2006-07-27 04:44:34 · 3 answers · asked by jujz 2 in Science & Mathematics ➔ Mathematics
start with left side of the equation
sinAsin2A = 1/2(cos A - cos 3A)
sin3Asin6A = 1/2(cos3A - cos9A)
sin4Asin13A = 1/2(cos9A - cos17A)
sinAcos2A = 1/2 (-sinA+sin3A)
sin3Acos6A = 1/2(-sin3A+sin9A)
sin4Acos13A = 1/2(-sin9A+sin17A)
(sinAsin2A + sin3Asin6A + sin4Asin13A) / (sinAcos2A + sin3Acos6A + sin4Acos13A) * 2 / 2
= (cos A - cos 3A + cos3A - cos9A + cos9A - cos17A) / (-sinA+sin3A -sin3A+sin9A -sin9A+sin17A)
= (cos A - cos 17A) / (sin 17A-sin A)
= -2 sin 9A sin 8A / 2 cos 9A sin 8A (cancel 2 and sin 8A)
= -sin 9A / cos 9A
= - tan 9A
2006-07-27 05:08:31 · answer #1 · answered by Imoet 2 · 1⤊ 0⤋
I'd start by converting all the sin and cos functions to tan functions.
2006-07-27 04:55:27 · answer #2 · answered by evokid 3 · 0⤊ 0⤋
I don't like trignometry
2006-07-27 05:02:38 · answer #3 · answered by bala 1 · 0⤊ 0⤋