Question on proving a trigonometric identity

[bkchng]

Hi!

I'm currently trying to work out (b), but am at a loss of how to even approach it. I've tried applying the Addition Formulae on some of the terms (e.g. change sin 5x to sin (3x + 2x) first, but don't seem to be getting anywhere.
Any pointers, please?

Thanks!

 

[Deleted member 4993]

View attachment 16806

Hi!

I'm currently trying to work out (b), but am at a loss of how to even approach it. I've tried applying the Addition Formulae on some of the terms (e.g. change sin 5x to sin (3x + 2x) first, but don't seem to be getting anywhere.
Any pointers, please?

Thanks!

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If I were to do this problem: I'll use:

cos(9x) - cos(x) = - 2 * sin(5x) * sin(4x) ..... and so on ...............edited (2/24)

 

[hoosie]

A slight change to Subhotosh’s suggestion:

(cos9x - cosx = -2sin5xsin4x)

If you are still not sure how to proceed further here are some steps you might like to follow:

write sin5xsin4x in terms of cos9x and cosx.

Complete:
cos(3x + 2x) =
cos(3x - 2x) =

write sin3xsin2x in terms of cos5x and cosx

Complete:
cos(8x + x) =
cos(8x - x) =

write sin8xsinx in terms of cos9x and cos7x

Simplify:
sin5xsin4x - sin3xsin2x - sin8xsinx
giving an expression in terms of cos5x and cos7x.

How can you use cos(A + B) and cos(A - B) to write your expression in terms of sin6x and sinx?

Can you use the given information (sin6x = 1/sinx) to complete the proof?

 

[bkchng]

Thank you for the suggestions. This is as far as I got:

What should I do from this point on?
And by the way, is there any other easier way? This is from a past exam question, and usually it's not supposed to be this complicated given the time constraints in that exam.
Thanks again!

 

[hoosie]

Simplify (1/2)[cos5x - cos7x] :

For 5x use 6x - x not 3x + 2x as we have to make use of fact sin6xsinx = 1
cos5x
= cos(6x - x)
= cos6xsinx + sin6xsinx

For 7x use 6x + x not 8x - x ( same reason as above).
cos7x
= cos(6x + x)
= cos6xsinx - sin6xsinx


(1/2)[cos5x - cos7x]
= (1/2)[cos6xsinx + sin6xsinx -(cos6xsinx - sin6xsinx)]
= (1/2)[2sin6xsinx]
= sin6xsinx
= 1 since sin6x = 1/sinx