Trigonometric Proof

[vanbeersj]

I need to show that the left side equals the ride side by transforming the left side of this equation

[(1 + cosø)/(1-cosø)] ^1/2 * sinø = 1 + cosø

This is my work....
= (1 + cosø)/(1-cosø)] * (sinø)^2
=(1 + cosø)*([(sinø)^2 - cosø(sinø)^2]
=(sinø)^2 + cosø(sinø)^2 - cosø(sinø)^2 - (cosø)^2(sinø)^2
=(sinø)^2- (cosø)^2(sinø)^2
=1 - cosø

I'm not sure where I'm making the mistake, but I keep getting the negative instead of the positive.

 

[Deleted member 4993]

I need to show that the left side equals the ride side by transforming the left side of this equation

[(1 + cosø)/(1-cosø)] ^1/2 * sinø = 1 + cosø

This is my work....
= (1 + cosø)/(1-cosø)] * (sinø)^2
=(1 + cosø)*([(sinø)^2 - cosø(sinø)^2]
=(sinø)^2 + cosø(sinø)^2 - cosø(sinø)^2 - (cosø)^2(sinø)^2
=(sinø)^2- (cosø)^2(sinø)^2
=1 - cosø

I'm not sure where I'm making the mistake, but I keep getting the negative instead of the positive.

\(\displaystyle \sqrt{\frac{1+cos(\phi)}{1-cos(\phi)}} * sin(\phi)\)

use

\(\displaystyle 1 - cos(\phi) = 2 sin^2(\frac{\phi}{2})\)

and

\(\displaystyle 1 + cos(\phi) = 2 cos^2(\frac{\phi}{2})\)

\(\displaystyle \sqrt{\frac{1+cos(\phi)}{1-cos(\phi)}} * sin(\phi) \, = \, \frac{cos(\frac{\phi}{2})}{sin(\frac{\phi}{2})} \cdot 2 sin(\frac{\phi}{2}) cos(\frac{\phi}{2}) \, = \, 2cos^2(\frac{\phi}{2}) \, = \, 1 + cos(\phi)\)

 

[soroban]

Hello, vanbeersj!

Another approach . . .

\(\displaystyle \text{Prove: }\;\sqrt{\frac{1 + \cos\theta}{1-\cos\theta}}\cdot\sin\theta \;=\; 1 + \cos\theta\)

\(\displaystyle \text{Under the radical, multiply by: }\:\frac{1+\cos\theta}{1+\cos\theta}\)

. . \(\displaystyle \sqrt{\frac{1+\cos\theta}{1-\cos\theta}\cdot\frac{1+\cos\theta}{1+\cos\theta}} \cdot\sin\theta \;\;=\;\;\sqrt{\frac{(1+\cos\theta)^2}{1-\cos^2\!\theta}}\cdot\sin\theta\)

. . . \(\displaystyle =\;\;\sqrt{\frac{(1+\cos\theta)^2}{\sin^2\!\theta}}\cdot\sin\theta \;\;=\;\; \frac{1+\cos\theta}{\sin\theta}\cdot\sin\theta \;\;=\;\;1+\cos\theta\)