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

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....