1. ## precalc problem help

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

where do i even start?

2. Is this want you would like to prove?

$cos (3\theta) = 4cos^3(\theta)-3cos(\theta)$

3. yess... thanks for making it the way it should!

4. Originally Posted by ohhhh
show that cos 3(theta) = 4cos^3(theta)-3cos(theta)

where do i even start?
$\cos{3\theta}\ =\ \cos{(2\theta+\theta)}$

Now apply the formula $\cos{(A+B)}=\cos{A}\cos{B}-\sin{A}\sin{B}$ as well as the double-angle formula for sine and for cosine.

5. 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?

6. Originally Posted by ohhhh
show that cos 3(theta) = 4cos^3(theta)-3cos(theta)
Complex numbers also work:

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

So $\cos 3\theta = \cos ^3 \theta - 3\cos \theta(1 - \cos ^2 \theta) = 4\cos ^3 \theta - 3\cos \theta\,\blacksquare$

7. thanks a lot for your help guys. got it down!

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

= cos^2a-sin^2a - 2sin(a)*cos(a)*sin(a)
You have lost a cos(a). It should read
$\cos (a)(\cos ^2(a)-\sin ^2(a))-2\sin (a) \cos (a) \sin (a)$