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I have been practicing a lot of pre-calculus equations, and currently, I am studying "Trigonometry"

I have stumbled across this following question:

  • If $\tan(2\theta) = \frac{4}{3}$, $\frac{\pi}{2} < \theta < \pi$, Find $\tan\theta$.

I have first got the formula of $\tan(2\theta)$ and equaled it to $\frac{4}{3}$, then I got solving until I was left with:

$$(2\tan\theta - 1)(\tan\theta + 2) = 0$$

So that means that $\tan\theta = \frac{1}{2}$, $\tan\theta = -2$. That's where I was stuck, so I checked for answers and said that $\frac{1}{2}$ was rejected? What does it mean? How do I know if something is rejected or not?

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  • $\begingroup$ You're missing another case. In particular, tangent is negative in the given range. $\endgroup$ – user223391 Oct 23 '17 at 20:48
  • $\begingroup$ What case? please be more specific. $\endgroup$ – Khalid Oct 23 '17 at 20:51
  • $\begingroup$ You did your algebra correct, but you got to incalculate the given domain, as indicated in the answer. $\endgroup$ – imranfat Oct 23 '17 at 20:59
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You're given that $\frac{\pi}{2}<\theta <\pi$.

$tan(\theta )$ is always negative in the given domain, and $\frac{1}{2}$ is a positive number. Therefore, $\frac{1}{2}$ is not an acceptable answer, and must be rejected.

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  • $\begingroup$ Thank you @高田航, this helped a lot! $\endgroup$ – Khalid Oct 23 '17 at 21:11
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You are told that $\pi/2<\theta<\pi$. For these angles $\tan(\theta)<0$ so $\tan\theta=1/2$ is not a valid solution See https://www.google.com/search?channel=fs&q=plot+tan%28x%29&ie=utf-8&oe=utf-8

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  • $\begingroup$ I am sorry Andrei, I am still confused. How can I prove that it is not a valid solution though? Sorry but this is new to me and I have been trying to understand how for quite some time $\endgroup$ – Khalid Oct 23 '17 at 20:59
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    $\begingroup$ You have two solutions for your equation, but only one falls in the required range. For example, if I have an equation like $x^2=1$ and I ask for the negative solution, $x=1$ is a solution for your equation, but it's not negative. Similarly here, for angles between $/pi/2$ and $/pi$, your $tan\theta$ must be negative. $\endgroup$ – Andrei Oct 23 '17 at 21:03
  • $\begingroup$ Oh, I get it now. This confused me. Thank you! $\endgroup$ – Khalid Oct 23 '17 at 21:11

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