# Trigonometric function...

• Jun 17th 2013, 07:27 AM
Trefoil2727
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)
• Jun 17th 2013, 07:46 AM
HallsofIvy
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?

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

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

Those, together with the fact that $\displaystyle tan(x)= \frac{sin(x)}{cos(x)}$, are what you need.
• Jun 18th 2013, 06:14 AM
Trefoil2727
Re: Trigonometric function...
Quote:

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?

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

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

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

yeah, I know that, I just don't know how to get tan(3x/2)
• Jun 18th 2013, 06:28 AM
Prove It
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).
• Jun 19th 2013, 07:34 AM
Trefoil2727
Re: Trigonometric function...
Quote:

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)
• Jun 19th 2013, 01:26 PM
topsquark
Re: Trigonometric function...
Quote:

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
• Jun 20th 2013, 08:34 PM
Trefoil2727
Re: Trigonometric function...
Quote:

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)..
• Jun 20th 2013, 10:15 PM
ibdutt
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
• Jun 21st 2013, 01:54 AM
Trefoil2727
Re: Trigonometric function...
Quote:

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 ?
• Jun 21st 2013, 04:19 AM
ibdutt
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
• Jun 27th 2013, 07:09 AM
Trefoil2727
Re: Trigonometric function...
• Jun 27th 2013, 09:27 PM
ibdutt
Re: Trigonometric function...
• Jun 28th 2013, 02:49 AM
Trefoil2727
Re: Trigonometric function...
Quote:

Originally Posted by ibdutt

I've solved that already, left the 1st question..
• Jun 28th 2013, 07:47 PM
Trefoil2727
Re: Trigonometric function...