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This may seem fairly straightforward, but I have been stuck on this for the past half-hour.

I need to use Double Angle Formulae such as the following:

  • $\sin2A ≡ 2\sin A \cos A$
  • $\cos2A ≡ \cos^2A - \sin^2A$
  • $\tan2A ≡ \frac{2\tan A}{1 - \tan^2A}$

and

  • $1 + \cos 2A ≡ 2\cos^2 A$
  • $1 - \cos 2A ≡ 2\sin^2A$

to solve this equation for all values of $\theta$ between $0^o < \theta <360^o$:

  • $3\tan\theta = 2\cos\theta$

I understand all of the identities above and how you get there, and I understand how to find the other values of $\theta$ between $0^o$ and $360^o$ once I have found one. I just get stuck solving this equation. Any help would be greatly appreciated.

Thanks!

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I'm not really sure why you need to use double angle formula (some exercise?), but I'd just do it in most straightforward way \begin{align} 3 \tan \theta &= 2 \cos \theta \\ 3 \sin \theta &= 2 \cos^2 \theta = 2 - 2\sin^2 \theta \end{align} $$ 2\sin^2 \theta + 3 \sin \theta - 2 = 0 \\ \sin \theta = \frac {-3 \pm \sqrt{9+16}}{4} = \frac {-3 \pm 5}{4} = \left [ \begin{array}{l} -2 \text{ (spurious solution)}\\ \frac 12 \text{ (correct solution)} \end{array}\right . $$ So, final answer is $\theta \in \left \{ \frac \pi 6, \frac {5 \pi}6\right \}$ or $\theta \in \left \{ 30^\circ, 150^\circ\right \}$ if operating with degrees.

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  • $\begingroup$ Thanks. Yes it was an exercise that required me to do it using double angle forumla $\endgroup$ – arch Oct 23 '13 at 17:42

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