Using de Moivre's Formula to Derive Trigonometric Identities

[DEMJ]

Homework Statement

Use de Moivre's formula to derive the following trigonometric identites:

$$(a) cos3\theta = cos^3\theta - 3cos\theta sin^2\theta$$

$$(b) sin3\theta = 3cos^2\theta sin\theta - sin^3\theta$$

Homework Equations

The Attempt at a Solution

The only way I have even figured out to solve this is by just doing
$$(cos\theta + isin\theta)^3 = (cos^3\theta - 3cos\theta sin^2\theta) + i(3cos^2\theta sin\theta - sin^3\theta) = cos3\theta + isin3\theta$$

but I fear that this is not what the problem is asking me to do. I think on (a) I should be factoring out $$cos^3\theta - 3cos\theta sin^2\theta = cos\theta(cos^2\theta - 3sin^2\theta)$$

should I then use the trig. formula that $$cos^2 - sin^2 = cos2\theta$$ but the 3 in front of sin is throwing me off. Anyone have a clue as how this problem is supposed to be done in the way the question is asking? Thank you

 

[rock.freak667]

$$(cos\theta + i sin \theta)^3 = cos 3 \theta+ i sin 3 \theta$$


Expand out the left side and notice that cos3θ is the real part.

 

[DEMJ]

Yea rock that's what I did I just do not know if that's what the question is asking for because I do not think it is but I was wondering if anyone else had any ideas on what to do besides what you mentioned.

 

[zcd]

All the terms with i would = sin3θ, while all the terms without i (reals) would = cos3θ