show that cos 3(theta) = 4cos^3(theta)-3cos(theta)

where do i even start? :o

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- Dec 19th 2007, 06:34 AMohhhhprecalc problem help
show that cos 3(theta) = 4cos^3(theta)-3cos(theta)

where do i even start? :o - Dec 19th 2007, 06:42 AMcolby2152
Is this want you would like to prove?

$\displaystyle cos (3\theta) = 4cos^3(\theta)-3cos(\theta)$ - Dec 19th 2007, 07:00 AMohhhh
yess... thanks for making it the way it should! :)

- Dec 19th 2007, 07:09 AMJaneBennet
- Dec 19th 2007, 07:29 AMohhhh
ok, so let me see.

cos (2a+a) = cos2(a)*cos(a) - sin2(a)*sin(a)

= cos^2a-sin^2a - 2sin(a)*cos(a)*sin(a)

= cos^2a-sin^2a - 2sin^2(a)*cos(a)

...is this right so far? - Dec 19th 2007, 07:50 AMKrizalid
Complex numbers also work:

$\displaystyle \cos 3\theta = \text{Re} \left( {\cos \theta + i\sin \theta} \right)^3 = \text{Re} \left( {\cos ^3 \theta + 3i\sin \theta\cos ^2 \theta - 3\sin ^2 \theta\cos \theta - i\sin ^3 \theta} \right).$

So $\displaystyle \cos 3\theta = \cos ^3 \theta - 3\cos \theta(1 - \cos ^2 \theta) = 4\cos ^3 \theta - 3\cos \theta\,\blacksquare$ - Dec 19th 2007, 04:26 PMohhhh
thanks a lot for your help guys. got it down! :)

- Dec 19th 2007, 04:31 PMbadgerigarQuote:

cos (2a+a) = cos2(a)*cos(a) - sin2(a)*sin(a)

Quote:

= cos^2a-sin^2a - 2sin(a)*cos(a)*sin(a)

$\displaystyle \cos (a)(\cos ^2(a)-\sin ^2(a))-2\sin (a) \cos (a) \sin (a)$