quick question

**johett** · January 12th 2008, 01:43 PM

Determine an equivalent sine function for each cosine function:
y=-cos2(x-pi/12) y=-1/2cos1/4x

Thank you

**Soroban** · January 12th 2008, 02:46 PM

Hello, johett!

Use the identity: . $\sin^2\!\theta \:=\:\frac{1-\cos2\theta}{2}\quad\Rightarrow\quad\cos2\theta \:=\:1-2\sin^2\!\theta$

Determine an equivalent sine function for each cosine function:

$1)\;\;y\:=\:-\cos\left[2\left(x-\frac{\pi}{12}\right)\right]$

Use the identity: . $y \;=\;-\left[1 - 2\sin^2\!\left(x-\frac{\pi}{12}\right)\right] \;=\;2\sin^2\!\left(x-\frac{\pi}{12}\right) - 1$

$2)\;\;y\:=\:-\frac{1}{2}\cos\left(\frac{x}{4}\right)$

We have: . $y \;=\;-\frac{1}{2}\cos\left(2\cdot\frac{x}{8}\right)$

Use the identity: . $y \;=\;-\frac{1}{2}\left[1 - 2\sin^2\!\left(\frac{x}{8}\right)\right] \;=\;\sin^2\!\left(\frac{x}{8}\right) - \frac{1}{2}$

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