Trig identity help?

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1 Answer
Jun 23, 2018

sin theta ( sin 4 theta + sin 6 theta) = cos theta(cos 4 theta - cos 6 theta)

Explanation:

We're getting into the sum of sines and product of sines identities, which are relatively obscure. Let's start by listing them; maybe I'll derive them at the end.

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

sin a sin b = 1/2 ( cos(a-b) - cos(a+b))

Let's apply them.

sin theta ( sin 4 theta + sin 6 theta)

= sin theta ( 2 sin (( 4 theta + 6 theta)/2) cos((4 theta - 6 theta )/2))

= 2 sin theta sin(5 theta) cos theta

= 2 cos theta ( sin theta sin 5 theta)

= 2 cos theta(1/2 ( cos(theta - 5 theta) - cos (theta + 5 theta)))

= cos theta(cos 4 theta - cos 6 theta)

I gotta go; derivations later maybe.

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OK, let's derive the two identities we used, first:

sin a sin b = 1/2 ( cos(a-b) - cos(a+b))

This comes from the cosine sum and difference angle formulas:

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

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

Subtracting,

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

sin a sin b = 1/2( cos(a-b) - cos(a+b) ) quad sqrt

Next,

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

This comes from the sine sum and difference angle formulas:

sin(x+y) = sin x cos y + cos x sin y

sin(x-y) = sin x cos y - cos x sin y

Adding,

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

Let a=x+y, b=x-y so a+b=2x or x=(a+b)/2 and y=(a-b)/2

sin a + sin b = 2 sin((a+b)/2) cos((a-b)/2) quad sqrt