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If $\tan \theta = \dfrac43$, find $\cos \theta$ and $\sin \theta$.

I have the answers, but I don't know what to do at all, is it going to be easy?

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    $\begingroup$ you have a triangle with 3 , 4 , 5. $\endgroup$
    – user147308
    Jun 9, 2014 at 8:14

6 Answers 6

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Geometric Approach

Make a (rough) sketch of a right triangle with one angle $\theta$. Then the opposite side and adjacent side are in the ratio $4:3$, as $\tan\theta = \dfrac{\text{Opposite side}}{\text{Adjacent side}}$. You can assume the sides to be exactly $4$ and $3$. Calculate the hypotenuse using Pythagorean theorem. Then calculate $\sin \theta$ and $\cos \theta$.

As 5xum points out, this is not very general. A more general approach is the algebraic one.


Algebraic Approach

$ \tan\theta = \dfrac{4}{3} \Rightarrow\\ \dfrac{\sin\theta}{\cos\theta} = \dfrac{4}{3} \Rightarrow\\ \cos\theta = \dfrac{3}{4}\sin\theta $

Now,
$ \cos^2\theta + \sin^2\theta = 1 \Rightarrow\\ \dfrac{9}{16}\sin^2\theta + \sin^2\theta = 1 \Rightarrow\\ \sin^2\theta = \dfrac{16}{25} \Rightarrow\\ \sin\theta = \pm \dfrac{4}{5} $

Then $\cos\theta = \dfrac{\sin\theta}{\tan\theta} = \pm\dfrac{3}{5}$ (with respective signs).

Thus $$\boxed{ \begin{matrix} \sin\theta = \dfrac{4}{5},\ \cos\theta = \dfrac{3}{5}\\ \text{or}\\ \sin\theta = -\dfrac{4}{5},\ \cos\theta = -\dfrac{3}{5} \end{matrix} } $$

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    $\begingroup$ This approach does not deal with the possibility that $\theta$ may not be in $[0,\pi/2]$ $\endgroup$
    – 5xum
    Jun 9, 2014 at 8:22
  • $\begingroup$ @5xum Fixed, I think? $\endgroup$
    – M. Vinay
    Jun 9, 2014 at 8:43
  • $\begingroup$ I agree. Just fix the last value of $\cos \theta$ to $-\frac35$. $\endgroup$
    – 5xum
    Jun 9, 2014 at 8:45
  • $\begingroup$ Damn sign. Thanks, @5xum! $\endgroup$
    – M. Vinay
    Jun 9, 2014 at 8:49
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Do you know that $$1+\tan^2\theta=\frac1{\cos^2\theta}$$ and $$\sin\theta=\tan\theta\cos\theta$$

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You use $$\sec^2 a=1+\tan^2 a$$ so you get value of $\sec a$, since $\cos a$ is reciprocal of $\sec a$, you can find $\cos a$. The rest is easy.

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  • $\begingroup$ @Fantini, thanks for your apt edit. $\endgroup$
    – SA-255525
    Jun 9, 2014 at 10:29
  • $\begingroup$ You are welcome. $\endgroup$ Jun 9, 2014 at 11:53
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I would try to visualize a right triangle. In other words draw a picture of a right triangle and label one of the angles $\theta$ and fill in the lengths of the sides that you know.

For this triangle we have:

 |\
4|  \
 |    \
 |_____ \ <- theta
   3

Now use the pythagorean theorem to find the final sides length and then finally calculate $\sin \theta, \cos \theta$.

EDIT: I've been corrected properly, you also need to consider a triangle with negative values on the legs, since $4/3=(-4)/(-3)$ i.e.

  |\
-4|  \
  |    \
  |_____ \ <- theta
    -3

and then go through the process as above to find the second set of possible values for $\sin \theta, \cos \theta$.

I'm going to go for a more visual approach to this, ignoring the triangle above. Consider a 2D cartesian graph, let's try to draw the possible triangles that the above condition represents. In order for $\tan \theta = \frac{4}{3}$ we either have a triangle in the first or third quadrant. Drawing these two triangles we either get

               /|
              / |
             /  |
            /   |
           /    | 4
          /     |
theta -> /______|
            3

or

         -3
   -------------- <- theta
   |            /
   |           /
-4 |          /
   |         /
   |        /
   |       /
   |      /
   |     /
   |    /
   |   /
   |  /
   | /
   |/

Now we can use the pythagorean theorem (taking the hypotenuse as positive since the slope here is clearly positive) to find the last sides of the triangles and from there find $\sin \theta, \cos \theta$.

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  • $\begingroup$ This approach does not deal with the possibility that $\theta$ may not be in $[0,\pi/2]$ $\endgroup$
    – 5xum
    Jun 9, 2014 at 8:23
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    $\begingroup$ @5xum for this problem it works, and it appears the OP wants help with problems similar to this one (i.e. where a geometrical approach is acceptable) hence I posted the solution I did (where $\theta \in (0,\pi/2)$. $\endgroup$
    – DanZimm
    Jun 9, 2014 at 8:30
  • $\begingroup$ I don't think it does work for this problem, as the solution $\cos \theta = -\frac35 \sin\theta = -\frac45$ cannot be found like this. $\endgroup$
    – 5xum
    Jun 9, 2014 at 8:41
  • $\begingroup$ @5xum are you saying because there is a possibility of having negative values for the cosine and sine? If so then you can be more precise with the geometric view and take the legs of the triangles to have both negative values as well and find the possible solutions from there as well. $\endgroup$
    – DanZimm
    Jun 9, 2014 at 8:42
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    $\begingroup$ I am sorry for the comment that insulted you, it was poorly worded. It is a poorly translated figure of speech from my own language. What I wanted to say is what I explained in the next comment. Anyway, I think we can safely say we agree to disagree on this topic. I still think your answer is wrong and that negative values of triangle lengths cause only confusion. Obviously, you disagree, but I suggest we leave it at that. $\endgroup$
    – 5xum
    Jun 9, 2014 at 9:10
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Since $\tan\theta = \frac{\sin\theta}{\cos\theta}$, you know that $$\frac{\sin\theta}{\cos\theta} = \frac{4}{3}.$$

Now, if we say that $\sin\theta$ and $\cos\theta$ are two variables, that means you have one equation for the two variables. Do you know of any other equation in which the two variables appear? (Hint: you do).

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Hint : $$ \tan\theta=\frac yx=\frac43 $$ then $$ \sin\theta=\frac y{\sqrt{x^2+y^2}} $$ and $$ \cos\theta=\frac x{\sqrt{x^2+y^2}}. $$

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  • $\begingroup$ These formulae annoy me, because they are the kind of formulae that students memorise before sitting an exam. They memorise these formulae as a substitution for understanding what is going on, or for getting to know one of the "standard" formula (e.g. $1+\tan^2\theta=\sec^2\theta$). $\endgroup$
    – user1729
    Jun 9, 2014 at 11:23

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