Given that the sin of 𝜃 is
negative one-third, where 𝜃 falls between 𝜋 and three 𝜋 over two, evaluate the
csc of two 𝜃 without using a calculator. Hint: take the csc of two 𝜃 to be
equal to one over two times sin of 𝜃 times cos of 𝜃.
It’s always good to start with what
we’re given. The sin of 𝜃 is negative
one-third, and 𝜃 is between 𝜋 and three 𝜋 over two. If we think about the quadrants
between 𝜋 and three 𝜋 over two, we’re dealing with an angle that falls in the
third quadrant. But we don’t know what angle it is,
so we can just sketch an angle 𝜃.
But before we go any further, it’s
also a good idea to think about what we’re trying to find. If we want to find the csc of two
𝜃, we need two pieces of information. We need to know the sin of 𝜃 and
the cos of 𝜃. But we were already given the sin
of 𝜃, which is negative one-third. And that means the only thing we
need in order to solve this problem is the cos of 𝜃.
If we’re given the sin of 𝜃, how
can we find the cos of 𝜃? Well, first, we think about these
relationships. The sine is equal to the opposite
over the hypotenuse, and the cosine is equal to the adjacent over the
hypotenuse. Since we have sin of 𝜃 equals
negative one-third, we can say the opposite side length is one and the hypotenuse
side length is three. We can sketch this right-angled
triangle on the coordinate grid where the opposite side length is one and the
hypotenuse side length is three.
When we look at this sketch, we
realize that it’s a bit off. Something like this is a bit more
to scale. We need to find that missing
side. So we use the Pythagorean
theorem. When we do that, we get 𝑏 equals
the square root of eight. And that can be simplified to 𝑏
equals two times the square root of two. Now that we know the adjacent side
length is two times the square root of two, we can write the cos of 𝜃 as negative
two times the square root of two over three.
We know that cosine will be
negative since our angle falls in the third quadrant. The CAST diagram tells us in the
third quadrant, the sine is negative, the cosine is negative, and the tangent is
positive. So we plug in our value for the cos
of 𝜃. We can rewrite it like this: one
divided by two times negative one-third times negative two times the square root of
two over three.
We want to multiply the three
numerators together, which will give us positive four times the square root of
two. And the denominators, three times
three is nine. To divide one by this value we
multiply by its reciprocal. And our final step is to
rationalize the denominator, which gives us nine times the square root of two over