Identity Issue...

• October 17th 2009, 09:35 AM
Savior_Self
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.

• October 17th 2009, 09:45 AM
ramiee2010
use identity
$1+\cos {2x} =2 \cos^2 x$
• October 17th 2009, 09:46 AM

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.
• October 18th 2009, 10:19 AM
Savior_Self
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)...
• October 18th 2009, 10:56 AM
Note that $\frac{x^2}{x} = x$ when x =/= 0
• October 18th 2009, 07:55 PM
pacman
http://www.mathhelpforum.com/math-he...5eb8319c-1.gif,

--------------------------------------------------------------
--------------------------------------------------------------

--------------------------------------------------------------

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 . . . .

• October 19th 2009, 06:03 AM
Savior_Self
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?
• October 20th 2009, 07:39 AM
pacman