1. [SOLVED] Deriving a Formula

How do you derive a formula for both $\cos {3\Theta}$ and $\cos {4\Theta}$? Do you just use the sum and difference formulas?

2. What do you mean by "deriving" the "formulas"? Are you supposed to restate the expressions in terms of $\sin{(\theta)}$ and $\cos{(\theta)}$?

Thank you!

3. Originally Posted by stapel
What do you mean by "deriving" the "formulas"? Are you supposed to restate the expressions in terms of $\sin{(\theta)}$ and $\cos{(\theta)}$?

Thank you!
It doesn't say specifically but I think so. So would you use the sum and difference formulas to derive?

4. Assuming you mean what stapel suggested, that is writing them in terms of $cos(\theta)$, $sin(\theta)$, then, yes. You use the sum formulas.

$cos(2\theta)= sin(\theta+ \theta)= cos(\theta)cos(\theta)- sin(\theta)sin(\theta)= cos^2(\theta)- sin^2(\theta)$
$sin(2\theta)= cos(\theta)sin(\theta)+ sin(\theta)cos(\theta)= 2 cos(\theta)sin(\theta)$

So
$sin(3\theta)= sin(2\theta+ \theta)= cos(2\theta)sin(\theta)+ sin(2\theta)cos(\theta)$
$cos(3\theta)= cos(2\theta+ \theta)= cos(2\theta)cos(\theta)- sin(2\theta)sin(\theta)$
and use the formulas above.

Similarly for
$cos(4\theta)= cos(3\theta+ \theta)$ and $sin(4\theta)= sin(3\theta+ \theta)$.

5. Originally Posted by HallsofIvy
Assuming you mean what stapel suggested, that is writing them in terms of $cos(\theta)$, $sin(\theta)$, then, yes. You use the sum formulas.

$cos(2\theta)= sin(\theta+ \theta)= cos(\theta)cos(\theta)- sin(\theta)sin(\theta)= cos^2(\theta)- sin^2(\theta)$
$sin(2\theta)= cos(\theta)sin(\theta)+ sin(\theta)cos(\theta)= 2 cos(\theta)sin(\theta)$

So
$sin(3\theta)= sin(2\theta+ \theta)= cos(2\theta)sin(\theta)+ sin(2\theta)cos(\theta)$
$cos(3\theta)= cos(2\theta+ \theta)= cos(2\theta)cos(\theta)- sin(2\theta)sin(\theta)$
and use the formulas above.

Similarly for
$cos(4\theta)= cos(3\theta+ \theta)$ and $sin(4\theta)= sin(3\theta+ \theta)$.
Ok thanx alot.