Question

sin of theta=9/20 Find cos(pi/2-theta)

Answer 1

I will do an example problem:

You are given that: \(\color{#000000 }{ \displaystyle \sin \theta = \frac{4}{5} }\) \(\tiny \\[1.2em]\)
You need to find: \(\color{#000000 }{ \displaystyle \cos \theta = \left(\theta+ \frac{4}{5}\right) }\)
\(\tiny \\[1.2em]\)
The rule is: \(\color{#000000 }{ \displaystyle \cos \left(a+b\right)=\cos(a)\cos(b)-\sin(a)\sin(b) }\)

You can tell that: \(\color{#000000 }{ \displaystyle a=\theta }\)
\(\color{#000000 }{ \displaystyle b=4/5 }\)

So, when you substitute in the rule accordingly you get,
\(\color{#000000 }{ \displaystyle \cos \left(\theta+ \frac{4}{5}\right)=\cos\left(\theta\right)\cos \left(\frac{4}{5}\right)-\sin\left(\theta\right)\sin \left(\frac{4}{5}\right) }\)

\(\color{#000000 }{ \displaystyle \cos \left(4/5\right) }\), \(\color{#000000 }{ \displaystyle \sin \left(4/5\right) }\) are just some (irrational) constants.
And you are given from the beginning that, \(\color{#000000 }{ \displaystyle \sin \left(\theta \right)=4/5 }\)

So, knowing this you get,
\(\color{#000000 }{ \displaystyle \cos \left(\theta+ \frac{4}{5}\right)=\cos\left(\theta\right)\cos \left(\frac{4}{5}\right)-\frac{4}{5}\sin \left(\frac{4}{5}\right) }\)

And now, to finish this up, we need to find \(\color{#000000 }{ \displaystyle \cos \left(\theta\right) }\).
WATCH THE TRICK!

You know that: \(\color{#000000 }{ \displaystyle \cos ^2\theta+\sin^2\theta =1 }\) (pythegorean)

And from that you can obtain the following:
\(\color{#000000 }{ \displaystyle \cos ^2\theta+\sin^2\theta =1 }\)
\(\color{#000000 }{ \displaystyle \cos ^2\theta+\sin^2\theta\color{red}{-\sin^2\theta} =1 \color{red}{-\sin^2\theta}}\)
\(\color{#000000 }{ \displaystyle \cos ^2\theta =1-\sin^2\theta }\)
\(\color{#000000 }{ \displaystyle \color{red}{\sqrt{\color{black}{\cos^2\theta} }}=\color{red}{\sqrt{\color{black}{1-\sin^2\theta} }} }\)
\(\color{#000000 }{ \displaystyle \cos \theta =\sqrt{1-\sin^2\theta } }\)

And therefore, you may write,
\(\color{#000000 }{ \displaystyle \cos \left(\theta+ \frac{4}{5}\right)=\left(\sqrt{1-\sin^2\theta}\right)\cos \left(\frac{4}{5}\right)-\frac{4}{5}\sin \left(\frac{4}{5}\right) }\)

\(\color{#000000 }{ \displaystyle \cos \left(\theta+ \frac{4}{5}\right)=\left(\sqrt{1-\left[\sin \theta \right]^2}\right)\cos \left(\frac{4}{5}\right)-\frac{4}{5}\sin \left(\frac{4}{5}\right) }\)

and then remember that you are given \(\sin\theta=4/5\).

\(\color{#000000 }{ \displaystyle \cos \left(\theta+ \frac{4}{5}\right)=\left(\sqrt{1-\left[\frac{4}{5}\right]^2}\right)\cos \left(\frac{4}{5}\right)-\frac{4}{5}\sin \left(\frac{4}{5}\right) }\)

\(\color{#000000 }{ \displaystyle \cos \left(\theta+ \frac{4}{5}\right)=\left(\sqrt{1-\frac{16}{25}}\right)\cos \left(\frac{4}{5}\right)-\frac{4}{5}\sin \left(\frac{4}{5}\right) }\)

and you found a constant value for \(\color{#000000 }{ \displaystyle \cos \left(\theta+ \frac{4}{5}\right) }\), now will apply algebra to simplify this:

\(\color{#000000 }{ \displaystyle \cos \left(\theta+ \frac{4}{5}\right)=\left(\sqrt{\frac{25}{25}-\frac{16}{25}}\right)\cos \left(\frac{4}{5}\right)-\frac{4}{5}\sin \left(\frac{4}{5}\right) }\)

\(\color{#000000 }{ \displaystyle \cos \left(\theta+ \frac{4}{5}\right)=\left(\sqrt{\frac{9}{25}}\right)\cos \left(\frac{4}{5}\right)-\frac{4}{5}\sin \left(\frac{4}{5}\right) }\)

\(\color{#000000 }{ \displaystyle \cos \left(\theta+ \frac{4}{5}\right)=\left(\frac{\sqrt{9}}{\sqrt{25}}\right)\cos \left(\frac{4}{5}\right)-\frac{4}{5}\sin \left(\frac{4}{5}\right) }\)

\(\color{#000000 }{ \displaystyle \cos \left(\theta+ \frac{4}{5}\right)=\frac{3}{5}\cos \left(\frac{4}{5}\right)-\frac{4}{5}\sin \left(\frac{4}{5}\right) }\) (this is it)

Answer 2

Note, in case you didn't know:

\(\color{#000000 }{ \displaystyle \sin^2\theta = (\sin \theta)^2 }\)
(and so for any power) \(\color{#000000 }{ \displaystyle \sin^k\theta = (\sin \theta)^k }\)

Also, in your case, the rule you are applying is a little different from the one I used in my example. Your rule is,

\(\color{#000000 }{ \displaystyle \cos(a-b)=\cos(a)\cos(b)+\sin(a)\sin(b) }\)