Video: Evaluating a Trigonometric Function Using a Trigonometric Identity

Given that sin(πœƒ) = βˆ’1/3, where πœ‹ < πœƒ < 3πœ‹/2, evaluate csc(2πœƒ) without using a calculator. Hint: take csc(2πœƒ) = 1/(2sin(πœƒ)cos(πœƒ)).

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Video Transcript

Given that the sin of πœƒ is negative one-third, where πœƒ falls between πœ‹ and three πœ‹ over two, evaluate the csc of two πœƒ without using a calculator. Hint: take the csc of two πœƒ to be equal to one over two times sin of πœƒ times cos of πœƒ.

It’s always good to start with what we’re given. The sin of πœƒ is negative one-third, and πœƒ is between πœ‹ and three πœ‹ over two. If we think about the quadrants between πœ‹ and three πœ‹ 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 πœƒ.

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 πœƒ, we need two pieces of information. We need to know the sin of πœƒ and the cos of πœƒ. But we were already given the sin of πœƒ, which is negative one-third. And that means the only thing we need in order to solve this problem is the cos of πœƒ.

If we’re given the sin of πœƒ, how can we find the cos of πœƒ? 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 πœƒ 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 πœƒ 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 πœƒ. 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 eight.

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