Compound Angle Identity

**DemonX01** · March 6th 2010, 03:16 PM

How would I prove these identities?

1) sin(A+B)*sin(A-B)=(sin A)^2*(sin B)^2

I've gotten to expanding it to
(sin(A)cos(B)+sin(B)cos(A))*(sin(A)cos(B)-sin(B)cos(A))
then

(sin(A)cos(B))^2-(sin(B)cos(A))^2, but I'm not sure what to do next

2) (cos W-sin 2W)/(cos 2W+sin W-1)=cot W

**Soroban** · March 6th 2010, 04:47 PM

GHello, DemonX01!

Please check the first one for typos . . .

$2)\;\frac{\cos x-\sin2x}{\cos2x+\sin x-1} \;=\;\cot x$

Note: If $x = \tfrac{\pi}{6}$ , the denominator is: . $\cos\tfrac{\pi}{3} + \sin\tfrac{\pi}{6} - 1 \:=\:\tfrac{1}{2} + \tfrac{1}{2}-1 \;=\;0$

Hence: . $x \,\neq\,\tfrac{\pi}{6}$

. . That is: . $\sin x \,\neq\, \tfrac{1}{2} \quad\Rightarrow\quad 2\sin x \,\neq\, 1 \quad\Rightarrow\quad 1 - 2\sin x\:\neq\:0$ . . We can cancel this

We have: . $\frac{\cos x - \sin2x}{\cos2x + \sin x - 1} \;=\;\frac{\cos x - 2\sin x\cos x}{(1-2\sin^2\!x) + \sin x - 1} \;=\;\frac{\cos x - 2\sin x\cos x}{\sin x - 2\sin^2\!x}$

Factor: . $\frac{\cos x(1 - 2\sin x)}{\sin x(1 - 2\sin x)}$

Reduce: . $\frac{\cos x}{\sin x} \;=\;\cot x$

**DemonX01** · March 6th 2010, 05:29 PM

Thanks for the help on the second one. But as far as I see, there aren't any typos in the first one that I notice. Is there something wrong?

**Soroban** · March 6th 2010, 09:06 PM

Hello again, DemonX01!

$1)\;\;\sin(A+B)\cdot\sin(A-B)\:=\:{\color{blue}\sin^2\!A \cdot\sin^2\!B}$ .??

Your work is correct!

$\bigg[\sin A\cos B+ \cos A\sin B\bigg]\cdot\bigg[\sin A\cos B -\cos A\sin B\bigg]$

. . . $=\qquad\sin^2\!A\cos^2\!B \quad - \quad \cos^2\!A\sin^2\!B$
. . . $=\;\sin^2\!A\overbrace{\left(1-\sin^2\!B\right)} - \overbrace{\left(1 - \sin^2\!A\right)}\sin^2\!B$

. . . $=\; \sin^2\!A - \sin^2\!A\sin^2\!B - \sin^2\!B + \sin^2\!A\sin^2\!B$

. . . $=\;\sin^2\!A - \sin^2\!B$

**DemonX01** · March 7th 2010, 03:51 AM

Thanks for the help! I just keep on forgetting about the basic identities

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