more trig help please!

**snowball** · September 17th 2008, 11:01 AM

1) 2sin^2 x+sin2x=1 find x for 0<x<pi

2) prove that (1+tan $\theta$ )(1+tan( $\pi$ /4- $\theta$ ))=2

3) prove $\frac{{sinx}}{{1-cosx}}=cot\frac{{x}}{{2}}$

I've been working on this all afternoon and my solution still doesn't match with the answers

**icemanfan** · September 17th 2008, 11:21 AM

For problem 2, I used the trigonometric identity $\tan(a - b) = \frac{\tan{a} - \tan{b}}{1 + \tan{a}\tan{b}}$ .

This results in the expression $(1 + \tan{\theta})(1 + \frac{1 - \tan{\theta}}{1 + \tan{\theta}})$ .

When you multiply it out you get this:

$1 + \tan{\theta} + \frac{1 - \tan{\theta}}{1 + \tan{\theta}} + \tan{\theta} \cdot \frac{1 - \tan{\theta}}{1 + \tan{\theta}}$ ,

which simplifies to $\frac{2 + 2 \tan{\theta}}{1 + \tan{\theta}} = 2$ .

**snowball** · September 17th 2008, 11:34 AM

$1 + \tan{\theta} + \frac{1 - \tan{\theta}}{1 + \tan{\theta}} + \tan{\theta} \cdot \frac{1 - \tan{\theta}}{1 + \tan{\theta}}$ ,

after this step I got $\frac{1 - \tan^2{\theta}}{{1+tan\theta}}+1+tan\theta$

am I right or wrong? and could u explain how to simplify down to 2+2tan $\theta$ /1+tan $\theta$ please?

sorry I'm not very familiar with the codes...

I just worked that out... thanks again... now anyone could tell me where to start on the other 2 questions pls?

**Moo** · September 17th 2008, 11:37 AM

Hello,

1) 2sin^2 x+sin2x=1 find x for 0<x<pi

This is the same as $\sin(2x)=1-2\sin^2(x)$

$\sin(2x)=\cos(2x)$
Now, you can think of several ways to solve it. Please show your work

Originally Posted by snowball

3) sinx/(1-cosx)=cotx/2

Is it $\cot \left(\frac x2\right)$ or $\frac{\cot(x)}{2}$ ?

Originally Posted by snowball

$1 + \tan{\theta} + \frac{1 - \tan{\theta}}{1 + \tan{\theta}} + \tan{\theta} \cdot \frac{1 - \tan{\theta}}{1 + \tan{\theta}}$ ,

after this step I got $\frac{1 - \tan^2{\theta}}/{1+tan\theta}+1+tan\theta$

am I right or wrong? and could u explain how to simplify down to 2+2tan $\theta$ /1+tan $\theta$ please?

sorry I'm not very familiar with the codes...

(you don't need to write / if you use \frac{...}{...})

Get the common denominator and simplify.

But it is easier to see it this way :

$(1+\tan \theta)\left(1+\tfrac{1-\tan \theta}{1+\tan \theta}\right)=(1+\tan \theta)+({\color{red}1+\tan \theta}) \cdot \frac{1-\tan \theta}{\color{red}1+\tan \theta}=(1+\tan \theta)+(1-\tan \theta)$

**snowball** · September 17th 2008, 11:57 AM

I finally know how the codes work... thanks to Moo too

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