Check out some similar questions!

b3cky2 · Oct 20, 2010, 12:02 PM

sinx+sin3x+sin5x+sin7x=4cosxcos2xsin4x

Pleeeaaaassseee help

Unknown008 · Oct 20, 2010, 12:26 PM

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

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

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

$=sin(x) + <br /> 2sin(x)cos^2(x) + (cos^2(x) - sin^2(x))sin(x) + <br /> (2sin(x)cos^2(x) + (cos^2(x) - sin^2(x))sin(x))(cos^2(x)-sin^2(x)) + 2sin(x)cos(x)cos(3x) + <br /> (cos^2(x) - sin^2(x))((2sin(x)cos^2(x) + (cos^2(x) - sin^2(x))sin(x))(cos^2(x)-sin^2(x)) + 2sin(x)cos(x)cos(3x)) + (cos^2(x) - sin^2(x))cos(5x)$

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

harum · Oct 21, 2010, 10:47 AM

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.