1. ## more trig help please!

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

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

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

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

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

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

When you multiply it out you get this:

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

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

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

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

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

4. Hello,
1) 2sin^2 x+sin2x=1 find x for 0<x<pi

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

$\displaystyle \sin(2x)=\cos(2x)$
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 $\displaystyle \cot \left(\frac x2\right)$ or $\displaystyle \frac{\cot(x)}{2}$ ?

Originally Posted by snowball
$\displaystyle 1 + \tan{\theta} + \frac{1 - \tan{\theta}}{1 + \tan{\theta}} + \tan{\theta} \cdot \frac{1 - \tan{\theta}}{1 + \tan{\theta}}$,

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

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

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

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