Lesson Video: Evaluating Trigonometric Ratios given the Value of Another Ratio Mathematics

In this video, we will learn how to find the value of a trigonometric function from a given value of another trigonometric function.

17:50

Video Transcript

In this lesson, we will learn how to find the value of a trigonometric function from a given value of another trigonometric function.

Before we get started, let’s think about some information we already should know. The trigonometric functions can be defined in terms of particular ratios of sides of right triangles. Let’s consider the angle at vertex 𝐴. We have the three primary relationships sin, cos, and tan of this angle. The sine relationship is the opposite side length over the hypotenuse. The cosine relationship is the adjacent side length over the hypotenuse. And the tangent relationship is the opposite side length over the adjacent side length.

In the triangle we’ve drawn here, if we use the angle 𝜃 at vertex 𝐴, the opposite side length will be lower case 𝑎. And the hypotenuse is always the side length opposite the right angle. In our case, that will be the lowercase 𝑐. To find the cosine relationship of this triangle, the adjacent side length is lowercase 𝑏 and the hypotenuse is still 𝑐, which makes the tangent relationship 𝑎 over 𝑏.

In addition to these three primary trig functions, we have the inverses cosecant, secant, and cotangent. The cosecant is the inverse of sine, and it is the hypotenuse over the opposite. Here, we would have 𝑐 over 𝑎. The secant is the inverse of the cosine, the hypotenuse over the adjacent. In this case, we would have 𝑐 over 𝑏. The cotangent relationship is the inverse of the tangent relationship. It’s the adjacent side length over the opposite side length. In this case, it’s 𝑏 over 𝑎.

Many people remember the sine, the cosine, and the tangent relationships with the phrase SOH CAH TOA. Sine is the opposite over the hypotenuse. Cosine is the adjacent over the hypotenuse. And tangent is the opposite over the adjacent.

When we’re dealing with right-triangle trigonometry, all of the angles are less than 90 degrees. And when all the angles are less than 90 degrees, all six of these ratios are positive. But sometimes, we’re dealing with angles that are larger than 90 degrees. And when that happens, some of these relationships end up being negative. Let’s remind ourselves what happens with these larger angles.

We can think about this in terms of a coordinate grid. Sometimes with these angles, we use degrees and sometimes we use radians. And when we measure an angle like this, we start at the right side of the 𝑥-axis and measure until we get to the line we’re interested in. The angle I’ve drawn here will fall somewhere between 90 degrees and 180 degrees. If we knew that this angle was 135 degrees, we could use a calculator to find the sin, cos, and tan of this angle. Our calculator will give us a value of positive 0.07 continuing for sin, negative 0.707 continuing for cos, and negative one for tan.

This pattern — a positive sine value, a negative cosine value, and a negative tangent value — will be true for all angles that fall between 90 and 180 degrees. If we drew a right-angled triangle here, the hypotenuse would have a length of one and each of the sides would measure the square root of two over two. We know this because this angle must be 45 degrees.

What this shows is that we can still use right-triangle trigonometry. We can still use side lengths to form ratios to calculate angles that are greater than 90. In order to do this, we memorize the positive and negative trig relationships on the coordinate grid. And for that, we use the CAST diagram, which looks like this.

If we start in the first quadrant with the A, it’s telling us that all the relationships are positive, which we’ve already discussed. Sine, cosine, and tangent relationships from zero to 90 degrees are all positive. Moving to the second quadrant, the S tells us that only the sine is positive. Any angles falling between 90 and 180 degrees will have a positive sine relationship and a negative cosine and tangent relationship.

In the third quadrant, the T tells us that the tangent relationship is positive. That means in the third quadrant, the sine relationship and the cosine relationship will be negative. And finally, in the fourth quadrant, the C tells us the cosine relationship is positive. The cosine relationship is positive, and the sine and tangent relationships of any angle in the fourth quadrant will be negative.

Knowing the six trig functions and the CAST diagram are the main tools we’ll use in solving the example problems. Let’s look at one now.

Find the csc of 𝜃 given the tan of 𝜃 equals 24 over seven and that the cos of 𝜃 is less than zero.

We’re given the tan of an angle is 24 over seven. And we need to find the csc of that same angle under the condition that the cos of 𝜃 is less than zero. That means our cos is negative and our tan is positive. This is really important information. Before we go any further, though, we remind ourselves of the sine, cosine, and tangent relationships and their reciprocals cosecant, secant, and cotangent. In addition to that, we can sketch a CAST diagram to help us determine which quadrant the angle we’re interested in will fall in.

Since we know that the tangent is positive, we can say that this will either fall in the first or third quadrant. However, because the cosine is negative, this angle cannot fall in the first quadrant. In the first quadrant, we have an 𝐴 because all three of the relationships are positive. In the third quadrant, only the tangent is positive and the sine and cosine are negative. And so we can say that our angle will fall in the third quadrant and sketch something like this. For our angle 𝜃, since we know the tan is 24 over seven and the tangent is the opposite over adjacent side lengths, we know two of the three sides.

What we want to do is sketch a right-angled triangle in the third quadrant. We’ll let this be our angle 𝜃. And then we can label the opposite side length 24 and the adjacent side length seven. Obviously, we haven’t drawn this completely to scale. In order to find the csc of 𝜃, we need the hypotenuse over the opposite. And at this point, we should have enough information to recognize to do that, we’re going to need to solve for the Pythagorean theorem.

If we let the hypotenuse be equal to 𝑐, we know that 𝑎 squared plus 𝑏 squared equals 𝑐 squared, which will give us 49 plus 576. 𝑐 squared is 625. Taking the square root of both sides, we find that 𝑐 equals 25. Now that we know the hypotenuse, we can write 25 over the opposite, 24. But this is not the final answer. We need to think carefully about whether this is a positive or negative solution.

Since it falls in the third quadrant, both the sine and the cosine values will be negative. And since the cosecant value is the reciprocal of the sine value, it will have the same sign as the sine function. The csc of this angle must be negative. Under these conditions, the csc of 𝜃 equals negative 25 over 24.

Let’s look at another example.

Given that the csc of 𝜃 is negative seven over six and tan of 𝜃 is greater than zero, find the cos of 𝜃.

Let’s think about the information we’re given. We know that the csc of this angle is negative seven over six and we know that the tan is positive. We want to remember the six trig functions. Since we were given the csc of 𝜃, we were given the hypotenuse over the opposite. And since we’re looking for the cos of this angle, we’re looking for the adjacent side length over the hypotenuse. But there’s one other bit of information we need to consider. And that is the location of this angle on a coordinate grid.

We know that the tangent is positive. And the tan of an angle is only positive in the first and third quadrants. But we also know that the cosecant is negative. And if the cosecant is negative, sine will be negative. In the first quadrant, all three of the relationships are positive. In the third quadrant, the tangent is positive, but both the sine and cosine relationships are negative. And that means we know that our relationship, our cosine relationship, must be negative.

So far, we know the hypotenuse is seven and the opposite side length is six. In order to solve our problem, we need to know the adjacent side length. If we think about this in terms of a right-angled triangle, the measures of the side lengths must be positive because distance is always measured with positive values. And that means we’ll need to use the absolute value for the hypotenuse. Instead of negative seven, it will be positive seven. And the opposite side length will be six.

We know we could use the Pythagorean theorem to solve for that third missing side length, the adjacent side length. So we say six squared plus 𝑏 squared equals seven squared. 36 plus 𝑏 squared equals 49. Subtracting 36 from each side, we get 𝑏 squared equals 13. From there, we take the square root of both sides to find our missing side length to be the square root of 13. Our cosine relationship is the adjacent side length, the square root of 13, over the hypotenuse, which we already know is seven. And because this falls in the third quadrant, this cosine relationship must be negative. And that makes our final answer negative square root of 13 over seven.

In this example, we’re given a range for the radian measure of the angle we’re using.

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

Before we do anything else, it will be helpful to identify the place where this 𝜃 must fall. It’s falling between 𝜋 over two and 𝜋. So we sketch a coordinate grid and label it with the radian measures. If 𝜃 falls between 𝜋 over two and 𝜋, the 𝜃 will fall somewhere in the second quadrant. If we sketch a line and our angle 𝜃, 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 𝜃. To do that, we’ll multiply sec of 𝜃 by sec of 𝜃.

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 𝜃 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 𝜃 is 13 over nine.

In our final example, we’re again given a range of values that our 𝜃 can be. In addition to that, we have a sine value, and we need to find the csc of two times that 𝜃 value.

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 eight.

The key points here are knowing the six trig functions and using the CAST diagram to identify their location on a coordinate grid.

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