how to prove that $\sin(90+v) = \cos(v)$

Question

I need to prove that $\sin(90+v) = \cos v$ and that $\cos(90+v) = -\sin v$

So I did the following steps to prove these statements

$\sin(90+v) = \sin(90-(-v)) = \cos(-v) = \cos(v)$

$\cos(90+v) = \cos(90-(-v)) = \sin(-v) = -\sin(v)$

Is this correct?

Thanks!!

You do not need to go through $v=-(-v)$. Just do it directly. — Claude Leibovici, Sep 21 '14 at 13:11
Yes, that is absolutely correct! If you already know the identities $\sin{(90-x)} = \cos{x}$ and $\cos{(90-x)} = \sin{x}$, this is a great way to prove it. — user795305, Sep 21 '14 at 13:14

amWhy · Accepted Answer · 2014-09-21T13:20:19.727

$$\sin (a \pm b) = \sin a \cos b \pm \sin b \cos a$$

$$\cos (a \pm b) = \cos a\cos b \mp \sin a \sin b$$

Your work ultimately appeals to the angle-difference identities, where one angle is $90^\circ$.

I'll base the following on the angle-sum identities.

From the first, we have $$\sin (90 + x) = \sin(90)\cos (x) + \sin x\cos(90) = \cos x + 0 = \cos x$$

For the second, we have $$\cos(90+x) = \cos(90)\cos(x) - \sin(90)\sin(x) = 0 - \sin(x) = -\sin x$$

score 0 · Answer 2 · answered Sep 21 '14 at 13:12

0

Use the following identities:

$$\sin(a+b)=\sin(a)\cos(b)+\cos(a)\sin(b) \\ \cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)$$

answered Sep 21 '14 at 13:12

Mary Star

2 Answers2