# How do you prove sin (π/2 – x) = sin (π/2 + x)?

Mar 17, 2016

What proofs are available depends on what facts we already have as given. One simple proof relies on the following:

• $\sin \left(x\right) = - \sin \left(x - \pi\right)$
-$\sin \left(- x\right) = - \sin \left(x\right)$

With those, we have

$\sin \left(\frac{\pi}{2} - x\right) = - \sin \left(\left(\frac{\pi}{2} - x\right) - \pi\right)$

$= - \sin \left(- x - \frac{\pi}{2}\right)$

$= - \sin \left(- \left(\frac{\pi}{2} + x\right)\right)$

$= - \left(- \sin \left(\frac{\pi}{2} + x\right)\right)$

$= \sin \left(\frac{\pi}{2} + x\right)$

Mar 18, 2016

Apply the sum identities.
$\sin \left(\frac{\pi}{2} - x\right) = \sin \left(\frac{\pi}{2}\right) . \cos x - \cos \left(\frac{\pi}{2}\right) . \sin x = \cos x$
because $\sin \left(\frac{\pi}{2}\right) = 1$ and $\cos \left(\frac{\pi}{2}\right) = 0$
$\sin \left(\frac{\pi}{2} + x\right) = \sin \left(\frac{\pi}{2}\right) . \cos x + \cos \left(\frac{\pi}{2}\right) \sin x = \cos x$
Therefor,
$\sin \left(\frac{\pi}{2} - x\right) = \cos x = \sin \left(\frac{\pi}{2} + x\right)$