cos3x over cosx=1-4sin^2 x verify the identities

Question

ujjwal · Answer

$\cos3x=4\cos^3x-3\cos x$

campbell_st · Answer

well you need to trig expansion for cos(3x) make it 

cos(2x + x) 

so you are looking at 

$$\frac{\cos(2x)\cos(x) - \sin(2x)(sinx)}{\cos(x)}$$

and the simplfy from here... substitute sin(2x) = 2sin(x)cos(x) 

and $$\cos(2x) = \cos^2(x) - \sin^2(x)$$

then its just the manipulation into your desired answer

anonymous · Answer

Sorry but I still don't get it I'm completely lost

campbell_st · Answer

ok.... do you know about 

cos(2x) and its expansion or    cos(a + b) and its expansion...

anonymous · Answer

Nope

campbell_st · Answer

wow... then it is virtually impossible for you to do the question using your trig skills

so use the first piece of information you have 

$$\cos(3x) = 4\cos^3(x) - 3\cos(x)$$

substitute it into your equation 

and you get 

$$\frac{4\cos^3(x) - 3\cos(x)}{\cos(x)}$$

which can be written as

$$\frac{4\cos^3(x)}{\cos(x)} - \frac{3\cos(x)}{\cos(x)}$$

do you think you could simplify the above fractions by removing the common factor?