Trig Identity

**IHuntMath** · February 17th 2011, 12:56 AM

I am failing to realize how do you go from

$cos(2t)+sin(2t)$

to

$\sqrt{2}cos(2t-\frac{\pi}{4})$

I was never good at trig.

**Prove It** · February 17th 2011, 01:14 AM

$\displaystyle \sqrt{2}\cos{\left(2t - \frac{\pi}{4}\right)} = \sqrt{2}\left[\cos{(2t)}\cos{\left(\frac{\pi}{4}\right)} + \sin{(2t)}\sin{\left(\frac{\pi}{4}\right)}\right]$

$\displaystyle = \sqrt{2}\left[\frac{\cos{2t}}{\sqrt{2}} + \frac{\sin{2t}}{\sqrt{2}}\right]$

$\displaystyle = \cos{2t} + \sin{2t}$ .

**IHuntMath** · February 17th 2011, 01:19 AM

Very much appreciate I am going to have analyze this for a few minutes to understand what identities were used and etc. because that radian number pi/4 really threw me off.

**Prove It** · February 17th 2011, 01:21 AM

$\displaystyle \frac{\pi^C}{4} = 45^{\circ}$

and

$\displaystyle \cos{(A\pm B)} \equiv \cos{A}\cos{B} \mp \sin{A}\sin{B}$ .

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