# Compound angles HELP!

• February 14th 2013, 03:33 PM
Kosky1
Compound angles HELP!
Hi I'm new to this, I've been given a question on compound angles and I really don't know how to go about it. Any help would be greatly appreciated:

Using compound angles prove that Cos(y-pi)+Sin(y+pi/2)=0

thanks Shaun
• February 14th 2013, 04:25 PM
Soroban
Re: Compound angles HELP!
Hello, Shaun!

You are expected to know these Compound Angle Identities:

. . $\sin(A \pm B) \:=\:\sin A\cos B \pm \cos A\sin B$

. . $\cos(A \pm B) \:=\:\cos A\cos B \mp \sin A\sin V$

Quote:

Using compound angles prove that: . $\cos(x-\pi)+\sin(x+\tfrac{\pi}{2})\:=\:0$

We have: . $\cos(x - \pi) \:=\:\cos(x)\cos(\pi) + \sin(x)\sin(\pi)$

. . . . . . . . . . . . . . . . $=\;\cos(x)\!\cdot\!(\text{-}1) + \sin(x)\!\cdot\!(0)$

. . . . . . . . . . . . . . . . $=\;\text{-}\cos(x)$

We have: . $\sin(x +\tfrac{\pi}{2}) \;=\;\sin(x)\cos(\tfrac{\pi}{2}) + \cos(x)\sin(\tfrac{\pi}{2})$

. . . . . . . . . . . . . . . . $=\;\sin(x)\!\cdot\!(0) + \cos(x)\!\cdot\!(1)$

. . . . . . . . . . . . . . . . $=\;\cos(x)$

Therefore: . $\cos(x-\pi) + \sin(x + \tfrac{\pi}{2}) \;=\;\text{-}\cos(x) + \cos(x) \;=\;0$

• February 14th 2013, 04:47 PM
Kosky1
Re: Compound angles HELP!
Ah thankyou. I had those identities right infront of me! Had a complete mind block
Thank you ever so much!