Does anyone have any idea how the solution changed $\displaystyle sin5\theta+sin\theta$ into $\displaystyle 2sin(3\theta)cos(2\theta)$ been trying for ages and just can't see it? thanks

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- May 19th 2009, 06:47 PMi_zz_y_illtrig manipulating
Does anyone have any idea how the solution changed $\displaystyle sin5\theta+sin\theta$ into $\displaystyle 2sin(3\theta)cos(2\theta)$ been trying for ages and just can't see it? thanks

- May 19th 2009, 06:59 PMAmer
ok this come from

sin(a+b) = sina cosb + sinb cosa

sin(a-b) = sina cosb - sinb cosa

find the sum of two

sin(a+b) + sin(a-b) = 2 sina cosb

let a+b = x and a-b =y ..........

solve this ( find the value of a , b with respect of x .y )

a = (x+y)/2

b = (x-y)/2

now we have

sinx + siny =2 sin((x+y)/2) cos ((x-y)/2)

finish