1. ## Double Angle Identity

Stuck on this, don't know where to start

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

2. 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!

3. 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?

4. 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

5. at 45

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

This is an equation which has solutions and not identity

8. 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?

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