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I have a value of a tangent. Is it possible to find the sine and/or cossine from that value? How?

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  • $\begingroup$ Hint: take a look at the graph of the tangent. $\endgroup$
    – user2055
    Commented Jun 11, 2013 at 16:34
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    $\begingroup$ from any trig ratio you can find out other 5 trig ratio.every ratio contain two sides in it so just using pythagoreous theorem you can find out 3rd side and when you have all 3 sides of an right angle triangle you can derive any trig ratio $\endgroup$ Commented Jun 11, 2013 at 16:50
  • $\begingroup$ Several poeple are saying "yes", but omitting to mention the "$\pm$" issue. See my answer below. $\endgroup$ Commented Jun 11, 2013 at 18:04

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Yes you can. Try drawing a right angle triangle with the angle you're interested in (it doesn't need to be to scale). What side lengths could you use as the opposite and adjacent given your value of tan of the angle? Now you can you Pythagoras's Theorem to determine the remaining side length and hence you can find what sin and cos of the angle would be.

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Yes, but you must know the range of the angle:

$$1+\tan^2x=\frac1{\cos^2x}\;\ldots$$

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Not completely: There's a "$\pm$" issue.

We have the standard identity $\sec^2\theta=1+\tan^2\theta$.

So $\cos^2\theta=\dfrac{1}{1+\tan^2\theta}$, and so $\cos\theta=\pm\dfrac{1}{\sqrt{1+\tan^2\theta}}$.

Once you know $\cos^2\theta$, you have $\sin^2\theta= 1-\cos^2\theta$, and then $\sin\theta=\pm\sqrt{1-\cos^2\theta}$.

Notice that if $\theta$ is in the first quadrant, then $\pi+\theta$ is in the third quadrant and both of those angles have the same tangent, but for one of them the sine and cosine are positive, whereas for the other they're both negative. If $\theta$ is in the second quadrant, then $\theta$ and $\pi+\theta$ again have the same tangent but for $\theta$ the sine is positive and the cosine is negative, whereas for $\pi+\theta$ the sine is negative and the cosine is positive.

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Hint:$$1+\tan^2(x)=\frac{1}{\cos^2(x)}$$$$1+\cot^2(x)=\frac{1}{\sin^2(x)}$$$$\tan(x)=\frac{1}{\cot(x)}$$

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If $\tan\theta=\frac ab$

$$\frac{\sin\theta}a=\frac{\cos\theta}b=\pm\sqrt{\frac{\sin^2\theta+\cos^2\theta}{a^2+b^2}}=\pm\frac1{\sqrt{a^2+b^2}}$$

If $b=1,$ $$\tan\theta=a,\frac{\sin\theta}a=\frac{\cos\theta}1=\pm\frac1{\sqrt{a^2+1}}$$

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  • $\begingroup$ why you always choose tough way to solve simple problems? $\endgroup$ Commented Jun 11, 2013 at 17:00
  • $\begingroup$ @iostream007, not sure why this method looked tough to you. Application of Right Angle is only valid for the angles in the first Quadrant, right? The application of this "Squaring & Adding" method in Trigonometry is often very useful as it directly gives you the values of both sin & cos with proper sign $\endgroup$ Commented Jun 11, 2013 at 18:36

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