1. ## Trigonometric function...

Proved the following identities:
1. tan(3x/2)-tan(x/2)=(2sinx)/(cosx+cos2x)
2. cos4a. tan2a-sin4a=(2tan a)/(tan^2 a -1)

2. ## Re: Trigonometric function...

Wouldn't it be better for you to proe them? Do you know identities for trig functions of multiple or fractional arguments?

$sin(2x)= 2sin(x)cos(x)$ and $cos(2x)= cos^2(x)- sin^2(x)$

In the second, we can replace $sin^2(x)$ by $1- cos^2(x)$ to get $cos(2x)= cos^2(x)- (1- cos^2(x))= 2cos^2(x)- 1$
so that $cos^2(x)= (1+ cos(2x))/2$, $cos(x)= \pm \sqrt{(1/2)(1+ cos(2x))}$ and letting $\theta= 2x$ that becomes $cos(\theta/2)= \pm\sqrt{(1/2)(1+ cos(2\theta))}$. Similarly, $sin(2x)= 2sin(x)cos(x)$ and $sin(\theta/2)= \pm\sqrt{(1/2)(1- cos(\theta))}$.

Those, together with the fact that $tan(x)= \frac{sin(x)}{cos(x)}$, are what you need.

3. ## Re: Trigonometric function...

Originally Posted by HallsofIvy
Wouldn't it be better for you to proe them? Do you know identities for trig functions of multiple or fractional arguments?

$sin(2x)= 2sin(x)cos(x)$ and $cos(2x)= cos^2(x)- sin^2(x)$

In the second, we can replace $sin^2(x)$ by $1- cos^2(x)$ to get $cos(2x)= cos^2(x)- (1- cos^2(x))= 2cos^2(x)- 1$
so that $cos^2(x)= (1+ cos(2x))/2$, $cos(x)= \pm \sqrt{(1/2)(1+ cos(2x))}$ and letting $\theta= 2x$ that becomes $cos(\theta/2)= \pm\sqrt{(1/2)(1+ cos(2\theta))}$. Similarly, $sin(2x)= 2sin(x)cos(x)$ and $sin(\theta/2)= \pm\sqrt{(1/2)(1- cos(\theta))}$.

Those, together with the fact that $tan(x)= \frac{sin(x)}{cos(x)}$, are what you need.
yeah, I know that, I just don't know how to get tan(3x/2)

4. ## Re: Trigonometric function...

Use the half angle identity for tangent, then everything will be in terms of tan(3x).

Then use the angle sum identity for tangent using tan(x + 2x).

5. ## Re: Trigonometric function...

Originally Posted by Prove It
Use the half angle identity for tangent, then everything will be in terms of tan(3x).

Then use the angle sum identity for tangent using tan(x + 2x).
but I still can't get rid of the tan^2 (3x/2)

6. ## Re: Trigonometric function...

Originally Posted by Trefoil2727
but I still can't get rid of the tan^2 (3x/2)
Prove It just told you how to do this. Tell you what...Use the 1/2 angle part and post what you've got. That will make things easier for us to know how to help you.

-Dan

7. ## Re: Trigonometric function...

Originally Posted by topsquark
Prove It just told you how to do this. Tell you what...Use the 1/2 angle part and post what you've got. That will make things easier for us to know how to help you.

-Dan
tan(3x/2)-tan(x/2)=(2sinx)/(cosx+cos2x)
LHS=(tan3x)(1-tan^2 3x/2)-(tanx)(1-tan^2 x/2)
=((tanx+tan2x)/(1-tanxtan2x))(1-tan^2 3x/2)-(tanx)(1-tan^2 x/2)..

8. ## Re: Trigonometric function...

The second one is quite simple
LHS = cos 4a tan 2a - sin 2a = ( cos 4a sin 2a ) / cos 2a - sin 2a = ( cos 4a sin 2a - sin 4a cos 2a )/ cos 2a = - sin ( 4a - 2a )/ cos 2a = - tan 2a
RHS = ( 2 tan a )/ ( tan^2 a - 1 ) = - ( 2 tan a )/ ( 1 - tan^2 a ) = - tan 2a
hence LHS = RHS

9. ## Re: Trigonometric function...

Originally Posted by ibdutt
The second one is quite simple
LHS = cos 4a tan 2a - sin 2a = ( cos 4a sin 2a ) / cos 2a - sin 2a = ( cos 4a sin 2a - sin 4a cos 2a )/ cos 2a = - sin ( 4a - 2a )/ cos 2a = - tan 2a
RHS = ( 2 tan a )/ ( tan^2 a - 1 ) = - ( 2 tan a )/ ( 1 - tan^2 a ) = - tan 2a
hence LHS = RHS
why ( cos 4a sin 2a - sin 4a cos 2a )/ cos 2a = - sin ( 4a - 2a )/ cos 2a = - tan 2a ?

10. ## Re: Trigonometric function...

remember the formula
sin ( A-B ) = sin A cosB - cos A sin B
Thus cos 4a sin 2a - sin 4a cos 2a = - ( sin 4a cos 2a - cos 4a sin 2a ) = - sin ( 4a - 2a ) = - sin 2a
also we know that sin A / cos A = tan A

13. ## Re: Trigonometric function...

Originally Posted by ibdutt
I've solved that already, left the 1st question..