Question

Let , 4 ,− 7 be a point on the terminal side of θ . find the exact values of cos θ , csc θ , and tan θ .

Answer 1

Ok, you are given a point and you need to find the exact values for $cos \theta csc \theta & tan \theta$

First thing first. We need to see if we are working with a unit circle and find the radius.

How to tell if we are working with a unit circle?
We know $x^2 + y^2 = r^2$ is a circle.

We know that to find the radius we can use the following formula:
$r^2 = \sqrt{x^2 + y^2}$

If $r^2 = \sqrt{x^2 + y^2} = 1$ we are working with a unit circle.

Lets see if it = 1.
$r^2 = \sqrt{4^2 + -7^2}$
$r^2 = \sqrt{16 + 49}$
$r^2 = \sqrt{65}$

Square both sides now
$\sqrt{r^2} = \sqrt{\sqrt{65}}$
$r = \pm 65^{\frac{1}{4}}}$

Since we squared, we have a + and a - but we disregard the - because we do not have - radii 
$r = 65^{\frac{1}{4}}}$
We can also say
$r = \sqrt[4]{65}$

Ok, since r does not equal 1, we are not working with a unit circle but we have found r, which is our radius.

Now that we know the value of r, which is $r = 65^{\frac{1}{4}}}$, we need to look at the identities of cos, csc and tan.


The identities:
$cos \theta = \frac{x}{r}$
$csc \theta = \frac{1}{y}$
$tan \theta = \frac{y}{x}$

Now that we know their identities and know the radius of our circle, we can find the exact values of cos, csc and tan.

$cos \theta = \frac{x}{r} = \frac{4}{65^{\frac{1}{4}}}$ 
$csc \theta = \frac{1}{y} = \frac{1}{-7}$
$tan \theta = \frac{y}{x} = \frac{-7}{4}$

The exact values for cos, csc and tan given the point (4,-7) are:
$cos \theta = \frac{4}{65^{\frac{1}{4}}}$ 
$csc \theta = \frac{1}{-7}$
$tan \theta = \frac{-7}{4}$