# Thread: Double Angle Question W/Picture.

1. ## Double Angle Question W/Picture.

Double Angle Question W/Picture. Attached

2. ## Re: Double Angle Question W/Picture.

Use the Pythagorean Identity to simplify the double angle identity you've been given so far...

\displaystyle \displaystyle \begin{align*} \sin^2{(x)} + \cos^2{(x)} \equiv 1 \end{align*}, so that means

\displaystyle \displaystyle \begin{align*} \cos{(2x)} &\equiv \cos^2{(x)} - \sin^2{(x)} \\ &\equiv 2\cos^2{(x)} - 1 \\ &\equiv 1 - 2\sin^2{(x)} \end{align*}

3. ## Re: Double Angle Question W/Picture.

Originally Posted by Melcarthus
Double Angle Question W/Picture. Attached
Do you understand that $\displaystyle \sin(2\alpha)=\frac{2}{5}~?$

So $\displaystyle 2\sin(\alpha)\cos(\alpha)=\frac{2}{5}~.$

4. ## Re: Double Angle Question W/Picture.

Actually I think you'll find \displaystyle \displaystyle \begin{align*} \sin{(2\alpha)} = \frac{4}{5} \end{align*}, not \displaystyle \displaystyle \begin{align*} \frac{2}{5} \end{align*}.

5. ## Re: Double Angle Question W/Picture.

i think Prove It is just right.