Proved the following identities:

1. tan(3x/2)-tan(x/2)=(2sinx)/(cosx+cos2x)

2. cos4a. tan2a-sin4a=(2tan a)/(tan^2 a -1)

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- Jun 17th 2013, 07:27 AMTrefoil2727Trigonometric 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 AMHallsofIvyRe: Trigonometric function...
Wouldn't it be better for

**you**to proe them? Do you know identities for trig functions of multiple or fractional arguments?

and

In the second, we can replace by to get

so that , and letting that becomes . Similarly, and .

Those, together with the fact that , are what you need. - Jun 18th 2013, 06:14 AMTrefoil2727Re: Trigonometric function...
- Jun 18th 2013, 06:28 AMProve ItRe: 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 AMTrefoil2727Re: Trigonometric function...
- Jun 19th 2013, 01:26 PMtopsquarkRe: Trigonometric function...
- Jun 20th 2013, 08:34 PMTrefoil2727Re: Trigonometric function...
- Jun 20th 2013, 10:15 PMibduttRe: 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 AMTrefoil2727Re: Trigonometric function...
- Jun 21st 2013, 04:19 AMibduttRe: 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 AMTrefoil2727Re: Trigonometric function...
- Jun 27th 2013, 09:27 PMibduttRe: Trigonometric function...
- Jun 28th 2013, 02:49 AMTrefoil2727Re: Trigonometric function...
- Jun 28th 2013, 07:47 PMTrefoil2727Re: Trigonometric function...