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 $\displaystyle \begin{align*} \frac{1 + \cos{x} + i\sin{x}}{\sin{x} - i\left(1 + \cos{x}\right)} \end{align*}$ and write it in $\displaystyle \begin{align*} r\left(\cos{\theta} + i\sin{\theta}\right) \end{align*}$ form. Anyway...

$\displaystyle \begin{align*} \frac{1 + \cos{x} + i\sin{x}}{\sin{x} - i\left(1 + \cos{x}\right)} &= \frac{1 + \cos{x} + i\sin{x}}{\sin{x} - i\left(1 + \cos{x}\right)}\cdot \frac{\sin{x} + i\left(1 + \cos{x}\right)}{\sin{x} + i\left(1 + \cos{x}\right)} \\ &= \frac{\left(1 + \cos{x} + i\sin{x}\right)\left[\sin{x} + i\left(1 + \cos{x}\right)\right]}{\left[\sin{x} - i\left(1 + \cos{x}\right)\right]\left[\sin{x} + i\left(1 + \cos{x}\right)\right]} \\ &= \frac{\sin{x}\left(1 + \cos{x}\right) + i\left(1 + \cos{x}\right)^2 + i\sin^2{x} - \sin{x}\left(1 + \cos{x}\right) }{\sin^2{x} + \left(1 + \cos{x}\right)^2} \\ &= \frac{\sin{x} + \sin{x}\cos{x} + i\left(1 + 2\cos{x} + \cos^2{x}\right) + i\sin^2{x} - \sin{x} - \sin{x}\cos{x}}{\sin^2{x} + 1 + 2\cos{x} + \cos^2{x}} \\ &= \frac{i\left(2 + 2\cos{x}\right)}{2 + 2\cos{x}} \\ &= i \\ &= \cos{\left(\frac{\pi}{2}\right)} + i\sin{\left(\frac{\pi}{2}\right)} \end{align*}$