Double-angle formulas with complex numbers

mitch_nufc · #1 $Old$ December 10th, 2008, 10:00 AM

use the equations cos(theta)=0.5(exp(i.theta)+exp(-i.theta)) and sin(theta)=0.5i(exp(i.theta)-exp(-i.theta)) to show that:

sin(theta-y)=sin(theta)cosy-sinycos(theta) and cos(tetha+y)=cos(theta)cosy - sin(theta)siny

Sorry about the symbols im not sure how to insert them

Please help, would be ever so greatful

mitch_nufc · #2 $Old$ December 10th, 2008, 11:05 AM

pleeeeeeeease help me with this im gettin totally stressed

chella182 · #3 $Old$ December 10th, 2008, 11:22 AM

The question is...

Using the equations $\cos\theta = \frac{1}{2}(e^{i\theta}+e^{-i\theta})$ and $\sin\theta = \frac{1}{2i}(e^{i\theta}-e^{-i\theta})$ show that...

$\sin(\theta-\phi)=\sin\theta\cos\phi-\cos\theta\sin\phi$

And

$\cos(\theta+\phi)=\cos\theta\cos\phi-\sin\theta\sin\phi$

Just to make it easier to read :].

Grandad · #4 $Old$ December 10th, 2008, 12:19 PM

Hello -

Quote:

Originally Posted by chella182 $View Post$

The question is...

Using the equations $\cos\theta = \frac{1}{2}(e^{i\theta}+e^{-i\theta})$ and $\sin\theta = \frac{1}{2i}(e^{i\theta}-e^{-i\theta})$ show that...

$\sin(\theta-\phi)=\sin\theta\cos\phi-\cos\theta\sin\phi$

And

$\cos(\theta+\phi)=\cos\theta\cos\phi-\sin\theta\sin\phi$

Just to make it easier to read :].

There's nothing very hard here (honest!). But you do need to be very accurate in removing brackets and simplifying powers.

Starting with $\sin\theta \cos\phi - \cos\theta \sin\phi$ :

Write each of these four trig functions in terms of $e$ .
Take out a factor $\frac{1}{4i}$ .
Carefully multiply out the brackets, noting that, for example, $e^{i\theta} \times e^{i\phi} = e^{i(\theta+\phi)}$ . (Watch the minus signs very carefully!
You should now have eight terms, four of which will cancel, and the other four will add together to give two.
Divide inside the bracket by 2, to leave $\frac{1}{2i}$ outside.

The result should be the expression for $\sin(\theta - \phi)$ in terms of $e$ .

$\cos(\theta+\phi)$ goes in a similar way (except that you take out a factor $\frac{1}{4}$ , noting that $i^2=-1$ ).

Grandad

chella182 · #5 $Old$ December 10th, 2008, 12:43 PM

I think I tried something similar to this, but failed obviously haha! It didn't look too hard, I just didn't know where to start with it. mitch_nufc has sorted it out now anyway though, but thank you