Question #4492c

2 Answers
Apr 20, 2017

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

Explanation:

You know that sin(pi/3) = sqrt3/2sin(π3)=32

-csc(pi/3) = -1/sin(pi/3)csc(π3)=1sin(π3)

-csc(pi/3) = -1/(sqrt3/2)csc(π3)=132

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

-csc(pi/3) = -2/sqrt3csc(π3)=23

We cannot leave the radical in the numerator so we multiply by sqrt3/sqrt333:

-csc(pi/3) = (-2sqrt3)/3csc(π3)=233

That same thing with pi/6π6

You know that sin(pi/6) = 1/2sin(π6)=12

-csc(pi/6) = -1/sin(pi/6)csc(π6)=1sin(π6)

-csc(pi/3) = -1/(1/2)csc(π3)=112

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

-csc(pi/3) = -2csc(π3)=2

Apr 20, 2017

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/sincsc=1sin, right?

So the question is: -1/sin(pi/3)=?1sin(π3)=?

If you look on a unit circle, sin(pi/3) = sqrt3/2sin(π3)=32.

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

-1/(sqrt3/2) = -2/sqrt3132=23

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

-2/sqrt3 * sqrt3/sqrt3 = -2sqrt3/32333=233