Question

Trig Identities
Sin [(pi/20 - x] / Cos [(pi/2) - x]

Answer 1

You should use,
\[\sin(a-b)=\sin a\cos b-\cos a \sin b \\
\cos(a-b)=\cos a\cos b+\sin a \sin b\] And,
\[\sin (\pi/2)=1\\
\cos(\pi/2)=0\]

Answer 2

use
Cos [(pi/2) - x] = sin x

Answer 3

how would you use the \[\sin (a - b) and \cos (a - b)\]

Answer 4

Let a=pi/2 and b=x, then,
\[\sin(\pi/2-x)=\sin(\pi/2)\cos x-\cos(\pi/2)\sin x\] Now, use
\[\sin(\pi/2)=1\\ \cos(\pi/2)=0\] Then,
\[\sin(\pi/2-x)=\cos x\] The same follows for the other expression.

Answer 5

Another possible approach:

Note that \[Sin [(\pi/20 - x] / Cos [(\pi/2) - x]\] could be re-written as \[\tan(\frac{ \pi }{ 2 }-x)\]... provided that your pi/20 is a typo and was meant to be pi/2.

Answer 6

Can this be reduced? Why or why not?