show that 8sin20sin40sin80 = root3

Hello differentiate

Quote:

Originally Posted by differentiate

show that 8sin20sin40sin80 = root3

how would I do this?

Thanks.

Start with $\sin3A = 3\sin A -4\sin^3 A$

Then $\sin 120$ $= \frac{\sqrt3}{2} = 3\sin40-4\sin^340$

$\Rightarrow\sqrt3= 2\sin40(3-4\sin^240)$

$=2\sin40(3-2[1-\cos80])$ , using $\sin^2A = \tfrac12(1-\cos2A)$

$=2\sin40(1+2\cos80)$

$=4\sin40(\tfrac12+\cos80)$

$=4\sin40(\cos60+\cos80)$

$=8\sin40\cos70\cos10$ , using $\cos A + \cos B=2\cos\tfrac12(A+B)\cos\tfrac12(A-B)$

$=8\sin40\sin20\sin80$ , using $\cos A = \sin(90-A)$

Grandad