Prove Trig Identity

[Anna]

Thanks for the Help on that other one. I have One more Identity that I'm having trouble with:

Sin2x = 2Tanx / (1 + tan^2x) (I think that is how I should write Tan squared x ...)

I know I need to work with the Right side.
I tried to use these identities: Tanx = sinx / cos x and tan^2x + 1 = sec^2x = (1 / cosx)^2
I get the following:

2Sinx / Cosx
_____________ How do I reduce this to Sin2x ?
(1 / Cos)^2

 

[lookagain]

I get the following:

2Sinx / Cosx
_____________ How do I reduce this to Sin2x ?
(1 / Cos)^2

Stay with the lower case or stay with the upper case. You left out the "x" in cos(x) in the
denominator of the larger fraction.

2sin(x)/cos(x)
----------------- =
[1/cos(x)]^2


$\displaystyle \frac{\frac{2sin(x)}{cos(x)}}{\frac{1}{cos^2(x)}} =$


Multiply the numerators of the two smaller fractions by $\displaystyle cos^2(x):$


$\displaystyle 2cos(x)cos(x)$


And what is a useful/needed identity that this expression can equal?

 

[soroban]

Hello, Anna!

Prove: . sin 2x .= .(2 tan x)/(1 + tan^2 x)

. . . . . . . . . . . . . . . . . .2 sin x cos x
The left side is: .sin 2x .= .---------------
. . . . . . . . . . . . . . . . . . . . .1



Divide top and bottom by cos^2 x:


. . . 2 sin x cos x . . . . . . . sin x
. . . --------------- - - - - . 2 -------
. . . . cos^2 x . . . . . . . . . cos x. . . . . . 2 tan x
. . ------------------- . = . ----------- . = . ---------------
. . . . . . 1 . . . . . . . . . .sec^2 x . . . . .1 + tan^2 x
. . . . -----------
. . . . cos^2 x

 

[Anna]

I'm seeing a (7) Can't run latex program ????? I can't see the results. Any suggestions?