# Lesson Video: Evaluating Trigonometric Functions Using Cofunction Identities Mathematics

In this video, we will learn how to use cofunction and odd/even identities to find the values of trigonometric functions.

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### Video Transcript

In this video, we will learn how to use cofunction and even odd identities to find the values of trigonometric functions. Trigonometric functions have many different properties and identities that help us simplify and solve equations. For this lesson, we want to review cofunction identities, even odd identities, and then use them to solve some problems.

First, let’s consider the cofunction identities, which tell us sin of 90 degrees minus 𝜃 equals the cos of 𝜃. cos of 90 degrees minus 𝜃 equals the sin of 𝜃. And the tan of 90 degrees minus 𝜃 equals the cot of 𝜃. We should also remember that if we’re working in radians, we’ll substitute 𝜋 over two in place of 90 degrees.

We can show these cofunction identities graphically using the unit circle. If we have an angle 𝜃 in the unit circle, we can say that it forms a right triangle with a hypotenuse of one and side lengths 𝑎 and 𝑏. We know the sin of 𝜃 will be equal to the opposite over the hypotenuse, which in this case would be 𝑎 over one. So sin of 𝜃 equals 𝑎. The cos of 𝜃 is the adjacent side length over the hypotenuse, 𝑏 over one, which is just 𝑏. And the tan of 𝜃 is the opposite over the adjacent side length, 𝑎 over 𝑏. And we want to explore a relationship between 𝜃 and 90 minus 𝜃.

In the unit circle on the coordinate grid, the space between our angle and 90 degrees can be written as 90 minus 𝜃. And so we can create a second right triangle in quadrant one, which will again have a hypotenuse of one and side lengths of 𝑎 and 𝑏. Looking at our new triangle in yellow, if we take the sin of 90 minus 𝜃, that’s the opposite over the hypotenuse, which in this case will be 𝑏 over one. Therefore, sin of 90 minus 𝜃 equals 𝑏. And the cos of 90 minus 𝜃 is the adjacent over the hypotenuse. In this case, that’s 𝑎. And the tan of 90 degrees minus 𝜃 will then be 𝑏 over 𝑎, and so we’ve demonstrated these cofunction pairs. sin of 𝜃 and cos of 90 degrees minus 𝜃 both equal 𝑎 here, which is what we should expect based on our cofunction identity.

We’ve also demonstrated that both cos of 𝜃 and sin of 90 degrees minus 𝜃 equal 𝑏. That is to say that sin of 90 degrees minus 𝜃 is equal to cos of 𝜃. The tangent pair, the tangent cofunction identity, is a little bit different. Notice that the tan of 90 degrees minus 𝜃 equals 𝑏 over 𝑎. But the tan of 𝜃 equals 𝑎 over 𝑏. If we take the reciprocal of the tan of 𝜃, that is, one over tan of 𝜃, it would be equal to 𝑏 over 𝑎. And then we recognize that the reciprocal of the tan of 𝜃 is equal to the cot of 𝜃 as the cot of 𝜃 is equal to the adjacent side length over the opposite side length, which demonstrates that third cofunction identity that the tan of 90 minus 𝜃 is equal to the cot of 𝜃.

Let’s now consider the odd and even trigonometric function identities. The odd and even identities are as follows. sin of negative 𝜃 is equal to negative sin of 𝜃, which makes sine an odd function. cos of negative 𝜃 is equal to cos of 𝜃, which makes cosine an even function. And the tan of negative 𝜃 is equal to the negative tan of 𝜃, which makes tangent an odd function. Again, we can demonstrate this with the unit circle. For an angle 𝜃, in the first quadrant of the unit circle, we can create a right-angled triangle with side lengths 𝑎 and 𝑏, and then there would be some angle negative 𝜃. Remember this negative tells us we’re moving clockwise instead of counterclockwise, making our angle fall in quadrant four. And again we’ll see that this angle can create a right triangle with side lengths 𝑎 and 𝑏 and the hypotenuse of one.

In order to find the sin of negative 𝜃, we’ll use our CAST diagram which helps us identify the signs of the trigonometric ratios in the four quadrants. In quadrant four, only the cosine function will be positive. The sign in tangent functions will be negative. Since the sine relationship is the opposite side length over the hypotenuse, in quadrant four, this will be negative 𝑏. The cosine relationship is the adjacent side length over the hypotenuse, which is 𝑎 over one. And in quadrant four, the cosine relationship is positive. So the cos of negative 𝜃 equals positive 𝑎. Similarly, the tan of negative 𝜃 will be the opposite over the adjacent side length and in quadrant four will be negative, which gives us the tan of negative 𝜃 is negative 𝑏 over 𝑎.

If we want to find the sin of 𝜃, in our diagram, that’s in quadrant one, it will be equal to 𝑏 over one. The cos of 𝜃 is equal to 𝑎 over one, and the tan of 𝜃 is equal to 𝑏 over 𝑎. Notice that the cosine relationships the cos of negative 𝜃 is equal to 𝑎 and the cos of positive 𝜃 is equal to 𝑎. We recognize this as our even function and our even function identity. But what about the other two relationships, the odd functions? If we multiply sin of 𝜃 and 𝑏 by negative one, we can say that the negative sin of 𝜃 equals negative 𝑏. Similarly, if we multiply tan by negative one, we get negative tan of 𝜃. And then we need to multiply the other side of the equation by negative one, which will give us negative 𝑏 over 𝑎. And so we would have to say that sin of negative 𝜃 is equal to negative sin of 𝜃 and that the tan of negative 𝜃 is equal to the negative tan of 𝜃.

Let’s now look at an example where we need to use these identities to help us solve a trigonometric equation.

Find the value of cos of 90 degrees plus 𝜃 given the sin of 𝜃 equals three-fifths, where 𝜃 is between zero and 90 degrees.

We’re given the sin of the angle of 𝜃, and we need to find the cos of 90 degrees plus 𝜃. We could do this with a graph. However, a much more straightforward approach is by rewriting this expression using cofunction identities. In fact, using cofunction identities, there’s a few different ways that we could rewrite this. We’ll consider one of those ways. If we want to use this cofunction identity on the cos of 90 degrees plus 𝜃, we can rewrite that as the cos of 90 degrees minus negative 𝜃. And then using our cofunction identity, we can say that the cos of 90 degrees minus negative 𝜃 must be equal to the sin of negative 𝜃.

Thinking about the sin of negative 𝜃, we recognize that the sine function is an odd function. And the sin of negative 𝜃 is equal to the negative sin of 𝜃. And we know what the sin of 𝜃 is equal to. The sin of 𝜃 in this case is three-fifths, which means the sin of negative 𝜃 equals negative three-fifths. We began with cos of 90 degrees plus 𝜃. And using cofunction identities and odd function identities, we were able to find that it will be equal to negative three-fifths. Before we move on, we can show one graphical representation of this. If we draw a unit circle with a measure of angle 𝜃, we know that 𝜃 falls between zero and 90 degrees. And if the sin of 𝜃 equals three-fifths, then the opposite over the hypotenuse must be equal to three-fifths.

We want to know the cos of 90 degrees plus 𝜃. If we add 90 degrees to this angle, we have a radius of one because this is the unit circle. And because we’ve rotated this 90 degrees counterclockwise, the distance from our point to the 𝑦-axis is three-fifths. And when we’re working with the unit circle, the cos of our angle will be equal to the 𝑥-coordinate, which in our case will be negative three-fifths, and confirms that the cos of 90 degrees plus 𝜃 is equal to negative three-fifths.

In our next example, we’re combining multiple trig functions. This makes the graphical approach much more difficult, so we’ll need to use a combination of different cofunction identities to simplify.

Find the value of sin of 180 degrees minus 𝑥 plus tan of 360 degrees minus 𝑥 plus seven times sin of 270 degrees minus 𝑥 given that sin of 𝑥 equals three-fifths, where 𝑥 is between zero and 90 degrees.

We have an expression we want to simplify that has three terms. And we know that sin of 𝑥 equals three-fifths. To simplify the whole expression, we’ll consider each of these terms in turn, beginning with sin of 180 degrees minus 𝑥. One of our cofunction identities tells us that the sin of 90 degrees minus 𝜃 equals cos of 𝜃. Now, our sine term doesn’t look exactly like this, but we could rewrite it to be sin of 90 degrees plus 90 degrees minus 𝑥. We want something in the form 90 degrees minus 𝜃. And that means we can rearrange this by calling it 90 degrees minus 𝑥 minus 90 degrees, where we’ll let 𝜃 be equal to 𝑥 minus 90 degrees. So cos would be equal to 𝜃, 𝑥 minus 90 degrees. This looks very close to the cos of 90 degrees minus 𝜃. But in order to rearrange this part of the function, we need to consider the fact that the cosine function is an even function.

We know that the cos of negative 𝜃 is equal to the cos of 𝜃. This means the cos of 𝑥 minus 90 will be equal to the cos of the negative of 𝑥 minus 90, which is the cos of 90 degrees minus 𝑥. The cos of 90 degrees minus 𝜃 equals the sin of 𝜃. And so we can say that the sin of 180 degrees minus 𝑥 is equal to the sin of 𝑥. That simplifies our first term. For our second term, instead of using cofunction identities, we’ll recall the periodic property of the tangent, which tells us tan of 𝜃 plus or minus 180 degrees equals the tan of 𝜃. We want to rewrite this in a form that’s usable to us, so we can say it’s tan of 180 degrees plus 180 degrees minus 𝑥. If we let 𝜃 be equal to 180 degrees minus 𝑥, then we’re now adding 𝜃 plus 180 degrees inside this tangent, which will be equal to tan of 180 degrees minus 𝑥.

We want to use this periodic property a second time, so we’ll let 𝜃 be equal to negative 𝑥. We’re saying the tan of 180 degrees minus 𝑥 is equal to the tan of negative 𝑥 plus 180 degrees. And that periodic property lets us simplify that to tan of negative 𝑥. Because we know the tangent function is odd, so the tan of negative 𝑥 will be equal to the negative tan of 𝑥. Our expression now is the sin of 𝑥 minus the tan of 𝑥. And we need to simplify this third term. This time, we’ll use a periodic property of sin that the sin of 𝜃 plus or minus 360 degrees equals sin of 𝜃. If we let 𝜃 be equal to 270 minus 𝑥, we can subtract 360 degrees from the argument inside. 270 minus 360 is negative 90. The argument inside the sine becomes negative 90 degrees minus 𝑥. Because sine is an odd function, seven times sin of negative 90 degrees minus 𝑥 is equal to negative seven sin of 90 degrees plus 𝑥.

To use our cofunction identity here, we’ll rewrite that to be 90 minus negative 𝑥, which simplifies to the negative seven cos of negative 𝑥. And since cosine is an even function, negative seven times cos of negative 𝑥 will be equal to negative seven times cos of 𝑥. Our new expression is then sin of 𝑥 minus tan of 𝑥 minus seven times cos of 𝑥. And since we know that 𝑥 is an acute angle, we can solve for the other values using right-triangle trigonometry. Sketching a right triangle with an angle of 𝑥, we know that sin is three-fifths. That is the opposite over the hypotenuse. At this point, we recognize that this is a three-four-five right triangle.

Of course, we could use the Pythagorean theorem to solve for that missing angle. But because it falls in the ratio with the sine relationship of three to five, we know that that third side must be equal to four. sin of 𝑥 equals three-fifths minus the tan of 𝑥 — which will be the opposite over the adjacent, three-fourths — minus seven times the cos, which will be four over five, the adjacent side over the hypotenuse. And three-fifths minus three-fourths minus seven times four-fifths will be equal to negative twenty-three fourths.

Let’s consider one final example where we’ll use cofunction identities to help us evaluate some relationships inside a triangle.

𝐴𝐵𝐶 is a right triangle at 𝐵. Find the cot of 𝛼 given that the cot of 𝜃 is four-thirds.

Because we know that 𝐴𝐵𝐶 is a right triangle, we can also say that 𝐴𝐵𝐷 is a right triangle. And this means we can identify angle 𝐴𝐷𝐵 as 90 degrees minus 𝜃. It also means we can say that 𝛼 plus 90 degrees minus 𝜃 is equal to 180 degrees as we know that 𝐵𝐶 forms a straight line. And then if we subtract 90 degrees from both sides of this equation, we see that 90 degrees equals 𝛼 minus 𝜃. And adding 𝜃 to both sides tells us that 𝛼 equals 90 degrees plus 𝜃, which means the cot of 𝛼 is equal to the cot of 90 degrees plus 𝜃. And now it seems like we’re getting closer because we know cot in terms of 𝜃. Based on our cofunction identity, we know that the tan of 90 degrees minus 𝜃 equals the cot of 𝜃, so we want to rearrange the cot of 90 degrees plus 𝜃.

We can rewrite it to be the cot of 90 degrees minus negative 𝜃. And then, we’ll rewrite this in terms of tangent because the cotangent is the reciprocal of tangent. We can say that this is equal to one over tan of 90 degrees minus negative 𝜃 and that tan of 90 degrees minus negative 𝜃 simplifies to the cot of negative 𝜃. But now we have one over cot of negative 𝜃, which will be equal to the tan of negative 𝜃. And since tan of negative 𝜃 is equal to the negative tan of 𝜃, it’s an odd function, which means we simplify to the negative tan of 𝜃.

Going back to our diagram, if the cot of 𝜃 is four-thirds, 𝐴𝐵 equals four, 𝐵𝐷 equals three, the tan of 𝜃 is the opposite over the adjacent side length, which here is three-fourths. And we need the negative tangent, which will be negative three-fourths. We’ve shown that the cot of 𝛼 will be equal to the negative tan of 𝜃 and it’s negative three-fourths.

Before we finish, let’s quickly review a few key points. Cofunction identities and even odd identities facilitate simplification of trigonometric expressions. These identities can be combined with other trigonometric identities and properties to help us evaluate expressions. Here are the three cofunction identities we explored and the three even odd identities we explored.