sinx+sin3x+sin5x+sin7x=4cosxcos2xsin4x

Pleeeaaaassseee help

Expand each of the terms on the left... =/

sin(x)+sin(3x)+sin(5x)+sin(7x) = sin(x) + sin(2x)cos(x) + cos(2x)sin(x) + sin(3x)cos(2x) + cos(3x)sin(2x) + sin(5x)cos(2x) + cos(5x)sin(2x)

=sin(x) + 2sin(x)cos^2(x) + (cos^2(x) - sin^2(x))sin(x) + sin(3x)(cos^2(x)-sin^2(x)) + 2sin(x)cos(x)cos(3x) + (cos^2(x) - sin^2(x))sin(5x) + (cos^2(x) - sin^2(x))cos(5x)

=sin(x) +
2sin(x)cos^2(x) + (cos^2(x) - sin^2(x))sin(x) +
(2sin(x)cos^2(x) + (cos^2(x) - sin^2(x))sin(x))(cos^2(x)-sin^2(x)) + 2sin(x)cos(x)cos(3x) +
(cos^2(x) - sin^2(x))((2sin(x)cos^2(x) + (cos^2(x) - sin^2(x))sin(x))(cos^2(x)-sin^2(x)) + 2sin(x)cos(x)cos(3x)) + (cos^2(x) - sin^2(x))cos(5x)

Well, I don't know many shortcuts, but I'm sure with some effort, you'll get the right side.

Easy one. You have to know or derive yourself (it is straightforward and you only need the definitions of trigonometric functions) the following equality:

sin(a) + sin(b) = 2*sin((a+b)/2)*cos((a-b)/2), here "a" and "b" are any angles and remember that cos(x) is an even function, i.e. cos(x) = cos(-x).

Apply this equality to each sum within the parentheses after regrouping the terms this way:

sin(x) + sin(3x) + sin(5x) + sin(7x) = ( sin(x) + sin(5x) ) + ( sin(3x) + sin(7x) );

I.e. apply the above equality to "sin(x) + sin(5x)", then to "sin(3x) + sin(7x)".
You will have a sum of two products. Then find the common factor in both resulting terms, get it outside the parentheses and apply the above equality once again to a two-term sum within the parentheses.

Here goes your answer.