1. ## Identity Issue...

Show: $\frac{1 + cos 2x}{csc x sin 2x} = cos x$

What I have so far:
(and I'm probably wrong)

$\frac{1 + cos^2x - sin^2x}{csc x 2sinx cos x}$

then csc x turns to $\frac{1}{sin x}$ in the denominator...
$\frac{1 + cos^2x - sin^2x}{\frac{2sin x cos x}{sin x}}$

cancels out one of the sin x...
$\frac{1 + cos^2x - sin^2x}{sin x cos x}$

then in the numerator I know I should use the " $cos^2x + sin^2x = 1$" rule somehow... but I can't figure out what to do next.

2. use identity
$1+\cos {2x} =2 \cos^2 x$

3. From these identies, I would choose the second one so that
cos2x + 1 = 2cos^2(x)

Also, the denominator should be 2cos(x) after you cancel out the sines. (the denominator will help in most cases to choose the identity that works best. you still used the first identity on the list correctly, but then it doesn't become as obvious.

EDIT:// what user above said.

4. Alright...

So that gets me to:

$\frac{2cos^2x}{2cosx}$

which i guess factors out to:

$\frac{cos^2x}{cosx}$

I'm afraid I'm still unclear on what to do next to manipulate it all to simply cos(x)...

5. Note that $\frac{x^2}{x} = x$ when x =/= 0

6. ,

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since the LHS is difficult, we may start there to prove it.

(1 + cos 2x)/(csc x sin 2x) = cos x,

NUMERATOR: [1 + cos 2x = (2 cos^2 x)]

DENOMENATOR: (csc x)(sin 2x) = (1/sin x)(2 sin x cos x) = 2cos x

can you finish it off?

if you can't still figure out . . . .

7. Thanks for the really clear explanation, pacman.

and from $\frac{2cos^2x}{2cosx}$

you get..

$\frac{cos^2 x}{cosx}$

then

$cos^2 x cos^{-1} x$
(dunno if that's actually correct... just trying to work out the "x^2/x = x" rule from MacstersUndead.)

ends up with...

$cos x$

pretty sure that's right.. look good?