, so we can use the subtraction formula for the tangent function where
$color{#4257b2}tan(a-b)=dfrac{tan a-tan b}{1+tan atan b}$.

$$
begin{align*}
cos left(x-dfrac{pi}{4}right)&=dfrac{tan x-tan dfrac{pi}{4}}{1+tan xtan dfrac{pi}{4}}
\ \
&=dfrac{tan x-1}{1+tan xcdot 1}
\ \
&=dfrac{tan x-1}{1+tan x}
\ \
&=dfrac{tan x-1}{tan x+1}
end{align*}
$$

Note that the angle $color{#4257b2}dfrac{pi}{4}$ is a special angle which we know the value of the tangent function for it where $color{#4257b2}tan dfrac{pi}{4}=1$.

So the expression $color{#4257b2}tan left(x-dfrac{pi}{4}right)$ can be simplified to $boxed{ dfrac{tan x-1}{tan x+1} }$

$$
text{color{#c34632}$(a) tan 2x$ $(b) sin x$ $(c) sin x$
\
\
$(d) cos x$ $(e) sqrt{2} left(cos x-sin xright)$ $(f) dfrac{tan x-1}{tan x+1}$}
$$

We would like to write an expression that is equivalent to $color{#4257b2}sqrt{3} cos x-3sin x$ using the information which say that the expression $color{#4257b2}acos x+bsin x$ can be expressed in the form $color{#4257b2}Rcos left(x-alpharight)$, where $color{#4257b2}R=sqrt{a^{2}+b^{2}}, cos alpha=dfrac{a}{R}$ and $color{#4257b2}sin alpha=dfrac{b}{R}$.

First, we note that our expression $color{#4257b2}sqrt{3} cos x-3sin x$ is on the same form of the expression $color{#4257b2}acos x+bsin x$ where $color{#4257b2}a=sqrt{3}$ and $color{#4257b2}b=-3$, so we can express our expression to the form $color{#4257b2}Rcos left(x-alpharight)$ but we need to evaluate $color{#4257b2}R$ and $color{#4257b2}alpha$. We can use the information $color{#4257b2}R=sqrt{a^{2}+b^{2}}, cos alpha=dfrac{a}{R}$ and $color{#4257b2}sin alpha=dfrac{b}{R}$ to evaluate $color{#4257b2}R$ and $color{#4257b2}alpha$.

For $color{#4257b2}R$

$$
begin{align*}
R=sqrt{a^{2}+b^{2}}&=sqrt{left(sqrt{3}right)^{2}+(-3)^{2}}
\ \
&=sqrt{3+9}
\ \
&=sqrt{12}=2sqrt{3}
end{align*}
$$

For $color{#4257b2}alpha$

$$
cos alpha=dfrac{a}{R}=dfrac{sqrt{3}}{2sqrt{3}}=dfrac{1}{2}
$$

Now we can take $color{#4257b2}cos^{-1}$ for each side to find the value of $color{#4257b2}alpha$.

$$
cos alpha=dfrac{1}{2}
$$

$$
cos^{-1}left(cos alpharight)=cos^{-1}left(dfrac{1}{2}right)
$$

But we know that $color{#4257b2}cos alpha$ is positive in quadrant $1$ and quadrant $4$, so we can find the values of $color{#4257b2}alpha$ where the reference angle is $color{#4257b2}60text{textdegree}$.

$$
alpha=60text{textdegree} text{or} alpha=360text{textdegree}-60text{textdegree}
$$

$$
alpha=60text{textdegree} text{or} alpha=300text{textdegree}
$$

Now we have two values for the angle $color{#4257b2}alpha$ but we know also that $color{#4257b2}sin alpha=dfrac{b}{R}=dfrac{-3}{2sqrt{3}}=-dfrac{sqrt{3}}{2}$, so we note that the sine value for the angle $color{#4257b2}alpha$ is negative which means that the angle $color{#4257b2}alpha$ must be in quadrant $3$ or quadrant $4$ and we determined $color{#4257b2}alpha$ from the cosine information that it will be in quadrant $1$ or quadrant $4$, so we can result that the angle $color{#4257b2}alpha$ will be in quadrant $4$.

So the solution of $color{#4257b2}alpha=60text{textdegree}$ is refused and the right solution is $color{#4257b2}alpha=300text{textdegree}$.

Now we evaluated the values of $color{#4257b2}R=2sqrt{3}$ and $color{#4257b2}alpha=300text{textdegree}$, so we can write the expression $color{#4257b2}sqrt{3} cos x-3sin x$ in the equivalent form
$$
boxed{ 2sqrt{3}cos left(x-300text{textdegree}right) }
$$

$$
color{#c34632}2sqrt{3}cos left(x-300text{textdegree}right)
$$

(a) We would like to evaluate the expression $color{#4257b2}2cos^{2}dfrac{2pi}{3}-1$. First, we note that our expression is on the form of the double angle formula for the cosine function $color{#4257b2}cos 2theta=2cos^{2}theta-1$ where $color{#4257b2}theta$ is considered to be $color{#4257b2}dfrac{2pi}{3}$ in our expression, so we can use this formula to simplify and evaluate our expression.

$$
begin{align*}
2cos^{2}dfrac{2pi}{3}-1&=cos left(2cdot dfrac{2pi}{3}right)
\ \
&=cos dfrac{4pi}{3}
\ \
&=cosleft(pi+dfrac{pi}{3}right)
\ \
&=-cos dfrac{pi}{3}
\ \
&=-dfrac{1}{2}
end{align*}
$$

Note that the angle $color{#4257b2}dfrac{pi}{3}$ is a special angle which we know the value of the cosine function for it where $color{#4257b2}cos dfrac{pi}{3}=dfrac{1}{2}$.

So the expression $color{#4257b2}2cos^{2}dfrac{2pi}{3}-1$ equals $boxed{ -dfrac{1}{2} }$

(b) We would like to evaluate the expression $color{#4257b2}2sin dfrac{11pi}{12}cos dfrac{11pi}{12}$. First, we note that our expression is on the form of the double angle formula for the sine function $color{#4257b2}sin 2theta=2sin thetacos theta$ where $color{#4257b2}theta$ is considered to be $color{#4257b2}dfrac{11pi}{12}$ in our expression, so we can use this formula to simplify and evaluate our expression.

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

Note that the angle $color{#4257b2}dfrac{pi}{6}$ is a special angle which we know the value of the sine function for it where $color{#4257b2}sin dfrac{pi}{6}=dfrac{1}{2}$.

So the expression $color{#4257b2}2sin dfrac{11pi}{12}cos dfrac{11pi}{12}$ equals $boxed{ -dfrac{1}{2} }$

(c) We would like to evaluate the expression $color{#4257b2}cos^{2}dfrac{7pi}{8}-sin^{2}dfrac{7pi}{8}$. First, we note that our expression is on the form of the double angle formula for the cosine function $color{#4257b2}cos 2theta=cos^{2}theta-sin^{2}theta$ where $color{#4257b2}theta$ is considered to be $color{#4257b2}dfrac{7pi}{8}$ in our expression, so we can use this formula to simplify and evaluate our expression.

$$
begin{align*}
cos^{2}dfrac{7pi}{8}-sin^{2}dfrac{7pi}{8}&=cos left(2cdot dfrac{7pi}{8}right)
\ \
&=cos dfrac{7pi}{4}
\ \
&=cosleft(2pi-dfrac{pi}{4}right)
\ \
&=cos dfrac{pi}{4}
\ \
&=dfrac{sqrt{2}}{2}
end{align*}
$$

Note that the angle $color{#4257b2}dfrac{pi}{4}$ is a special angle which we know the value of the cosine function for it where $color{#4257b2}cos dfrac{pi}{4}=dfrac{sqrt{2}}{2}$.

So the expression $color{#4257b2}cos^{2}dfrac{7pi}{8}-sin^{2}dfrac{7pi}{8}$ equals $boxed{ dfrac{sqrt{2}}{2} }$

(d) We would like to evaluate the expression $color{#4257b2}1-2sin^{2}dfrac{pi}{2}$. First, we note that our expression is on the form of the double angle formula for the cosine function $color{#4257b2}cos 2theta=1-2sin^{2}theta$ where $color{#4257b2}theta$ is considered to be $color{#4257b2}dfrac{pi}{2}$ in our expression, so we can use this formula to simplify and evaluate our expression.

$$
begin{align*}
1-2sin^{2}dfrac{pi}{2}&=cos left(2cdot dfrac{pi}{2}right)
\ \
&=cos pi
\ \
&=-1
end{align*}
$$

Note that the angle $color{#4257b2}pi$ is a special angle which we know the value of the cosine function for it where $color{#4257b2}cos pi=-1$.

So the expression $color{#4257b2}1-2sin^{2}dfrac{pi}{2}$ equals $boxed{ -1 }$

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

(a) We would like to determine $color{#4257b2}sin x$. First, we note that $color{#4257b2}cos^{2}x=dfrac{10}{11}$, so we can use the Pythagorean identity $color{#4257b2}sin^{2}theta+cos^{2}theta=1$ to find $color{#4257b2}sin x$.

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

$$
sin^{2}x+dfrac{10}{11}=1
$$

$$
sin^{2}x=1-dfrac{10}{11}=dfrac{1}{11}
$$

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

$$
sin x=pm sqrt{dfrac{1}{11}}=pm dfrac{1}{sqrt{11}}
$$

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

$$
boxed{ sin x=-dfrac{1}{sqrt{11}} }
$$

(b) We would like to determine $color{#4257b2}sin x$. First, we note that $color{#4257b2}cos^{2}x=dfrac{10}{11}$, so we can take the square root for each side to find $color{#4257b2}cos x$.

$$
cos^{2}x=dfrac{10}{11}
$$

$$
cos x=pm sqrt{dfrac{10}{11}}
$$

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

$$
boxed{ cos x=-sqrt{dfrac{10}{11}} }
$$

(c) We would like to determine $color{#4257b2}sin 2x$. First, we note that $color{#4257b2}sin 2x$ is a sine function with double angle of $color{#4257b2}x$, so we can use the double angle formula for the sine function $color{#4257b2}sin 2x=2sin xcos x$.

$$
begin{align*}
sin 2x&=2sin xcos x
\ \
&=2cdot left(-dfrac{1}{sqrt{11}}right)cdot left(-sqrt{dfrac{10}{11}}right)
\ \
&=dfrac{2sqrt{10}}{11}
end{align*}
$$

Note that we used the values of $color{#4257b2}sin x$ and $color{#4257b2}cos x$ from the first two requirements.

So $boxed{ sin 2x=dfrac{2sqrt{10}}{11} }$

(d) We would like to determine $color{#4257b2}cos 2x$. First, we note that $color{#4257b2}cos 2x$ is a cosine function with double angle of $color{#4257b2}x$, so we can use the double angle formula for the cosine function $color{#4257b2}cos 2x=cos^{2}x-sin^{2}x$.

$$
begin{align*}
cos 2x&=cos^{2}x-sin^{2}x
\ \
&=left(-sqrt{dfrac{10}{11}}right)^{2}-left(-dfrac{1}{sqrt{11}}right)^{2}
\ \
&=dfrac{10}{11}-dfrac{1}{11}
\ \
&=dfrac{9}{11}
end{align*}
$$

Note that we used the values of $color{#4257b2}sin x$ and $color{#4257b2}cos x$ from the first two requirements.

So $boxed{ cos 2x=dfrac{9}{11} }$

$$
text{color{#c34632}$(a) -dfrac{1}{sqrt{11}}$ $(b) -sqrt{dfrac{10}{11}}$ $(c) dfrac{2sqrt{10}}{11}$ $(d) dfrac{9}{11}$}
$$

In this problem we would like to determine $color{#4257b2}sin 2x$ and $color{#4257b2}cos 2x$, so we need to know the values of $color{#4257b2}sin x$ and $color{#4257b2}cos x$ and then substitute in the double angle formulas for the sine and cosine functions. First, we note that $color{#4257b2}sin x=dfrac{3}{5}$, so we can use the Pythagorean identity $color{#4257b2}sin^{2}theta+cos^{2}theta=1$ to find $color{#4257b2}cos x$.

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

$$
left(dfrac{3}{5}right)^{2}+cos^{2}x=1
$$

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

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

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

$$
cos x=pm sqrt{dfrac{16}{25}}=pm dfrac{4}{5}
$$

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

$$
boxed{ cos x=dfrac{4}{5} }
$$

Now we have the values of $color{#4257b2}sin x$ and $color{#4257b2}cos x$, so we can use the double angle formulas for the sine and cosine functions $color{#4257b2}sin 2x=2sin xcos x$ and
$color{#4257b2}cos 2x=cos^{2}x-sin^{2}x$.

$$
begin{align*}
sin 2x&=2sin xcos x
\ \
&=2cdot left(dfrac{3}{5}right)cdot left(dfrac{4}{5}right)
\ \
&=dfrac{24}{25}
end{align*}
$$

So $boxed{ sin 2x=dfrac{24}{25} }$

$$
begin{align*}
cos 2x&=cos^{2}x-sin^{2}x
\ \
&=left(dfrac{4}{5}right)^{2}-left(dfrac{3}{5}right)^{2}
\ \
&=dfrac{16}{25}-dfrac{9}{25}
\ \
&=dfrac{7}{25}
end{align*}
$$

So $boxed{ cos 2x=dfrac{7}{25} }$

$$
text{color{#c34632}$(a) sin 2x=dfrac{24}{25}$ $(b) cos 2x=dfrac{7}{25}$}
$$

In this problem we would like to determine $color{#4257b2}sin 2x$, so we need to know the values of $color{#4257b2}sin x$ and $color{#4257b2}cos x$ and then substitute in the double angle formulas for the sine and cosine functions. First, we note that $color{#4257b2}sin x=dfrac{5}{13}$, so we can use the Pythagorean identity $color{#4257b2}sin^{2}theta+cos^{2}theta=1$ to find $color{#4257b2}cos x$.

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

$$
left(dfrac{5}{13}right)^{2}+cos^{2}x=1
$$

$$
dfrac{25}{169}+cos^{2}x=1
$$

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

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

$$
cos x=pm sqrt{dfrac{144}{169}}=pm dfrac{12}{13}
$$

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

$$
boxed{ cos x=dfrac{12}{13} }
$$

Now we have the values of $color{#4257b2}sin x$ and $color{#4257b2}cos x$, so we can use the double angle formula for the sine function $color{#4257b2}sin 2x=2sin xcos x$.

$$
begin{align*}
sin 2x&=2sin xcos x
\ \
&=2cdot left(dfrac{5}{13}right)cdot left(dfrac{12}{13}right)=dfrac{120}{169}
end{align*}
$$

So $boxed{ sin 2x=dfrac{120}{169} }$

$$
color{#c34632}sin 2x=dfrac{120}{169}
$$

In this problem we would like to determine $color{#4257b2}tan 2x$, so we need to know the values of $color{#4257b2}tan x$ and then substitute in the double angle formulas for the tangent function. First, we note that $color{#4257b2}cos x=-dfrac{4}{5}$, so we can use the identity $color{#4257b2}sec theta=dfrac{1}{cos theta}$ to find $color{#4257b2}sec x$.

$$
sec x=dfrac{1}{cos x}
$$

$$
sec x=dfrac{1}{-dfrac{4}{5}}=-dfrac{5}{4}
$$

Now we can use the Pythagorean identity $color{#4257b2}1+tan^{2}theta=sec^{2}theta$ to find $color{#4257b2}tan x$.

$$
1+tan^{2}x=sec^{2}x
$$

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

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

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

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

$$
tan x=pm sqrt{dfrac{9}{16}}=pm dfrac{3}{4}
$$

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

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

Now we have the value of $color{#4257b2}tan x$, so we can use the double angle formula for the tangent function $color{#4257b2}tan 2x=dfrac{2tan x}{1-tan^{2}x}$.

$$
begin{align*}
tan 2x&=dfrac{2cdot dfrac{3}{4}}{1-left(dfrac{3}{4}right)^{2}}
\ \
&=dfrac{dfrac{6}{4}}{1-dfrac{9}{16}}
\ \
&=dfrac{left(dfrac{6}{4}right)cdot left(dfrac{4}{4}right)}{dfrac{7}{16}}
\ \
&=dfrac{dfrac{24}{cancel{16}}}{dfrac{7}{cancel{16}}}
\ \
&=dfrac{24}{7}
end{align*}
$$

So $boxed{ tan 2x=dfrac{24}{7} }$

$$
color{#c34632}tan 2x=dfrac{24}{7}
$$