Question

(i) Using the expansions of cos(3x − x) and cos(3x + x), prove that
1/2(cos 2x − cos 4x) ≡ sin 3x sin x.

Answer 1

do you know the expansions of cos(3x − x) and cos(3x + x) ?

Answer 2

sum/difference formula for cosine...

Answer 3

cos(a-b) = (cosa)(cosb) + (sina)(sinb)

Answer 4

cos(a+b) = (cosa)(cosb) - (sina)(sinb)

Answer 5

try using these formulas here...

Answer 6

cos(3x − x) = cos3xcosx + sin3xsinx
- cos(3x + x) =cos3xcosx - sin3xsinx
-----------------------------------
cos 2x-cos 4x =sin3xsinx-(-sin3xsinx)
cos 2x - cos4x = 2sin3xsinx
1/2 (cos 2x - cos4x) = sin3xsinx

Answer 7

It's just the concept only, write a better proof :)

Answer 8

yup.. thats what you had to do
or simply replace cos 2x by cos3xcosx + sin3xsinx and cos 4x by cos3xcosx - sin3xsinx and it can be proved...

Answer 9

I'll look at this soon. I need to do something... Thanks for helping!! I'm getting some of it

Answer 10

LS = 1/2(cos 2x − cos 4x)
= (1/2) [ cos(3x − x) - cos(3x + x)]
= (1/2) [ (cos3xcosx + sin3xsinx) - (cos3xcosx - sin3xsinx)]
= (1/2) (2sin3xsinx)
= sin3xsinx
= RS
Therefore, 1/2(cos 2x − cos 4x) ≡ sin 3x sin x