# Why 2sinxsin3x-cos2x=cos4x?I tried with this 2sinx*sin3x=2*1/2(cos2x-cos4x)=cos2x-cos4x...

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I think you wrote the problem down wrong.

cos(2x) - 2 sin(x)sin(3x) = cos(4x)

First use the cos addition and subtraction formulas

cos(a+b) = cos(a)cos(b) - sin(a)sin(b)

cos(a-b) = cos(a)cos(b) + sin(a)sin(b)

Subtract cos(a-b) from cos(a+b) and the cos(a)cos(b) term disappears.

cos(a+b) - cos(a-b) = -2sin(a)sin(b)

For our problem set a = 3x, b = x

cos(3x + x) - cos(3x - x) = -2sin(3x)sin(x)

cos(4x) - cos(2x) = -2 sin(x) sin(3x)

cos(4x) = cos(2x) - 2 sin(x) sin(3x)

If you use pi/4 you get

cos(4pi/4) = cos(2pi/4) - 2 sin(pi/4) sin(3pi/4)

cos(pi) = -1, cos(pi/2) = 0, sin(pi/4) = sqrt(2)/2 and sin(3pi/4) = sqrt(2)/2

so -1 = 0 - 2(sqrt(2)/2)(sqrt(2)/2)

-1 = - 2(2/4)

-1 = -1 checks....

Does not work with your formula....

Use the formula for the triple of the argument:

`sin 3x = 3sin x - 4 sin^3 x`

Use the formula for the double of the argument:

`cos 2x = 1 - 2 sin^2 x`

Write the left side of the equation:

`2sin x*(3sin x - 4 sin^3 x) - 1 + 2 sin^2 x = cos 4x`

`6sin^2 x - 8sin^4 x - 1 + 2 sin^2 x = cos 4x`

`8sin^2 x - 8sin^4 x - 1 = cos 2*(2x)`

`8sin^2 x - 8sin^4 x - 1 = 1 - 2sin^2 (2x)`

`8sin^2 x - 8sin^4 x -2 = - 2sin (2x)*sin(2x)`

`8sin^2 x - 8sin^4 x -2 = -8 sin^2 x*cos^2 x`

Using the basic formula of trigonometry yields:

`8sin^2 x - 8sin^4 x -2 = -8 sin^2 x*(1 - sin^2 x)`

`8sin^2 x - 8sin^4 x -2 = -8 sin^2 x+ 8sin^4 x`

**This last line proves that `2sin x*sin 3x-cos 2x != cos 4x` .**

cos(4x) = cos (3x+x) =-sin(x)sin(3x) + cos (x)cos(3x)

=-sin(x) sin(3x) +cos(x) (cos(x) cos(2x) - sin(x) sin(2x)

= -sin (x) sin(3x) + cos(x)^2 cos(2x) - cos(x) sin(x) sin(2x)

= -sin (x) sin(3x) + (1 - sin^2(x) ) cos(2x) - cos(x) sin(x) sin(2x)

= -sin (x) sin(3x) + cos(2x) - sin(x) (sin(x) cos(2x) + cos(x) sin(2x) )

= -sin(x) sin(3x) + cos(2x) - sin(x) sin(2x+x)

= -2sin(x)sin(3x) + cos(2x)

The question has the sign reversed.