solutions of trigonometric equations..

ebaines · Jul 24, 2012, 10:11 AM

Remember that the sine and cosine functions have periods of $2 \pi$ , so that $\sin(\theta) = \sin (\theta \pm 2n \pi)$ and $\cos(\theta) = \cos(\theta \pm 2n \pi)$ . Any solution to an equation involving sines and/or cosines will have general solutions with $\pm 2 n \pi$ . Also recall that $\sin(\theta) = - \sin (\theta \pm \pi)$ , and similarly $\cos(\theta) = - \cos (\theta \pm \pi )$ , and hence the tangent function is periodic with period $\pi$ , since the negative signs cancel out:

$<br /> \tan \theta = \frac {\sin (\theta \pm n \pi)}{\cos (\theta \pm n \pi)} = \tan (\theta \pm n \pi)$ .

By the way, you wrote that the solution is

$\phi = 2 n \pi \pm \frac n 4 + \frac n {12}$

but I believe it's actually $\phi = \frac n 4 + \frac n {12} \pm 2 n \pi$

subhan2010 · Jul 24, 2012, 09:32 PM

thank you....!

solutions of trigonometric equations..

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