# Double Angle Question W/Picture.

• Jul 2nd 2013, 01:24 PM
Melcarthus
Double Angle Question W/Picture.
Double Angle Question W/Picture. Attached
• Jul 2nd 2013, 05:23 PM
Prove It
Re: Double Angle Question W/Picture.
Use the Pythagorean Identity to simplify the double angle identity you've been given so far...

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

\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*}
• Jul 2nd 2013, 05:55 PM
Plato
Re: Double Angle Question W/Picture.
Quote:

Originally Posted by Melcarthus
Double Angle Question W/Picture. Attached

Do you understand that $\sin(2\alpha)=\frac{2}{5}~?$

So $2\sin(\alpha)\cos(\alpha)=\frac{2}{5}~.$
• Jul 2nd 2013, 06:14 PM
Prove It
Re: Double Angle Question W/Picture.
Actually I think you'll find \displaystyle \begin{align*} \sin{(2\alpha)} = \frac{4}{5} \end{align*}, not \displaystyle \begin{align*} \frac{2}{5} \end{align*}.
• Jul 3rd 2013, 02:18 AM
ibdutt
Re: Double Angle Question W/Picture.
i think Prove It is just right.
Attachment 28718