[SOLVED] Deriving a Formula

**chrozer** · February 28th 2009, 01:39 PM

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

**stapel** · February 28th 2009, 01:57 PM

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!

**chrozer** · February 28th 2009, 02:07 PM

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?

**HallsofIvy** · February 28th 2009, 02:08 PM

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 $sin(4\theta)= sin(3\theta+ \theta)$ .

**chrozer** · February 28th 2009, 02:16 PM

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.

Thread: [SOLVED] Deriving a Formula

LinkBack

Thread Tools

Display

[SOLVED] Deriving a Formula

Similar Math Help Forum Discussions

Deriving a trig formula

Deriving a generic formula:

Help in deriving formula

Deriving the Quadratic Formula

deriving the formula

Search Tags