How do you show that:
2sinxcosx=
(4tan{x/2})/(1+tan^{2}{x/2}) . (1-tan^{2}{x/2})/(1+tan^{2}{x/2})
Thanks a lot for helping!!
$1+\tan^2(\frac x 2) = \sec^2(\frac x 2)$
$1-\tan^2(\frac x 2) = \dfrac{2 \tan(\frac x 2)}{\tan(x)}$
$\sin(2x)=2\sin(x)\cos(x)$
-------------------------
$\dfrac{4 \tan(\frac x 2)}{1+\tan^2(\frac x 2)}\dfrac{1-\tan^2(\frac x 2)}{1+\tan^2(\frac x 2)}=$
$\dfrac{4 \tan(\frac x 2)}{\sec^2(\frac x 2)}\dfrac{ \dfrac{2 \tan(\frac x 2)}{\tan(x)}}{\sec^2(\frac x 2)}=$
$4\cos(\frac x 2)\sin(\frac x 2)\dfrac{2\sin(\frac x 2)\cos(\frac x 2)}{\tan(x)}=$
$2\sin(x)\dfrac{\sin(x)}{\tan(x)}=$
$2\sin(x)\cos(x)$
Hello, JiaGeng!
I was wondering if there is a way to do it from LHS to get RHS.
Would that be even harder?
If you would rather not run the steps in reverse,
it can be done ... with a little imagination.
. . . . . . . . .
In both fractions, divide numerator and denominator by
. . . .