how to evaluate trig functions with complementary identities

when we flip an angle over the line $y=x$, we have equations called complementary relationships

from the figure above, we can see that
$\cos (\theta )=\sin (\frac{\pi }{2}-\theta )$
$\sin (\theta )=\cos (\frac{\pi }{2}-\theta )$

apply this for a general rule, we see that for angle $\theta$, and $c$ is an odd integer:
$\sin (\theta )=\cos (c\cdot \frac{\pi }{2}\pm \theta )$
$\cos (\theta )=\sin (c\cdot \frac{\pi }{2}\pm \theta )$

practice complementary relationships

worked example: if $\cos (a)=0.7$ and $\sin (b)=0.4$ and they are both in the first quadrant, evaluate $\sin (\frac{\pi }{2}-a)$
$|\sin (\frac{\pi }{2}-a)|=\cos (a)=0.7$
applying signs
since angle $a$ is in the first quadrant, $\frac{\pi }{2}-a$ is also in the first quadrant, where sine and cosine are both positive
$\sin (\frac{\pi }{2}-a)=0.7$

worked example: if $\cos (a)=0.7$ and $\sin (b)=0.4$ and they are both in the second quadrant, evaluate $\cos (\frac{\pi }{2}+b)$
$|\cos (\frac{\pi }{2}+b)|=\sin (b)=0.4$
applying signs
since angle $b$ is in the second quadrant, $\frac{\pi }{2}+b$ is in the third quadrant, where sine and cosine are both negative
$\cos (\frac{\pi }{2}+b)=-0.4$

\worked example: if $\cos (a)=0.7$ and $\sin (b)=0.4$ and they are both in the third quadrant, evaluate $\sin (\frac{3\pi }{2}+a)$
$|\sin (\frac{3\pi }{2}+a)|=\cos (a)=0.7$
applying signs
since angle $a$ is in the third quadrant, $\frac{3\pi }{2}+a$ is in the second quadrant, where sine is positive and cosine is negative
$\sin (\frac{3\pi }{2}-a)=0.7$

worked example: if $\cos (a)=\frac{\sqrt{3}}{2}$ and $\sin (a)=\frac{1}{2}$ and angle $a$ is in the first quadrant, evaluate $\tan (\frac{3\pi }{2}-a)$
$\tan (\frac{3\pi }{2}-a)=\frac{\sin (\frac{3\pi }{2}-a)}{\cos (\frac{3\pi }{2}-a)}$
$|\sin (\frac{3\pi }{2}-a)|=\cos (a)=\frac{\sqrt{3}}{2}$
$|\cos (\frac{3\pi }{2}-a)|=\sin (a)=\frac{1}{2}$
applying signs
since angle $a$ is in the first quadrant, $\frac{3\pi }{2}-a$ is in the third quadrant, where sine and cosine are both negative, and tangent is positive
$\tan (\frac{3\pi }{2}-a)=\frac{1}{\sqrt{3}}$