*(a)* Note that the cosine function is a periodic function with a period $2pi$. That means that the behavior of the function repeats in the same way for every $thetapm 2kpi, kin N$. Translating the given parent function horizontally for $pm2kpi$ in any direction will result in the function of the same graph.

Hence, the function $y=cos theta$ will have the same graph as the function $y=cos (theta-2pi)$, $y=cos(theta+2pi)$ or $y=cos(theta-6pi)$. The answer can be any other cosine function that is translated horizontally either to the left or to the right for the even number of $pi$.

*(b)* We can use the cofunction identity
$$
y=cos theta=sin left(dfrac{pi}{2}-theta right)tag{1}$$
to rewrite the cosine function as a sine function. This can be one solution. Furthermore, sine function is also a periodic function with a period $2pi$ and translating the function given in Eq. $(1)$ horizontally for $pm2kpi$ in any direction will result in the function of the same graph.

Hence, as in the previous step, we can add or subtract an even number of $pi$ to the angle $frac{pi}{2}-theta$ and obtain the function of the same graph. Some possible solutions are written below:
$$
begin{align*}
y=sin left(dfrac{pi}{2}-theta +2piright)&=sin left(dfrac{5pi}{2}-theta right)\
y=sin left(dfrac{pi}{2}-theta-2pi right)&=sin left(-dfrac{3pi}{2}-theta right).\
end{align*}$$

a) $y=cos (theta-2pi)$, $y=cos(theta+2pi)$ or $y=cos(theta-6pi)$
b) $y=sin left(frac{pi}{2}-theta right)$, $y=sin left(frac{5pi}{2}-theta right)$, $y=sin left(-frac{3pi}{2}-theta right)$

*(a)* The reciprocal cosine function is secant. Since the cosine function is even, the secant function is even as well. Thus, we can write:
$$
sec (-theta)=sec theta$$

The reciprocal sine function is cosecant. Since the sine function is odd, the cosecant function is odd as well. Thus, we can write:
$$
csc (-theta)=-csc theta$$

The reciprocal tangent function is cotangent. Since the tangent function is odd, the cotangent function is odd as well. Thus, we can write:
$$
cot (-theta)=-cot theta$$

*(b)* Note the graph of a secant function depicted below.

This function is symmetric about the $y$-axis. When we set $theta=-theta$ the graph $y=sec theta$ is reflected across the $y$-axis, and the function $y=sec(- theta)$ has the same graph as the function $y=sec theta$.
$$
sec (-theta)=sec theta$$

Note the graph of a cosecant function depicted below.

This function is symmetric about the origin. When we set $theta=-theta$ the graph $y=csc theta$ is reflected across the $y$-axis, but the graph function $y=csc (-theta)$ is still not the same as the graph of the function $y=csc theta$. If we now reflect the graph function $y=csc theta$ across the $x$-axis, we obtain the function graph function $y=-csc theta$, which is now the same as the graph of the function $y=csc (-theta)$. Hence,
$$
csc (-theta)=-csc theta$$

Note the graph of a cotangent function depicted below.

This function is symmetric about the origin. When we set $theta=-theta$ the graph $y=cot theta$ is reflected across the $y$-axis, but the graph function $y=cot (-theta)$ is still not the same as the graph of the function $y=cot theta$. If we now reflect the graph function $y=cot theta$ across the $x$-axis, we obtain the function graph function $y=-cot theta$, which is now the same as the graph of the function $y=cot (-theta)$. Hence,
$$
cot (-theta)=-cot theta$$

a) $sec (-theta)=sec theta$, even
$csc (-theta)=-csc theta$, odd
$cot (-theta)=-cot theta$, odd
b) See the explanation.

The cofunction identities describe trigonometric relationships between the complementary angles $theta$ and $frac{pi}{2}-theta$ in a right triangle are
$$
begin{align*}
sin theta&= cos left(dfrac{pi}{2}-theta right)quadquad &&(1)\
cos theta&= sinleft(dfrac{pi}{2}-theta right)quadquad &&(2)\
tan theta&= cot left(dfrac{pi}{2}-theta right)quadquad &&(3)\
end{align*}
$$

*(a)* Hence, using Eq. $(1)$ we can rewrite the given function as follows:
$$
sin dfrac{pi}{6}= cos left(dfrac{pi}{2}-dfrac{pi}{6}right)= cos left(dfrac{3pi}{6}-dfrac{pi}{6}right)=cos left(dfrac{2pi}{6}right)=cosdfrac{pi}{3}.$$

*(b)* Hence, using Eq. $(2)$ we can rewrite the given function as follows:
$$
cos dfrac{5pi}{12}= sin left(dfrac{pi}{2}-dfrac{5pi}{12}right)= sinleft(dfrac{6pi}{12}-dfrac{5pi}{12}right)=sindfrac{pi}{12}.$$

*(c)* Hence, using Eq. $(3)$ we can rewrite the given function as follows:
$$
tan dfrac{3pi}{8}= cot left(dfrac{pi}{2}-dfrac{3pi}{8}right)= cotleft(dfrac{4pi}{8}-dfrac{3pi}{8}right)=cotdfrac{pi}{8}.$$

*(d)* Hence, using Eq. $(2)$ we can rewrite the given function as follows:
$$
cos dfrac{5pi}{16}= sin left(dfrac{pi}{2}-dfrac{5pi}{16}right)= sinleft(dfrac{8pi}{16}-dfrac{5pi}{16}right)=sindfrac{3pi}{16}.$$

*(e)* Hence, using Eq. $(1)$ we can rewrite the given function as follows:
$$
sin dfrac{pi}{8}= cos left(dfrac{pi}{2}-dfrac{pi}{8}right)= cos left(dfrac{4pi}{8}-dfrac{pi}{8}right)=cosdfrac{3pi}{8}.$$

*(f)* Hence, using Eq. $(3)$ we can rewrite the given function as follows:
$$
tan dfrac{pi}{6}= cot left(dfrac{pi}{2}-dfrac{pi}{6}right)= cotleft(dfrac{3pi}{6}-dfrac{pi}{6}right)=cotdfrac{2pi}{6}=cotdfrac{pi}{3}.$$

a) $cos frac{pi}{3}$
b) $sin frac{pi}{12}$
c) $cot frac{pi}{8}$
d) $sinfrac{3pi}{16}$
e) $cosfrac{3pi}{8}$
f) $cotfrac{pi}{3}$

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 $frac{7pi}{8}$. 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:
$$
frac{7pi}{8}=pi-thetaimplies theta=pi-frac{7pi}{8}=frac{pi}{8}
$$
Using the equation for the sine function and plugging in $theta=frac{pi}{8}$, we obtain:
$$
sinfrac{7pi}{8}=sinleft(pi-frac{pi}{8} right)=sin frac{pi}{8}$$

*(b)* The given angle is $frac{13pi}{12}$. 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:
$$
frac{13pi}{12}=pi+thetaimplies theta=frac{13pi}{12}-pi=frac{pi}{12}
$$
Using the equation for the cosine function from the set $(2)$ and plugging in $theta=frac{pi}{12}$, we obtain:
$$
cosfrac{13pi}{12}=cosleft(pi+frac{pi}{12} right)=-cos frac{pi}{12}$$

*(c)* The given angle is $frac{5pi}{4}$. 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:
$$
frac{5pi}{4}=pi+thetaimplies theta=frac{5pi}{4}-pi=frac{pi}{4}
$$
Using the equation for the tangent function from the set $(2)$ and plugging in $theta=frac{pi}{4}$, we obtain:
$$
tanfrac{5pi}{4}=tanleft(pi+frac{pi}{4} right)=tan frac{pi}{4}$$

*(d)* The given angle is $frac{11pi}{6}$. Note that this angle falls in the interval $frac{3pi}{2}le thetale2pi$. Hence, we need to refer to quadrant IV and set of Eq. $(3)$. The given angle we need to rewrite in the form $2pi-theta$ and compute $theta$. It yields:
$$
frac{11pi}{6}=2pi-thetaimplies theta=2pi-frac{11pi}{6}=frac{pi}{6}
$$
Using the equation for the cosine function from the set $(3)$ and plugging in $theta=frac{pi}{6}$, we obtain:
$$
cosfrac{11pi}{6}=cosleft(2pi-frac{pi}{6} right)=cos frac{pi}{6}$$

*(e)* The given angle is $frac{13pi}{8}$. Note that this angle falls in the interval $frac{3pi}{2}le thetale2pi$. Hence, we need to refer to quadrant IV and set of Eq. $(3)$. The given angle we need to rewrite in the form $2pi-theta$ and compute $theta$. It yields:
$$
frac{13pi}{8}=2pi-thetaimplies theta=2pi-frac{13pi}{8}=frac{3pi}{8}
$$
Using the equation for the sine function from the set $(3)$ and plugging in $theta=frac{3pi}{8}$, we obtain:
$$
sinfrac{13pi}{8}=sinleft(2pi-frac{3pi}{8} right)=-sin frac{3pi}{8}$$

*(f)* The given angle is $frac{5pi}{3}$. Note that this angle falls in the interval $frac{3pi}{2}le thetale2pi$. Hence, we need to refer to quadrant IV and set of Eq. $(3)$. The given angle we need to rewrite in the form $2pi-theta$ and compute $theta$. It yields:
$$
frac{5pi}{3}=2pi-thetaimplies theta=2pi-frac{5pi}{3}=frac{pi}{3}
$$
Using the equation for the tangent function from the set $(3)$ and plugging in $theta=frac{pi}{3}$, we obtain:
$$
tanfrac{5pi}{3}=tanleft(2pi-frac{pi}{3} right)=-tan frac{pi}{3}$$

a) $sin frac{pi}{8}$
b) $-cos frac{pi}{12}$
c) $tan frac{pi}{4}$
d) $cos frac{pi}{6}$
e)$-sin frac{3pi}{8}$
f)$-tan frac{pi}{3}$

*(a)* Let the given circle be a unit circle whose radius is $r=1$. Note that point $P$ is point $Q$ reflected about the line $y=x$. Hence, if we denote the coordinates of the point $Q$ to be $(x,y)$ and reflect them about the line $y=x$, we obtain that the coordinates of point $P$ are $(y,x)$. Note the right-angled triangle that the point $Q$ makes with the $x$-axis and radius of the circle at this point as depicted in *Figure 1*.

Figure 1.

Note that the radius makes an angle $theta$ with the positive $x$-axis and that the sine function of this angle is equal to the ratio of the opposite leg of a triangle and hypotenuse $r=1$
$$sin theta=dfrac{y}{1}=y.tag{1}$$

Now we can set the line from the point $P$ that is perpendicular to the $x$-axis, and note the right angled triangle depicted in *Figure 2*.

Figure 2.

The angle $alpha$ that the hypotenuse of this right angled triangle makes with the positive $x$ axis is
$$
alpha =dfrac{pi}{2}-theta,$$
while the cosine of this angle is equal to the ratio of the adjacent side of a triangle and the hypotenuse and hypotenuse $r=1$, or
$$
cos alpha = dfrac{y}{1}=y.$$

Rewriting the angle $alpha$ as $frac{pi}{2}-theta$, we obtain
$$
cos left(dfrac{pi}{2}-theta right)=y.tag{2}$$

Note that the right hand sides of the Eq. $(1)$ and $(2)$ are equal. Hence, the left hand sides of these equations have to be equal as well, and we obtain
$$sintheta=cos left(dfrac{pi}{2}-theta right).$$

*(b)* Let the given circle be a unit circle whose radius is $r=1$.If we denote the coordinates of the point $Q$ to be $(x,y)$ and rotating this point about the origin counterclockwise for an angle $frac{pi}{2}$, we obtain that the coordinates of point $P$ are $(-y,x)$. Note the right-angled triangle that the point $Q$ makes with the $x$-axis and radius of the circle at this point as depicted in *Figure 3*.

Figure 3.

Note that the radius makes an angle $theta$ with the positive $x$-axis and that the sine function of this angle is equal to the ratio of the opposite leg of a triangle and hypotenuse $r=1$
$$sin theta=dfrac{y}{1}=y,
$$
while
$$-sintheta=-y.tag{1}$$

Now we can set the line from the point $P$ that is perpendicular to the $x$-axis, and note the angle $frac{pi}{2}+theta$ depicted in *Figure 4*.

Figure 4.

Note that the cosine function of this angle is equal to the coordinate of the point $P$ on the $x$-axis, that is
$$
cos left(dfrac{pi}{2}+theta right)=-y.tag{2}$$

Note that the right hand sides of the Eq. $(1)$ and $(2)$ are equal. Hence, the left hand sides of these equations have to be equal as well, and we obtain
$$-sintheta=cos left(dfrac{pi}{2}+theta right).$$