How to prove: $1-\cos 2\alpha = 2\sin^2\alpha$

Question

I am taking a preperation course for college.

I got this as homework.

Prove that:

$1-\cos2\alpha = 2\sin^2\alpha$

Thank you!

Answer 1

Hint: $$1 - \cos(2\alpha) = 1 -(\cos^2 \alpha - \sin^2 (\alpha)) $$

Now use the fact that $\cos^2 \alpha = 1-\sin^2 \alpha$, and be careful with the signs of the terms.

Answer 2

Let me describe an alternative proof that uses only some Euclidean geometry plus the definitions of cos and sin --- no double-angle formulas or summation formulas. I realize that this should be done with a figure, which I can't draw on a computer, so I'll describe it in words and hope you can draw the picture.

Start with a circle of radius 1, and call its center C. Let AB be a horizontal diameter, with A at the left end and B at the right. From C, draw a ray upward and to the right, making an angle $2\alpha$ with the horizontal ray CB. Let D be the point where this new ray meets the circle, and let E be the point on CB that lies directly below D. Also, draw the line segments AD and BD. The first fact I need from elementary geometry is that the angle CAD is $\alpha$. (For a proof, if you don't already know this, notice that ACD is an isosceles triangle, so its angles at A and D are equal, and their sum is the external angle $2\alpha$ at C.)

Looking at the right triangle CED whose hypotenuse CD is 1, we see that the length of CE is $\cos 2\alpha$. Therefore, the length of BE is $1-\cos 2\alpha$, the left side of your equation.

Now look at triangle ADB. The second fact I need from elementary geometry is that it has a right angle at D, because this is an angle inscribed in a semicircle. Since this right triangle has hypotenuse AB of length $2$, we see that the length of BD is $2\sin\alpha$.

Finally, notice that BED is a right triangle, whose hypotenuse we just computed as $2\sin\alpha$. Its angle at D is $\alpha$ (because the sides of this angle, BD and ED, are perpendicular to AD and AB, respectively). Therefore, the side BE opposite this angle $\alpha$ is $2\sin^2\alpha$, which is the right side of your equation.