1. ## help req/d proving an identity please?

hi there
can anyone get me a little bit further with this question?
prove the identity:

cos 3x = 4cos^3x -3cosx cos 3x can also =cos(2x+x)

so far i have used the formula

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

so this i have equated to

cos(2x+x)=cos2x.cosx-sin2x.sinx

back to the orig equation cos2x.cosx-sin2x.sinx=4cos^3x-3cosx

so thats where i end up and i am now confused as how to proceed next!

any help appreciated.

many thanks in advance sure its nearly there.

2. You need to expand your cosine identities a bit. Use the one you stated to produce:

$\cos(2x) = \cos^{2}(x) - \sin^{2}(x) = 1 - 2\sin^{2}(x) = 2\cos^{2}(x) - 1$

This should get you where you are going.

3. You're on the right path, you need to use the double angle identities to expand further

$cos(3x) = cos(2x)cos(x)-sin(2x)sin(x)$. Useful identities are below

$cos(2x) = 2cos^2(x)-1$

$sin(2x) = 2sin(x)cos(x)$

$sin^2(x) = 1-cos^2(x)$

4. ## many thanks

hi again,firstly many thanks for the very prompt replies.

i'm sorry to say i not quite getting there which is annoying me.

i know the lhs needs to equal the rhs and that the identities can represent other identities and thats how it is all simplified but i cant seem to convert to get the appropriate identities and the theory is so simple!

my ongoing attempt, is this getting there or not?

cos2x.cosx-sin2x.sinx changes to the following?

2cos^2-1.cosx-sin2x.sinx

2cos^3x-1-sinx.sinx

2cos^3-1-sin^2 2x?

so its all going a bit pete tong as they say

not seeing how this will equate to the rhs i'm afraid and i have honestly been trying for hours now.

hi again,firstly many thanks for the very prompt replies.

i'm sorry to say i not quite getting there which is annoying me.

i know the lhs needs to equal the rhs and that the identities can represent other identities and thats how it is all simplified but i cant seem to convert to get the appropriate identities and the theory is so simple!

my ongoing attempt, is this getting there or not?

cos2x.cosx-sin2x.sinx changes to the following?

2cos^2-1.cosx-sin2x.sinx

2cos^3x-1-sinx.sinx

2cos^3-1-sin^2 2x?

so its all going a bit pete tong as they say

not seeing how this will equate to the rhs i'm afraid and i have honestly been trying for hours now.
Take it one step at a time and see if you can follow my steps:

$cos(2x)cos(x)-sin(2x)sin(x)$

As we know $cos(2x) = 2cos^2(x)-1$ and $sin(2x) = 2sin(x)cos(x)$

$[2cos^2(x)-1]cos(x) - 2sin(x)cos(x)sin(x)$

Expand and simplify

$2cos^3(x)-cos(x) - 2cos(x)sin^2(x)$

As $sin^2(x) = 1-cos^2(x)$

$2cos^3(x)-cos(x)-2cos(x)[1-cos^2(x)]$

Expand again

$2cos^3(x)-cos(x)-2cos(x)+2cos^3(x)$

Collect like terms: $4cos^3(x)-3cos(x)$ = RHS

6. ## many thanks

it is actually incredibly straight forward isnt it?

i think my problem was i was struggling with the multiplications believe it or not how stupid can you be.

it is a great feeling when it all comes together and you can see where you went wrong ,thankyou again its very clear now.