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$$\cos\theta \cos3\theta -\sin\theta \sin3\theta =0$$

I know that this uses the cosine addition formula and could be rewritten as $\cos(\theta +3\theta )=0$

and then as $\cos(4\theta )=0$ However , I don't know where to go from here. I'd like to be guided so that I can possibly arrive at the correct answer by myself.

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$$\cos4\theta=0\implies4\theta=(2n+1)\frac\pi2,\theta=(2n+1)\frac\pi8$$ where $n$ is any integer

We need $0\le(2n+1)\dfrac\pi8<2\pi \implies0\le2n+1<16\implies n=? $

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  • $\begingroup$ $cos(4\theta )=0$, so $4\theta =\frac { \pi }{ 2 } +2\pi k$ So now I divide it by $4$ to get $\theta $ itself. This comes out to $4\theta =\frac { \pi }{ 8 } +\frac { 2\pi k }{ 4 } $ And now I just add right before I reach a full rotation? $\endgroup$ – Cherry_Developer Dec 9 '14 at 10:41
  • $\begingroup$ yep, starting with k=0, then k=1, etc $\endgroup$ – Alan Dec 9 '14 at 10:41
  • $\begingroup$ When I add it the first time, I already reach $\frac { 5\pi }{ 8 } $, yet $\frac { 3\pi }{ 8 }$ should be a solution $\endgroup$ – Cherry_Developer Dec 9 '14 at 10:45
  • $\begingroup$ @Alan, $2n<15\implies n<7.5\implies n\le7$ $\endgroup$ – lab bhattacharjee Dec 9 '14 at 10:45
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    $\begingroup$ @Cherry_Developer $$\cos(4\theta) = 0 \Rightarrow 4\theta = \frac{\pi}{2} + k\pi, k \in \mathbb{Z}$$ so you should obtain $$\theta = \frac{\pi}{8} + k\frac{\pi}{4}$$ which is equivalent to the correct answer provided by lab bhattacharjee. $\endgroup$ – N. F. Taussig Dec 9 '14 at 12:32

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