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Basically I'm trying to find the exact value of $\tan{\theta}$ when $\theta = \arctan{(8/3)}$.

I'm not exactly sure where to start. I know that $\arctan$ is the inverse of $\tan$, but I can't really figure out how to do the inverse of this one.

I also have to find $\sin{\theta}$, but I feel that will be easier once I find $\tan{\theta}$.

Thanks in advance.

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    $\begingroup$ $\tan(\arctan x)=x$, of course… $\endgroup$
    – Bernard
    Jun 13, 2016 at 1:09
  • $\begingroup$ To get $ \ \sin \theta \ $ , construct a right triangle with an angle $ \ \theta \ $ having an "opposite side" of $ \ 8 \ $ and an "adjacent side" of $ \ 3 \ $ : find the hypotenuse and you can get any of the trig values for $ \ \theta \ $ that you may want. $\endgroup$ Jun 13, 2016 at 1:11
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    $\begingroup$ Do tell your quadrant, or all will come to grief. $\endgroup$ Jun 13, 2016 at 1:14

1 Answer 1

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$$\tan(\theta) = \tan(\arctan{(8/3)} ) = 8/3.$$

$$\sin(\theta) = \pm\frac{\tan(\theta)}{\sqrt{1+\tan^2(\theta)}} = \pm \frac{8}{\sqrt{3^3+8^2}} = \pm \frac{8}{\sqrt{73}}.$$

Quadrant information will tell you whether it is +ve or -ve.

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  • $\begingroup$ Thanks, one more question..... what about cosθ. Where are you getting these equations from? $\endgroup$
    – fruit
    Jun 13, 2016 at 2:11
  • $\begingroup$ dummies.com/how-to/content/… . For $\cos(\theta)$, you will have $\pm \sqrt{1-\sin^2(\theta)} = \pm \frac{3}{\sqrt{73}}$. $\endgroup$
    – Vaneet
    Jun 13, 2016 at 2:24

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