0
$\begingroup$

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.

  • 1
    What do you get?2012-08-03

2 Answers 2

3

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?

  • 1
    @MTurgeon I had some mis-placed parenthesis! I just re-ran the computation to verify it is indeed correct.2012-08-03
0

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 $-, +, +, -$