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This is question 1c of a list of related items.

State the value(s) of $\theta$ in the range $0^\circ$ to $360^\circ$ so that the following is true: $$\tan\theta = \tan 20^\circ$$

Here is the answer (from the list of answers):

$$\theta = 20^\circ;\quad \theta=180^\circ+20^\circ=200^\circ$$

I am using the trig identity for tan, the one where $$\tan\theta = \tan(\theta-180^\circ)$$ If $\theta = 20^\circ$ for the question, then $\tan(\theta-180^\circ)$ is $\tan(20^\circ-180^\circ)$, which is $\tan(-160^\circ)$, which is taking me to a completely different direction than the solution.

I would appreciate it if someone could explain the steps of using this trig identity to determine which other angles have the same tan ratio as $20^\circ$.

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  • $\begingroup$ Hi Bushra, welcome to MSE! As a general tip, it is better to typeset the question in full than to link to images. In answer to your question, you do indeed want to use that the $\tan$ graph repeats every 180 deg, but plugging $\theta = 20$ directly into that formula is going the wrong way. What if you set $\theta - 180 = 20$? $\endgroup$ – bounceback Aug 26 '20 at 18:53
  • $\begingroup$ In which two quadrants, do both $\sin$ and $\cos$ have the same sign? Why do you think you are away from the answer? $-\theta$ can be written as $2\pi - \theta$. So what will $-160^0$ be? $\endgroup$ – Math Lover Aug 26 '20 at 19:00
  • $\begingroup$ Hi @bounceback. Thank you for your help. Whilst I understood your comment and applied it to this question, I don't seem to understand how the solution was derived for question 1(f) using the same logic. It would be an incredible help if you explained the solution for this one. Thank you. $\endgroup$ – Bee Aug 26 '20 at 21:15
  • $\begingroup$ Use that $\tan$ is an odd function: $\tan(-x) = -\tan(x)$. Then argue as before $\endgroup$ – bounceback Aug 26 '20 at 23:15
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Hint:period of tangent function is $180^\circ, \tan 20^\circ=\tan(180+20)^\circ, \tan(20-180)^\circ =\tan(-160)^\circ, 200^\circ$ is anticlockwise rotation, $-160^\circ$ is clockwise rotation.

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  • $\begingroup$ Hi. Thank you for your comment. So is the period of the sine and cosine function 180 also? $\endgroup$ – Bee Aug 26 '20 at 19:15
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    $\begingroup$ Period of sine and cosine is 360, tangent and cotangent is180 $\endgroup$ – Lion Heart Aug 26 '20 at 19:17
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Since $\tan \theta = \tan (n\pi + \theta)$, where $n$ is any integer and $\theta=\dfrac{\pi}{9}$ in our case. So, we have $0 \leq n\pi+\dfrac{\pi}{9} \leq 2\pi \implies 0 \leq (9n+1)\pi \leq 18\pi$. Now you can clearly see that only $n=0$ and $n=1$ satisfy this bound, hence the only answers are $\theta=\dfrac{\pi}{9},\dfrac{10\pi}{9}$.

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$$\theta = 20^\circ;\quad \theta=20^\circ\pm n\cdot 180^\circ;$$

Other angles are

$$ =20^\circ,200^\circ,380^\circ,460^\circ $$

We can add or subtract $ n\cdot \pi$ ( $n$ positive or negative integer) to $ 20^\circ $ getting the tip of radius vector from first to third quadrant. The radius vector can be rotated indefinitely around the origin adding $ 180^\circ$.

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You suggest we use that $$ \tan \theta = \tan(\theta - 180^\circ). $$ Equivalently, if we set $\phi = \theta - 180^\circ$, the above equation reads $$ \tan(\phi+180^\circ) = \tan\phi. $$ To summarize: the tangent function is $180^\circ$-periodic: it repeats every $180^\circ$. So the possible answers are $\dots,-340^\circ,-160^\circ, 20^\circ, 200^\circ, 380^\circ, \dots$

Narasimham's answer shows how to express this in a single equation with $n$ varying, ABCD does the same in the language of radians ($180^\circ = \pi$ radians).

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