Question

How do you sketch the angle whose terminal side in standard position passes through (-3,1) and how do you find sin and cos?

Answer 1

Please see below.

Explanation:

To sketch the angle in standard position,

One side of the angle is the positive `x` axis (the right side of the horizontal axis). That is the initial side.

The terminal side has one end at the origin (the point `(0,0)`, also the intersection of the two axes) and goes through the point `(-3,1)`. So locate the point `(-3,1)`. Starting at the origin and count `3` to the left and up `1`. That will get you to the point `(-3,1)`. Put a dot there.
Now draw a line from the origin through the point `(-3,1)`. Your sketch should look a lot like this:

If we knew how the angle was made (which direction and how many times around the circle), we would show that also.

Give the angle a name. I will use `theta` (that is the Greek letter "theta")

Memorize this

If the point `(a,b)` lies on the terminal side of an angle in standard position, then let `r = sqrt(a^2+b^2)`

The angle has sine `b/r` and it has cosine `a/r`

For this question

We have `(a,b) = (-3,1)`, so

`r = sqrt((-3)^2+(1)^2) = sqrt(9+1) = sqrt10`.

So the sine of `theta` is

`sin(theta) = 1/sqrt10`

Many trigonometry teachers will insist that you write your answer with a rational number in the denominator. If this is something you have to do, multiply the answer by `sqrt10/sqrt10` to look like this:

`1/sqrt10 * sqrt10/sqrt10 = (1*sqrt10)/(sqrt10 * sqrt10) = sqrt10/10`

So we should answer

`sin(theta) = sqrt10/10`

The cosine of `theta` is `a/r` so we have

`cos(theta) = (-3)/sqrt10 = (-3)/sqrt10 * sqrt10/sqrt10 = -(3sqrt10)/10`

Our answer is
`cos(theta) = -(3sqrt10)/10`