1. ## Trigonometric Identity Problem

How do you show that:
2sinxcosx=
(4tan{x/2})/(1+tan2{x/2}) . (1-tan2{x/2})/(1+tan2{x/2})
Thanks a lot for helping!!

2. ## Re: Trigonometric Identity Problem

Originally Posted by JiaGeng
How do you show that:
2sinxcosx=
(4tan{x/2})/(1+tan2{x/2}) . (1-tan2{x/2})/(1+tan2{x/2})
$1+\tan^2(\frac x 2) = \sec^2(\frac x 2)$

$1-\tan^2(\frac x 2) = \dfrac{2 \tan(\frac x 2)}{\tan(x)}$

$\sin(2x)=2\sin(x)\cos(x)$

-------------------------
$\dfrac{4 \tan(\frac x 2)}{1+\tan^2(\frac x 2)}\dfrac{1-\tan^2(\frac x 2)}{1+\tan^2(\frac x 2)}=$

$\dfrac{4 \tan(\frac x 2)}{\sec^2(\frac x 2)}\dfrac{ \dfrac{2 \tan(\frac x 2)}{\tan(x)}}{\sec^2(\frac x 2)}=$

$4\cos(\frac x 2)\sin(\frac x 2)\dfrac{2\sin(\frac x 2)\cos(\frac x 2)}{\tan(x)}=$

$2\sin(x)\dfrac{\sin(x)}{\tan(x)}=$

$2\sin(x)\cos(x)$

3. ## Re: Trigonometric Identity Problem

Hello, JiaGeng!

$\text{Prove: }\:2\sin x\cos x\:=\:\frac{4\tan\frac{x}{2}}{1+\tan^2\frac{x}{2}} \cdot \frac{1-\tan^2\frac{x}{2}}{1 + \tan^2\frac{x}{2}}$

The first fraction is:

. . $\frac{4\tan\frac{x}{2}}{1 + \tan^2\frac{x}{2}} \;=\;\frac{4\frac{\sin\frac{x}{2}}{\cos\frac{x}{2} }}{\sec^2\frac{x}{2}} \;=\;\frac{4\frac{\sin\frac{x}{2}}{\cos\frac{x}{2} }}{\frac{1}{\cos^2\frac{2}{2}}} \;=\;4\sin\tfrac{x}{2}\cos\tfrac{x}{2} \;=\;2\sin x$

The second fraction is:

. . $\frac{1-\tan^2\frac{x}{2}}{1 + \tan^2\frac{x}{2}} \;=\;\frac{1-\frac{\sin^2\frac{x}{2}}{\cos^2\frac{x}{2}}}{\sec^ 2\frac{x}{2}} \;=\; \frac{1-\frac{\sin^2\frac{x}{2}}{\cos^2\frac{x}{2}}}{\frac {1}{\cos^2\frac{x}{2}}} \;=\;\cos^2\tfrac{x}{2} - \sin^2\tfrac{x}{2} \;=\;\cos x$

Therefore, the RHS becomes: . $2\sin x\cos x$.

4. ## Re: Trigonometric Identity Problem

Thanks romsek and soroban!
But I was wondering if there is a way to do it from LHS to get RHS because that's what the question requires.... Would that be even harder?

5. ## Re: Trigonometric Identity Problem

Originally Posted by JiaGeng
Thanks romsek and soroban!
But I was wondering if there is a way to do it from LHS to get RHS because that's what the question requires.... Would that be even harder?
Equality works in both directions. If you want, write the equations in reverse order and equality still holds at every step.

6. ## Re: Trigonometric Identity Problem

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.

$2\sin x\cos x \;=\;\frac {2(2\sin \frac{x}{2}\cos\frac{x}{2})}{1}\cdot \frac{\cos^2\!\frac{x}{2} - \sin^2\!\frac{x}{2}}{1}$

. . . . . . . . . $=\;\frac{4\sin\frac{x}{2}\cos\frac{x}{2}}{\cos^2\! \frac{x}{2} + \sin^2\!\frac{x}{2}} \cdot \frac{\cos^2\!\frac{x}{2} - \sin^2\!\frac{x}{2}}{\cos^2\!\frac{x}{2} + \sin^2\!\frac{x}{2}}$

In both fractions, divide numerator and denominator by $\cos^2\!\tfrac{x}{2}\!:$

$\dfrac{\frac{4\sin\frac{x}{2} \cos\frac{x}{2}}{\cos^2\!\frac{x}{2}}} {\frac{\cos^2\!\frac{x}{2}}{\cos^2\!\frac{x}{2}} + \frac{\sin^2\!\frac{x}{2}}{\cos^2\!\frac{x}{2}}} \cdot \dfrac{\frac{\cos^2\!\frac{x}{2}}{\cos^2\!\frac{x} {2}} - \frac{\sin^2\!\frac{x}{2}}{\cos^2\!\frac{x}{2}}} {\frac{\cos^2\!\frac{x}{2}}{\cos^2\!\frac{x}{2}} + \frac{\sin^2\!\frac{x}{2}}{\cos^2\!\frac{x}{2}}} \;=\; \dfrac{4\frac{\sin\frac{x}{2}}{\cos\frac{x}{2}}} {1 + \left(\frac{\sin\frac{x}{2}}{\cos\frac{x}{2}} \right)^2} \cdot \dfrac{1 - \left(\frac{\sin\frac{x}{2}}{\cos\frac{x}{2}} \right)^2} {1 + \left(\frac{\sin\frac{x}{2}}{\cos\frac{x}{2}} \right)^2}$

. . . . $=\;\frac{4\tan\frac{x}{2}}{1 + \tan^2\!\frac{x}{2}} \cdot \frac{1-\tan^2\!\frac{x}{2}}{1 + \tan^2\!\frac{x}{2}}$