# Simple trigonometry....But I cant solve it agian :(

• Jul 27th 2012, 08:20 AM
Tiffany
Simple trigonometry....But I cant solve it agian :(
Here:
(1+cosx+isinx) divided by (sinx-i(1+cosx))
Write in trigonometry form.
• Jul 27th 2012, 08:42 AM
Prove It
Re: Simple trigonometry....But I cant solve it agian :(
Quote:

Originally Posted by Tiffany
Here:
(1+cosx+isinx) divided by (sinx-i(1+cosx))
Write in trigonometry form.

It's difficult to know exactly what you want. I expect you're asked to simplify \displaystyle \displaystyle \begin{align*} \frac{1 + \cos{x} + i\sin{x}}{\sin{x} - i\left(1 + \cos{x}\right)} \end{align*} and write it in \displaystyle \displaystyle \begin{align*} r\left(\cos{\theta} + i\sin{\theta}\right) \end{align*} form. Anyway...

\displaystyle \displaystyle \begin{align*} \frac{1 + \cos{x} + i\sin{x}}{\sin{x} - i\left(1 + \cos{x}\right)} &= \frac{1 + \cos{x} + i\sin{x}}{\sin{x} - i\left(1 + \cos{x}\right)}\cdot \frac{\sin{x} + i\left(1 + \cos{x}\right)}{\sin{x} + i\left(1 + \cos{x}\right)} \\ &= \frac{\left(1 + \cos{x} + i\sin{x}\right)\left[\sin{x} + i\left(1 + \cos{x}\right)\right]}{\left[\sin{x} - i\left(1 + \cos{x}\right)\right]\left[\sin{x} + i\left(1 + \cos{x}\right)\right]} \\ &= \frac{\sin{x}\left(1 + \cos{x}\right) + i\left(1 + \cos{x}\right)^2 + i\sin^2{x} - \sin{x}\left(1 + \cos{x}\right) }{\sin^2{x} + \left(1 + \cos{x}\right)^2} \\ &= \frac{\sin{x} + \sin{x}\cos{x} + i\left(1 + 2\cos{x} + \cos^2{x}\right) + i\sin^2{x} - \sin{x} - \sin{x}\cos{x}}{\sin^2{x} + 1 + 2\cos{x} + \cos^2{x}} \\ &= \frac{i\left(2 + 2\cos{x}\right)}{2 + 2\cos{x}} \\ &= i \\ &= \cos{\left(\frac{\pi}{2}\right)} + i\sin{\left(\frac{\pi}{2}\right)} \end{align*}
• Jul 31st 2012, 03:27 AM
Tiffany
Re: Simple trigonometry....But I cant solve it agian :(
Thank you so much. that is exactly what I want. :))
• Aug 4th 2012, 10:19 PM
reflection_009
Re: Simple trigonometry....But I cant solve it agian :(
(1+cosx+isinx) /(sinx-i(1+cosx))=((2Cos^2(x/2)+i2sin(x/2)cos(x/2))/((2sin(x/2)cos(x/2))-i2Cos^2(x/2))
=(cos(x/2)+isin(x/2))/(sin(x/2)-icos(x/2))=(cos(x/2)+isin(x/2))/[(-i)((cos(x/2)+isin(x/2))]
=(-1)/i= i (proved in 3 steps)