Lesson Explainer: Evaluating Trigonometric Functions Using Cofunction Identities Mathematics

Example 1: Finding the Value of a Trigonometric Function Using Cofunction Identities

Find the value of given , where .

Answer

In this question, we are told that the sine of angle is equal to and that . We need to use this information to evaluate . We could do this graphically by using the definitions of the sine and cosine functions; however, it is actually far easier to do this by rewriting this expression using the cofunction identities.

There are a lot of different ways of evaluating this expression using the cofunction identities; we will only demonstrate one of these.

First, recall that one of the cofunction identities tells us that, for any angle ,

We want to use this on the expression , so we need to find a way to subtract our angle from . One way of doing this is to recall that adding is the same as subtracting :

Then, we apply the cofunction identity, with :

Finally, we know that the sine function is an odd function, which means

Therefore, .

We can also see why this is true graphically by noticing that our angle is in the first quadrant and the fact that the coordinates of a point on the unit circle centered at the origin are .

To use this to find , the angle will be with an extra counterclockwise rotation.

Then, the angle of this line segment with the positive -axis is , so is the -coordinate of this rotated point.

We can then see that the -coordinate can be found from the original triangle.

This confirms our answer that, for our angle ,

Example 2: Using Cofunctions Identities and Periodic Identities to Evaluate Expressions

Find the value of given , where .

Answer

In this question, we are given an expression involving several compound trigonometric terms and asked to evaluate this using the fact that and .

We might be tempted to do this by using a graphical approach, but this would mean we need to sketch three diagrams (one for each term). Instead, we will try and simplify this expression by using trigonometric identities.

We will simplify each of the three terms separately; let’s start with .

We can simplify this term using our cofunction identity:

However, the angles inside our parentheses do not match. We can get around this by rewriting as :

Now, we can apply our cofunction identity with :

We cannot directly evaluate this expression, but we can simplify this by using another cofunction identity and the fact that cosine is an even function.

Since cosine is an even function,

Then, we want to apply the following cofunction identity:

We can do this by setting , which gives us

We have successfully evaluated the first term. Let’s now evaluate .

To do this, we might want to try using the cofunction identities; however, in this case, it’s easier to use the periodic property of the tangent function:

We might be worried since we are subtracting our angle ; however, we can apply this identity with :

Then, we apply our identity again, this time with :

We can simplify this further by remembering that the tangent function is an odd function:

Hence, we need to find the value of . To do this, we will use the fact that is in the first quadrant and that . Remember, the sine of an angle is the ratio between the length of the side opposite that angle and the hypotenuse in a right triangle.

is the ratio between the length of the opposite side and adjacent side in this right triangle. We can find the missing side by using the Pythagorean theorem:

This then gives us .

It is also worth reiterating that we know this is positive because we are working in the first quadrant.

Therefore, we have shown , which means the second term in our expression evaluates to give us .

Now, we only need to evaluate the last term in this expression, .

We will do this by first using the fact that the sine function is periodic with a period of , meaning that we can subtract this value from our argument:

Next, we will use the fact that the sine function is odd:

We can then use our cofunction identity by subtracting negative :

We can simplify further by using the fact that cosine is an even function:

Then, to evaluate , we will use our triangle:

Therefore, we have shown

Finally, we can then use these three results to evaluate our entire expression:

Example 3: Using Cofunction Identities to Evaluate a Cosecant Function

Find the value of given , where is the smallest positive angle.

Answer

We want to evaluate , given that and that is the smallest positive angle. We have a few options to try and evaluate this expression; we could try a graphical approach or we could try using trigonometric identities to make this expression easier to evaluate. There are multiple ways of using trigonometric identities to solve this question; we will go through two of these.

Method 1:

Since the angles involved are similar to the cofunction identities, we will try and rewrite this expression.

First, we will use the fact that the cosecant is the reciprocal of the sine function:

Next, we want to use the cofunction identity:

We can do this by rewriting our argument:

We can then apply our cofunction identity with :

To simplify this expression further, we want to apply our other cofunction identity. However, if we did this now, we would end up with a result that is not useful. Instead, we want to simplify the argument by using the fact that cosine is an even function:

We can then apply our other cofunction identity in a very similar way:

We rewrite the argument:

Then, we use our cofunction identity with :

We might want to further simplify our argument but, remember, in the question we are told . We can write our denominator in this way by using the fact that the sine function is odd:

Method 2:

Alternatively, we can start by rewriting cosecant using its definition:

Then, using the fact that the sine function is periodic gives us

We can simplify further by using the fact that the sine function is odd:

Now, we want to apply the cofunction identity

Next, we use the fact that the cosine function is even to simplify further:

We can then rewrite this in terms of by using the cofunction identity ; this gives us

Finally, we are told that ; therefore,

Hence, if is the smallest positive angle, where , then .

Example 4: Using Periodic and Cofunction Identities to Evaluate a Trigonometric Function of a Given Angle

is a right triangle at . Find given that .

Answer

We want to find the value of , from the fact that and the diagram. To do this, we want to first find an expression for in terms of . We can find this from the diagram. First, because the sum of the angles in triangle must be .

Next, we can see that and lie on a straight line, so these angles sum to give us . This means

Therefore,

To evaluate this expression, we will use the cofunction identities. To use the cofunction identities, we can start by rewriting our argument:

We then write this in terms of the tangent:

One way of evaluating this expression is to use the cofunction identity for the tangent function, which tells us

Applying this to our expression, we get

We then want to write this in terms of ; we can do this by using the fact that cotangent is an odd function:

Hence, we have shown that .

Example 5: Evaluating Trigonometric Expressions Using the Relations between Trigonometric Functions of Complementary Angles

Find the value of .

Answer

We cannot evaluate this expression directly, which means we will need to simplify it first. Looking at the arguments of the trigonometric expression given, we can see that and . In other words, these angles are complementary.

When we are dealing with complementary angles, it is a good idea to try using the cofunction identities.

Let’s start with ; we can write this as

We can then use our cofunction identity with :

We can then substitute this into the expression given to us in the question:

We can cancel the shared factor of , since we know .

We can do the same to rewrite . First, we rewrite the argument:

Then, we want to use our cofunction identity:

We set :

Finally, we substitute this into our expression:

Remember, it is important to check when we cancel this shared factor. To do this, we can recall that, in the CAST diagram, tangent is positive in the first quadrant, so , which means its reciprocal is also positive.

Therefore, we have shown that