Question

help to prove sinAsin(60-A)sin(60+A) = 1/4sin3A

Answer 1

use some identity trigonometry formula here, like :
sin(A+B) = ...
(cosA)^2 = ...
sinAsinB = ....
etc.

Answer 2

for sin(3A) = sin (A+2A) = ...
can u expand that ?

Answer 3

sinAcos2A+cosAsin2A

Answer 4

ok, right.. from that convert sin2A = 2sinAcosA, what happend next ?

Answer 5

sinAcos2A+cosA(2sinAcosA)
sinA [cos2A+2cos^2A]
sinA[cos2A+2(1-sin^2A)]
sinA[cos2A+2-2sin^2A]
sinA[ 1-2sin^2A +2-2sin^2A]
sinA [3-4 sin^2A]

Answer 6

OK, correct..
so, to right side we have (1/4)sin3A = (1/4)sinA [3-4 sin^2A], right ?

Answer 7

I get the way!!!
sin A sin (60-A)sin(60+A)
= sinA[-1/2(cos120-cos2A)]
= sinA{-1/2[-1/2-(1-2sin^2A)]}
= sin A[3/4-sin^2 A]
= 1/4 sin A [ 3-4sin^2 A]
= 1/4sinA[ 1-2sin^2A +2-2sin^2A]
= 1/4sinA[cos2A+2-2sin^2A]
= 1/4sinA[cos2A+2(1-sin^2A)]
= 1/4sinA [cos2A+2cos^2A]
= 1/4sinAcos2A+cosA(2sinAcosA)
= 1/4sinAcos2A+cosAsin2A
= 1/4sin (A+2A)
= 1/4sin3A (proven)

THANKS SO MUCHHH!!!

Answer 8

woww, congrate..
my pleasure ^^