# Double Angle Identity

• Feb 25th 2009, 06:33 AM
Beard
Double Angle Identity
Stuck on this, don't know where to start

$\displaystyle 2\cos^{2}x = 2\sin(x)\cos(x) + 1,\ for\ 0^{\circ} \leq x < 360^{\circ}$
• Feb 25th 2009, 06:41 AM
HallsofIvy
Quote:

Originally Posted by Beard
Stuck on this, don't know where to start

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

Sorry, I thought you were trying to prove an identity!
• Feb 25th 2009, 06:43 AM
Chop Suey
Quote:

Originally Posted by Beard
Stuck on this, don't know where to start

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

I'll start you off:

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

When does the cosine of an angle equals the sine of an angle?
• Feb 25th 2009, 06:44 AM
Quote:

Originally Posted by Beard
Stuck on this, don't know where to start

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

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

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

EDIT: Far too late :D
• Feb 25th 2009, 06:45 AM
Beard
at 45
• Feb 25th 2009, 06:47 AM
Chop Suey
I kind of automatically assumed that Beard mistitled this thread and meant it as a solve the trigonometric equation.
• Feb 25th 2009, 06:53 AM
This is an equation which has solutions (Nod) and not identity(Shake)
• Feb 25th 2009, 06:53 AM
HallsofIvy
Quote:

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?
• Feb 25th 2009, 07:44 AM
Beard
Quote:

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