I understand that $\cos(\theta) = \sin(\pi/2 – \theta)$ holds true. But,
Does $\cos(\theta) = \sin(\pi/2 +\theta)$ always hold true?
I am asking this question because I encountered the following question in my workbook.
If $h(x) = \cos x$, $g(x) = \sin x$, and $h(x) = g(f(x))$, which of the following can be $f(x)$?
(a) $-x$
(b) $\pi/2 + x$
(c) $\pi – x$
(d) $3\pi/2 – x$
(e) $3\pi/2 + x$
My book says the correct answer is (b), and I am a bit baffled by this.
I can see that this holds true by plugging in certain values for $x$. But is there a mathematical proof for $\cos(\theta) = \sin(\pi/2 + \theta)$?
Best Answer
$\sin (\pi/2+x)=\sin (\pi/2)\cos (x)+\cos (\pi/2)\sin (x)=1\times\cos (x)+0\times\sin (x)=\cos (x) $
Or, you can apply what you already know, but to $-x $: $\sin (\pi/2+x)=\sin (\pi/2-(-x))=\cos (-x)=\cos (x) $. ($\cos $ is an "even function".)