Double Angle Question W/Picture. Attached

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- Jul 2nd 2013, 01:24 PMMelcarthusDouble Angle Question W/Picture.
Double Angle Question W/Picture. Attached

- Jul 2nd 2013, 05:23 PMProve ItRe: 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

$\displaystyle \displaystyle \begin{align*} \cos{(2x)} &\equiv \cos^2{(x)} - \sin^2{(x)} \\ &\equiv 2\cos^2{(x)} - 1 \\ &\equiv 1 - 2\sin^2{(x)} \end{align*}$ - Jul 2nd 2013, 05:55 PMPlatoRe: Double Angle Question W/Picture.
- Jul 2nd 2013, 06:14 PMProve ItRe: Double Angle Question W/Picture.
Actually I think you'll find $\displaystyle \displaystyle \begin{align*} \sin{(2\alpha)} = \frac{4}{5} \end{align*}$, not $\displaystyle \displaystyle \begin{align*} \frac{2}{5} \end{align*}$.

- Jul 3rd 2013, 02:18 AMibduttRe: Double Angle Question W/Picture.
i think Prove It is just right.

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