SOLUTION: 2sin(3x)cos(3x)(2cos^2 3x-1)

Question 1144138: 2sin(3x)cos(3x)(2cos^2 3x-1)

Answer by ikleyn(52933) (Show Source): You can put this solution on YOUR website!
.


The product of the first three factors (terms) is


    2*sin(3x)*cos(3x) = sin(6x).


The fourth term is


    2*cos^2(3x)-1 = cos(6x).


Therefore


    2*sin(3x)*cos(3x)*(2cos^2 3x-1) = sin(6x)*cos(6x) = (1/2)*sin(12x).    ANSWER

Solved.

RELATED QUESTIONS
Solve the equation for x, on the interval 0 is less than or equal to x and 2pi is greater (answered by Alan3354)

2cos (3x+24°) = 1 (answered by ikleyn)

Please help me solve these 3 equations for solutions from [0, 2pi]: 1.)... (answered by ikleyn)

prove the following identity 1. cos^2 3x - cos6x = sin^2 3x find the general... (answered by ikleyn)

2cos(3x)-square root 2 =... (answered by lwsshak3)

The identity of: sin^4x+cos^4x= (sin^2x+cos^2x)(sin^2x+cos^2x) sin^2x+cos^2=1 This... (answered by scott8148)

Solve each equation on the interval (0,2pi): a.) 2sin^(2)x-cos x-1=0, b.) sin 3x cos 2x + (answered by lwsshak3)

amplitude of... (answered by stanbon)

solve :... (answered by Fombitz)