Video: Evaluating a Trigonometric Function Using a Trigonometric Identity

Given that cot(πœƒ) = βˆ’3/2, where πœ‹/2 < πœƒ < πœ‹, evaluate secΒ²(πœƒ) without using a calculator.

02:50

Video Transcript

Given that cot of πœƒ equals negative three-halves, where πœ‹ over two is less than πœƒ which is less than πœ‹, evaluate sec squared πœƒ without using a calculator.

Before we do anything else, it will be helpful to identify the place where this πœƒ must fall. It’s falling between πœ‹ over two and πœ‹. So we sketch a coordinate grid and label it with the radian measures. If πœƒ falls between πœ‹ over two and πœ‹, the πœƒ will fall somewhere in the second quadrant. If we sketch a line and our angle πœƒ, we could then make a sketch of a right-angled triangle.

From there, we’ll need to remember our trig relationships. Since we know that cot of our angle is negative three over two, we know that the three represents the adjacent side length and the two represents the opposite side length. We can use that information to label this sketch. Our goal is to find sec squared πœƒ. To do that, we’ll multiply sec of πœƒ by sec of πœƒ.

That secant relationship is the hypotenuse over the adjacent side length. But in order to solve this problem, we’ll need to figure out what the hypotenuse is. We can use the Pythagorean theorem to do this. Two squared plus three squared equals the hypotenuse squared. That will be four plus nine, 13, equals the hypotenuse squared, which means the hypotenuse equals the square root of 13.

But we need to be really careful here. We have to consider, because our angle falls in this second quadrant, if the secant is going to be positive or negative. To do that, we can use a CAST diagram. We already know we’re interested in the second quadrant. And because there’s an S here, it tells us that the sine relationship is positive, but the cosine and tangent relationships will be negative.

secant is the inverse of cosine. And that means if the cosine value is negative, the secant value will be negative. Since the secant is the hypotenuse over the adjacent side length, we have the square root of 13 over three, but we know that this value must be negative. And that makes the sec of πœƒ the negative square root of 13 over three. sec squared will be negative square root of 13 over three times negative square root of 13 over three. The two negatives become positive. The square root of 13 multiplied by the square root of 13 equals 13. And three times three is nine. Under these conditions, sec squared πœƒ is 13 over nine.

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