Another simple trig identity

**youngb11** · November 21st 2010, 07:19 AM

This looks simple of paper but for the life of me I can't seem to figure it out

LS=cos(x-y)cosy-sin(x-y)siny

RS= cosx

Greatly appreciated!

**Soroban** · November 21st 2010, 07:36 AM

Hello, youngb11!

Do you know your Compound Angle Identities?

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

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

$\cos(x-y)\cos y- \sin(x-y)\sin y\;=\;\cos x$

The left side is:

. . $(\cos x\cos y + \sin x\sin y)\cos y - (\sin x\cos y - \cos x\sin y)\sin y$

. . . . $=\; \cos x\cos^2\!y + \underbrace{\sin x\sin y\cos y - \sin x\sin y\cos y}_{\text{These cancel}} + \cos x\sin^2\!y$

. . . . $=\;\cos x\cos^2\!y + \cos x\sin^2\!y$

. . . . $=\; \cos x\underbrace{(\cos^2\!y + \sin^2\!y)}_{\text{This is 1}}$
. . . . $=\; \cos x$

**youngb11** · November 21st 2010, 07:43 AM

Thanks a lot! I can't believe I missed that.

Thanks again and have a great day

!

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