Question

Question #4492c

Answer 1

Use the identity `csc(theta) = 1/sin(theta)`
Therefore, `-cos(theta)=-1/sin(theta)`

Explanation:

You know that `sin(pi/3) = sqrt3/2`

`-csc(pi/3) = -1/sin(pi/3)`

`-csc(pi/3) = -1/(sqrt3/2)`

We know that when we divide by a fraction we can flip and multiply:

`-csc(pi/3) = -2/sqrt3`

We cannot leave the radical in the numerator so we multiply by `sqrt3/sqrt3`:

`-csc(pi/3) = (-2sqrt3)/3`

That same thing with `pi/6`

You know that `sin(pi/6) = 1/2`

`-csc(pi/6) = -1/sin(pi/6)`

`-csc(pi/3) = -1/(1/2)`

We know that when we divide by a fraction we can flip and multiply:

`-csc(pi/3) = -2`

Answer 2

Your HW is correct. Even though you don't need a unit circle, solving is easier with one. You could solve the equation with a calculator, but you need to remember to put it in radian mode.

Explanation:

`csc = 1/sin`, right?

So the question is: `-1/sin(pi/3)=?`

If you look on a unit circle, `sin(pi/3) = sqrt3/2`.

So when you put that into the original equation, you get:

`-1/(sqrt3/2) = -2/sqrt3`

Now you need to rationalize it, so you end up with:

`-2/sqrt3 * sqrt3/sqrt3 = -2sqrt3/3`