trigonometric

• March 16th 2009, 08:57 PM
n7167113
trigonometric
I have 2 questions:

1. find for z ; where z^3=sqrt[imaginary]
2. Simplify: sin x/(1-cos x)
can somebody help me with these 2 questions showing working out?
• March 16th 2009, 09:28 PM
Soroban
Hello, n7167113!

Quote:

Find $z\!:\;\;z^3\:=\:\sqrt{i}$
We have: . $z \:=\:i^{\frac{1}{6}}$

We know: . $i \:=\:\cos\tfrac{\pi}{2} + i\sin\tfrac{\pi}{2}$,

Use DeMoivre's Formula for the six roots of: . $\left(\cos\tfrac{\pi}{2} + i\sin\tfrac{\pi}{2}\right)^{\frac{1}{6}}$

Quote:

Simplify: . $\frac{\sin x}{1-\cos x}$
This doesn't simplify very much, but here goes . . .

Multiply by $\frac{1 +\cos x}{1+\cos x}\!:\;\;\frac{\sin x}{1 - \cos x}\cdot\frac{1+\cos x}{1+\cos x} \;=\;\frac{\sin x(1 + \cos x)}{1-\cos^2\!x} \;=\;\frac{\sin x(1 + \cos x)}{\sin^2\!x}$

. . $= \;\frac{1+\cos x}{\sin x} \;=\;\frac{1}{\sin x} + \frac{\cos x}{\sin x} \;=\;\csc x + \cot x$

• March 16th 2009, 09:33 PM
coolguy99
For the simplification one, try multiplying both the numerator and denominator by (1+cos(x)), and from there you should be able to get a nice answer

edit> I was beat to the punch :p