# How do you prove cosx=pi/2 - sinx ?

Mar 1, 2016

The closest true identities to the given equality would be

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

(These are equivalent, as you could, for example, substitute $x = y + \frac{\pi}{2}$ into the first to obtain the second)

To prove the above, we can use the identity

$\cos \left(\alpha + \beta\right) = \cos \left(\alpha\right) \cos \left(\beta\right) - \sin \left(\alpha\right) \sin \left(\beta\right)$

along with sine being an odd function and cosine being an even function. That is
$\sin \left(- x\right) = - \sin \left(x\right)$
and
$\cos \left(- x\right) = \cos \left(x\right)$

With these, we have

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

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

$= \cos \left(x\right) \cdot 0 + \sin \left(x\right) \cdot 1$

$= \sin \left(x\right)$