[SOLVED] trig question

**Burger king** · June 5th 2009, 02:31 AM

Hi, got a problem that I need some help with.

How can I express cos 4x in cos x?

Help appreciated.

**great_math** · June 5th 2009, 02:50 AM

$\cos 4x=\cos(3x+x)=\cos3x\cdot\cos x - \sin 3x\cdot \sin x$

$=(4\cos^3x-3\cos x)\cos x-(3\sin x-4\sin^3 x)\cdot \sin x$

$=4\cos^4x-3\cos^2x-3\sin^2x+4\sin^4x$

$=4\cos^4x+4\sin^4x-3$

$=4\cos^4x+4(1-\cos^2 x)^2-3$

$=4\cos^4x+4(1+\cos^4x-2\cos^2x)-3$

$=4\cos^4x+4+4\cos^4x-8\cos^2x-3$

$=8\cos^4x-8\cos^2x+1$

**great_math** · June 5th 2009, 03:26 AM

Another way:

$\cos 4x=2\cos^22x-1$

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

$=2(4\cos^4x+1-4\cos^2x)-1$

$=8\cos^4x-8\cos^2x+1$

**Burger king** · June 5th 2009, 03:34 AM

Great, the other one was a bit easier

Does the answer satisfy the question in the sense that it requires the answer to be expressed in cosx, whereas the answer you calculated has a + 1 term?

**great_math** · June 5th 2009, 03:37 AM

1 is a constant so it does satisfy the question

**great_math** · June 5th 2009, 03:43 AM

$8\cos^4x-8\cos^2x+1$

This can also be written as $(8\cos^4x-8\cos^2x+1)\frac{\cos x}{\cos x}$

$=\frac{8\cos^5x-8\cos^3x+\cos x}{\cos x}$

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