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 #