Trigonometry $\tan \theta = \dfrac43$ 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?
 A: 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
$$\boxed{
\begin{matrix}
\sin\theta = \dfrac{4}{5},\ \cos\theta = \dfrac{3}{5}\\
\text{or}\\
\sin\theta = -\dfrac{4}{5},\ \cos\theta = -\dfrac{3}{5}
\end{matrix}
}
$$
A: Do you know that
$$1+\tan^2\theta=\frac1{\cos^2\theta}$$
and 
$$\sin\theta=\tan\theta\cos\theta$$
A: 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 $\sin \theta, \cos \theta$.
A: 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.    
A: 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).
A: Hint :
$$
\tan\theta=\frac yx=\frac43
$$
then
$$
\sin\theta=\frac y{\sqrt{x^2+y^2}}
$$
and
$$
\cos\theta=\frac x{\sqrt{x^2+y^2}}.
$$
