Sin2x + _sin2x = 2
1-cosx 1+cosx

i used sin^2x + cos^2 x = 1
then got (sin^2x)/(1-cos^2x)=0
and thats a diff of 2 squares
so its
(sin^2x)/(1-cosx)(1+cosx) = 0

but where does the two come from
what did i do wrong??

2. Hello deej813
Originally Posted by deej813
Sin2x + _sin2x = 2
1-cosx 1+cosx

i used sin^2x + cos^2 x = 1
then got (sin^2x)/(1-cos^2x)=0
and thats a diff of 2 squares
so its
(sin^2x)/(1-cosx)(1+cosx) = 0

but where does the two come from
what did i do wrong??
I'm not clear what the question is here. My best guess is that you have to solve the equation
$\frac{\sin2x}{1-\cos x}+\frac{\sin2x}{1+\cos x}=2$
If that's it then you'd write:
$\Rightarrow \frac{\sin2x(1+\cos x + 1-\cos x)}{(1-\cos x)(1+\cos x)}=2$

$\Rightarrow \frac{4\sin x \cos x}{1-\cos^2 x}=2$, using $\sin 2x = 2\sin x\cos x$

$\Rightarrow \frac{2\sin x\cos x}{\sin^2x}=1$

$\Rightarrow 2\cot x = 1$

$\Rightarrow \tan x = 2$

$\Rightarrow x = \arctan(2) + n\pi$

3. nah i've gotta prove that the LHS = RHS = 2
so rearrange the LHS and get it to equal two
but i keep getting 0

4. Hello deej813

Sorry, but the LHS just isn't identically equal to 2. If you followed my working you'd see that it's $4\cot x$, and that's as far as you can go. So here it is once again:

$\frac{\sin2x}{1-\cos x}+\frac{\sin2x}{1+\cos x}$
$=\frac{\sin2x(1+\cos x + 1-\cos x)}{(1-\cos x)(1+\cos x)}$

$=\frac{2\sin2x}{1-\cos^2 x}$
$= \frac{4\sin x\cos x}{\sin^2x}$
$=4\cot x$

5. sorry its sin^2x on top of both fractions.
does that make a difference??

6. Hello deej813
Originally Posted by deej813
sorry its sin^2x on top of both fractions.
does that make a difference??
It sure does! Instead of $\sin2x$ on the top then, we get:
$\frac{\sin^2x}{1-\cos x}+\frac{\sin^2x}{1+\cos x}$
$=\frac{\sin^2x(1+\cos x + 1-\cos x)}{(1-\cos x)(1+\cos x)}$

$=\frac{2\sin^2x}{1-\cos^2 x}$
$= \frac{2\sin^2 x}{\sin^2x}$
$=2$