Question

How do you sketch the angle whose terminal side in standard position passes through (6,8) 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 `(6,8)`. So to locate the point `(6,8)`, starting at the origin and count `6` to the right and up `8`. That will get you to the point `(6,8)`. Put a dot there.
Now draw a line from the origin through the point `(6,8)`. 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) = (6,8)`, so

`r = sqrt((6)^2+(8)^2) = sqrt(36+64) = sqrt100=10`.

So the sine of `theta` is

`sin(theta) = 8/10`

When we think we have finished, we should make sure that fracions are reduced.

`8/10 = (2*4)/(2*5) = (cancel(2) * 4)/(cancel(2) * 5) = 4/5`

Our answer is:

`sin theta = 4/5`

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

`cos(theta) = 6/10 = (2*3)/(2*5) = (cancel(2) * 3)/(cancel(2) * 5) = 3/5`

Our answer is
`cos(theta) = 3/5`