Double Angle Identity

**Beard** · February 25th 2009, 06:33 AM

Stuck on this, don't know where to start

$2\cos^{2}x = 2\sin(x)\cos(x) + 1,\ for\ 0^{\circ} \leq x < 360^{\circ}$

**HallsofIvy** · February 25th 2009, 06:41 AM

Originally Posted by Beard

Stuck on this, don't know where to start

$2\cos^{2}x = 2\sin(x)\cos(x) + 1,\ for\ 0^{\circ} \leq x < 360^{\circ}$

Sorry, I thought you were trying to prove an identity!

**Chop Suey** · February 25th 2009, 06:43 AM

Originally Posted by Beard

Stuck on this, don't know where to start

$2\cos^{2}x = 2\sin(x)\cos(x) + 1,\ for\ 0^{\circ} \leq x < 360^{\circ}$

I'll start you off:

$2\cos^{2}x = 2\sin(x)\cos(x) + 1 \implies 2\cos^{2}x -1 = 2\sin(x)\cos(x) \implies \cos{2x} = \sin{2x}$

When does the cosine of an angle equals the sine of an angle?

**ADARSH** · February 25th 2009, 06:44 AM

Originally Posted by Beard

Stuck on this, don't know where to start

$2\cos^{2}x = 2\sin(x)\cos(x) + 1,\ for\ 0^{\circ} \leq x < 360^{\circ}$

First of all this is not an identity
Put x= 0
In case you need to solve the equation

Put $1 = sin^2(x) + cos^2(x)$

EDIT: Far too late

**Beard** · February 25th 2009, 06:45 AM

at 45

**Chop Suey** · February 25th 2009, 06:47 AM

I kind of automatically assumed that Beard mistitled this thread and meant it as a solve the trigonometric equation.

**ADARSH** · February 25th 2009, 06:53 AM

Read the initial lines
This is an equation which has solutions

and not identity

**HallsofIvy** · February 25th 2009, 06:53 AM

Originally Posted by Beard

at 45

Sine and cosine are equal for 45 degrees and at 180+ 45= 225 degrees.

That, of course, would be 2x. Now what values does x have?

**Beard** · February 25th 2009, 07:44 AM

Originally Posted by Chop Suey

I kind of automatically assumed that Beard mistitled this thread and meant it as a solve the trigonometric equation.

Sorry about this

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