# Sum and difference of sines and cosines

How would I solve the following question?

Show that

$$2\sin(127.5)\sin(97.5)=(\sqrt{3}+\sqrt{2})/2$$

My work is I know

$$\sin A\sin B=(-1/2)(\cos(A+B)-\cos(A-B))$$

So I did

$$(-1/2)(\cos(127.5+97.5)-\cos(127.5-97.5))$$

but I do not get the correct answer.

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What do you get? –  M Turgeon Aug 3 '12 at 19:20

I assume that 127.5 and 97.5 are given in degrees, not radians, and will write $127.5^\circ$ and $97.5^\circ$ instead.

The step $-\frac{1}{2} (\cos (127.5^\circ + 97.5^\circ) - \cos(127.5^\circ - 97.5^\circ))$ is correct. Now, note that $127.5 + 97.5 = 225 = 180 + 45$ and $127.5 - 97.5 = 30$. Since $\cos 180^\circ + \alpha = -\cos \alpha$, this reduces to $-\frac{1}{2} (-\cos 45^\circ -\cos 30^\circ)$. Can you do the rest yourself?

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@EdGorcenski Well, my computations give the right result. Maybe you forgot the factor of 2 in front of the signs...? –  M Turgeon Aug 3 '12 at 19:18
@MTurgeon I had some mis-placed parenthesis! I just re-ran the computation to verify it is indeed correct. –  Arkamis Aug 3 '12 at 19:20
The equation you have for $\sin A \cdot \sin B$ is incorrect (sign flip). It should be: $$\sin A \cdot \sin B = \frac{1}{2}\cos(A-B) - \frac{1}{2}\cos(A+B)$$ or $$\sin A \cdot \sin B = -\frac{1}{2}\cos(A+B) + \frac{1}{2}\cos(A-B)$$ $+, -, -, +$ OR $-, +, +, -$