Question

4 sin(x) sin(pi/3+x) sin(pi/3-x) = sin(3 x)

Answer 1

2sinx(2sinAsinB) => 2sinx(cos(A-B) - cos(A+B))
A-B = pi/3 + x - pi/3 + x = 2x
A+B = pi/3 +x + pi/3 - x = 2pi/3 => cos(pi - pi/3) => -cos pi/3 => -1/2
2sinx(cos2x + 1/2) => 2sinxcos2x + sinx => 2sinx(2cos^2x - 1) + sinx
=> 4sinxcos^2x - sinx => 4sinx(1-sin^2x) - sinx => 3sinx - 4sin^3x =>sin3x

Hope You Understood

Answer 2

thank you i have been stuck on that one forever.

Answer 3

Haha... You're Welcome :)

Answer 4

sorry to bother you again but what identity formulas did you use.

Answer 5

2sinAsinB = cos(A-B) - cos(A+B)
cos2x = 2cos^2x - 1 = 1- 2sin^2x
cos^2x = 1 - sin^2x
sin3x = 3sinx - 4sin^3x

These are the things I used.

Answer 6

thanks