Lesson Explainer: Evaluating Trigonometric Functions Using Cofunction Identities Mathematics

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

We have seen a number of different identities and properties for the trigonometric functions that we can use to help us simplify and solve equations. Before we see how we can apply these properties and identities, we will start by recapping the results we have shown so far.

First, we have the definitions for some of our trigonometric functions.

Definition: Trigonometric Identities

For any angle πœƒ measured in degrees or radians,

  • tansincosπœƒβ‰‘πœƒπœƒ,
  • seccosπœƒβ‰‘1πœƒ,
  • cscsinπœƒβ‰‘1πœƒ,
  • cottanπœƒβ‰‘1πœƒ.

Next, we have the fact that the trigonometric functions are periodic.

Definition: Periodic Identities

For any angle πœƒ measured in degrees,

  • sinsin(πœƒΒ±360)β‰‘πœƒβˆ˜,
  • coscos(πœƒΒ±360)β‰‘πœƒβˆ˜,
  • tantan(πœƒΒ±180)β‰‘πœƒβˆ˜.

We could also write similar identities for angles measured in radians.

We also have some identities we can derive by using the Pythagorean theorem and the definitions of our trigonometric ratios.

Definition: Pythagorean Identities

For any angle πœƒ measured in degrees or radians,

  • sincosοŠ¨οŠ¨πœƒ+πœƒβ‰‘1,
  • 1+πœƒβ‰‘πœƒtansec,
  • 1+πœƒβ‰‘πœƒcotcsc.

We can show that the sine function is odd and the cosine function is even by considering reflections of points on the unit circle, giving us the following identities.

Definition: Odd/Even Trigonometric Function Identities

For any angle πœƒ measured in degrees or radians,

  • sinsin(βˆ’πœƒ)β‰‘βˆ’πœƒ, where sine is an odd function;
  • coscos(βˆ’πœƒ)β‰‘πœƒ, where cosine is an even function;
  • tantan(βˆ’πœƒ)β‰‘βˆ’πœƒ, where tangent is an odd function.

We can find the following identities either by considering rotations of points on the unit circle or by considering the corresponding angle of πœƒ in a right triangle.

Definition: Cofunction Identities

For any angle πœƒ measured in degrees,

  • sincos(90βˆ’πœƒ)β‰‘πœƒβˆ˜,
  • cossin(90βˆ’πœƒ)β‰‘πœƒβˆ˜,
  • tancot(90βˆ’πœƒ)β‰‘πœƒβˆ˜.

We could also write these identities in terms of radians by using the fact that 90=πœ‹2∘.

There are in fact many more identities we can use by combining all of these results together. By combining these identities, we are able to rewrite equations into a form that is easier to solve.

Let’s start with an example where we need to use the cofunction identities to help us solve a trigonometric equation.

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

Find the value of cos(90+πœƒ)∘ given sinπœƒ=35, where 0<πœƒ<90∘∘.

Answer

In this question, we are told that the sine of angle πœƒ is equal to 35 and that 0<πœƒ<90∘∘. We need to use this information to evaluate cos(90+πœƒ)∘. 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 π‘₯, cossin(90βˆ’π‘₯)≑π‘₯.∘

We want to use this on the expression cos(90+πœƒ)∘, so we need to find a way to subtract our angle from 90∘. One way of doing this is to recall that adding πœƒ is the same as subtracting βˆ’πœƒ: coscos(90+πœƒ)=(90βˆ’(βˆ’πœƒ)).∘∘

Then, we apply the cofunction identity, with π‘₯=βˆ’πœƒ: cossin(90βˆ’(βˆ’πœƒ))≑(βˆ’πœƒ).∘

Finally, we know that the sine function is an odd function, which means sinsin(βˆ’πœƒ)=βˆ’πœƒ=βˆ’35.

Therefore, cos(90+πœƒ)=βˆ’35∘.

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

To use this to find cos(90+πœƒ)∘, the angle 90+πœƒβˆ˜ will be πœƒ with an extra 90∘ counterclockwise rotation.

Then, the angle of this line segment with the positive π‘₯-axis is 90+πœƒβˆ˜, so cos(90+πœƒ)∘ is the π‘₯-coordinate of this rotated point.

We can then see that the π‘₯-coordinate can be found from the original triangle.

This confirms our answer that, for our angle πœƒ, cos(90+πœƒ)=βˆ’35.∘

In the previous example, we combined a cofunction identity and the fact that the sine function was odd to show that coscossinsin(90+πœƒ)=(90βˆ’(βˆ’πœƒ))=(βˆ’πœƒ)=βˆ’πœƒ.∘∘

This gives us a new identity; in fact, we can combine any of the cofunction identities with the parity of the function to construct the following identities.

Definition: Alternate Cofunction Identities

For any angle πœƒ measured in degrees,

  • sincos(90+πœƒ)β‰‘πœƒβˆ˜,
  • cossin(90+πœƒ)β‰‘βˆ’πœƒβˆ˜,
  • tancot(90+πœƒ)β‰‘βˆ’πœƒβˆ˜.

We could also write these identities in terms of radians by using the fact that 90=πœ‹2∘.

We also get similar identities if we use the reciprocal trigonometric functions.

In our next example, we will see how we may need to combine multiple different cofunction identities togethers, making a graphical approach much more difficult.

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

Find the value of sintansin(180βˆ’π‘₯)+(360βˆ’π‘₯)+7(270βˆ’π‘₯)∘∘∘ given sinπ‘₯=35, where 0<π‘₯<90∘∘.

Answer

In this question, we are given an expression involving several compound trigonometric terms and asked to evaluate this using the fact that sinπ‘₯=35 and 0<π‘₯<90∘∘.

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 sin(180βˆ’π‘₯)∘.

We can simplify this term using our cofunction identity: sincos(90βˆ’πœƒ)β‰‘πœƒ.∘

However, the angles inside our parentheses do not match. We can get around this by rewriting 180∘ as 90+90∘∘: sinsinsinsin(180βˆ’π‘₯)=(90+90βˆ’π‘₯)=(90+[90βˆ’π‘₯])=(90βˆ’[π‘₯βˆ’90]).∘∘∘∘∘∘∘

Now, we can apply our cofunction identity with πœƒ=π‘₯βˆ’90∘: sinsincos(180βˆ’π‘₯)=(90βˆ’[π‘₯βˆ’90])=(π‘₯βˆ’90).∘∘∘∘

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, coscoscos(π‘₯βˆ’90)=(βˆ’(90βˆ’π‘₯))=(90βˆ’π‘₯).∘∘∘

Then, we want to apply the following cofunction identity: cossin(90βˆ’πœƒ)β‰‘πœƒ.∘

We can do this by setting πœƒ=π‘₯, which gives us sincossin(180βˆ’π‘₯)=(90βˆ’π‘₯)=π‘₯=35.∘∘

We have successfully evaluated the first term. Let’s now evaluate tan(360βˆ’π‘₯)∘.

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: tantan(πœƒΒ±180)β‰‘πœƒ.∘

We might be worried since we are subtracting our angle πœƒ; however, we can apply this identity with πœƒ=βˆ’π‘₯+180∘: tantantantan(360βˆ’π‘₯)=(βˆ’π‘₯+360)=([βˆ’π‘₯+180]+180)=(βˆ’π‘₯+180).∘∘∘∘∘

Then, we apply our identity again, this time with πœƒ=βˆ’π‘₯: tantan(βˆ’π‘₯+180)=(βˆ’π‘₯).∘

We can simplify this further by remembering that the tangent function is an odd function: tantantan(360βˆ’π‘₯)=(βˆ’π‘₯)=βˆ’(π‘₯).∘

Hence, we need to find the value of tan(π‘₯). To do this, we will use the fact that π‘₯ is in the first quadrant and that sinπ‘₯=35. 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.

tan(π‘₯) 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: 5=3+π‘Žπ‘Ž=√25βˆ’9=4.

This then gives us tanoppositeadjacent(π‘₯)==34.

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

Therefore, we have shown βˆ’(π‘₯)=βˆ’34tan, which means the second term in our expression evaluates to give us βˆ’34.

Now, we only need to evaluate the last term in this expression, 7(270βˆ’π‘₯)sin∘.

We will do this by first using the fact that the sine function is periodic with a period of 360∘, meaning that we can subtract this value from our argument: 7(270βˆ’π‘₯)=7(270βˆ’π‘₯βˆ’360)=7(βˆ’90βˆ’π‘₯).sinsinsin∘∘∘∘

Next, we will use the fact that the sine function is odd: 7(βˆ’90βˆ’π‘₯)=7(βˆ’[90+π‘₯])=βˆ’7(90+π‘₯).sinsinsin∘∘∘

We can then use our cofunction identity by subtracting negative π‘₯: βˆ’7(90+π‘₯)=βˆ’7(90βˆ’(βˆ’π‘₯))=βˆ’7(βˆ’π‘₯).sinsincos∘∘

We can simplify further by using the fact that cosine is an even function: βˆ’7(βˆ’π‘₯)=βˆ’7(π‘₯).coscos

Then, to evaluate cos(π‘₯), we will use our triangle:

cosadjacenthypotenuse(π‘₯)==45.

Therefore, we have shown 7(270βˆ’π‘₯)=βˆ’7(π‘₯)=βˆ’7ο€Ό45=βˆ’285.sincos∘

Finally, we can then use these three results to evaluate our entire expression: sintansin(180βˆ’π‘₯)+(360βˆ’π‘₯)+7(270βˆ’π‘₯)=35βˆ’34βˆ’285=βˆ’234.∘∘∘

In the previous example, we were able to use the fact the sine function is periodic, the parity of the trigonometric functions, and a cofunction identity to show sinsinsinsinsincoscos(270βˆ’π‘₯)=(270βˆ’π‘₯βˆ’360)=(βˆ’90βˆ’π‘₯)=βˆ’(90+π‘₯)=βˆ’(90βˆ’(βˆ’π‘₯))=βˆ’(βˆ’π‘₯)=βˆ’(π‘₯).∘∘∘∘∘∘

We can follow the same process to show the following identities.

Definition: More Alternate Cofunction Identities

For any angle πœƒ measured in degrees,

  • sincos(270βˆ’πœƒ)β‰‘βˆ’πœƒβˆ˜,
  • cossin(270βˆ’πœƒ)β‰‘βˆ’πœƒβˆ˜,
  • tancot(270βˆ’πœƒ)β‰‘βˆ’πœƒβˆ˜.

We could also write these identities in terms of radians by using the fact that 90=πœ‹2∘.

Once again, we could construct similar identities for the reciprocal trigonometric functions in the same way or by using these three identities.

In a similar way, sinsinsinsincos(270+π‘₯)=(270+π‘₯βˆ’360)=(π‘₯βˆ’90)=βˆ’(90βˆ’π‘₯)=βˆ’(π‘₯).∘∘∘∘∘

We can follow the same method to show the following.

Definition: Further Alternate Cofunction Identities

For any angle πœƒ measured in degrees,

  • sincos(270+πœƒ)β‰‘βˆ’πœƒβˆ˜,
  • cossin(270+πœƒ)β‰‘πœƒβˆ˜,
  • tancot(270+πœƒ)β‰‘βˆ’πœƒβˆ˜.

We could also write these identities in terms of radians by using the fact that 90=πœ‹2∘.

We can find similar identities to the above for the reciprocal trigonometric functions. We can also find many more identities by combining these properties in different ways.

In our next example, we will show that it is also possible to use the cofunction identities when we are dealing with the reciprocal trigonometric functions.

Example 3: Using Cofunction Identities to Evaluate a Cosecant Function

Find the value of csc(270βˆ’πœƒ)∘ given sin(90βˆ’πœƒ)=27∘, where πœƒ is the smallest positive angle.

Answer

We want to evaluate csc(270βˆ’πœƒ)∘, given that sin(90βˆ’πœƒ)=27∘ and that πœƒ 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: cscsin(270βˆ’πœƒ)=1(270βˆ’πœƒ).∘∘

Next, we want to use the cofunction identity: sincos(90βˆ’π‘₯)=π‘₯.∘

We can do this by rewriting our argument: 1(270βˆ’πœƒ)=1(90βˆ’(βˆ’180)βˆ’πœƒ)=1(90βˆ’(βˆ’180+πœƒ)).sinsinsin∘∘∘∘∘

We can then apply our cofunction identity with π‘₯=βˆ’180+πœƒβˆ˜: 1(90βˆ’(βˆ’180+πœƒ))=1(βˆ’180+πœƒ).sincos∘∘∘

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: 1(βˆ’180+πœƒ)=1(βˆ’(180βˆ’πœƒ))=1(180βˆ’πœƒ).coscoscos∘∘∘

We can then apply our other cofunction identity in a very similar way: cossin(90βˆ’πœƒ)≑π‘₯.∘

We rewrite the argument: 1(180βˆ’πœƒ)=1(90βˆ’(βˆ’90)βˆ’πœƒ)=1(90βˆ’(βˆ’90+πœƒ)).coscoscos∘∘∘∘∘

Then, we use our cofunction identity with π‘₯=βˆ’90+πœƒβˆ˜: 1(90βˆ’(βˆ’90+πœƒ))=1(βˆ’90+πœƒ).cossin∘∘∘

We might want to further simplify our argument but, remember, in the question we are told sin(90βˆ’πœƒ)=27∘. We can write our denominator in this way by using the fact that the sine function is odd: 1(βˆ’90+πœƒ)=1(βˆ’(90βˆ’πœƒ))=1βˆ’(90βˆ’πœƒ)=βˆ’1(90βˆ’πœƒ)=βˆ’1=βˆ’72.sinsinsinsin∘∘∘∘

Method 2:

Alternatively, we can start by rewriting cosecant using its definition: cscsin(270βˆ’πœƒ)=1(270βˆ’πœƒ).∘∘

Then, using the fact that the sine function is periodic gives us 1(270βˆ’πœƒ)=1(270βˆ’πœƒβˆ’360)=1(βˆ’90βˆ’πœƒ).sinsinsin∘∘∘∘

We can simplify further by using the fact that the sine function is odd: 1(βˆ’90βˆ’πœƒ)=1(βˆ’(90βˆ’πœƒ))=1βˆ’(90+πœƒ)=βˆ’1(90+πœƒ).sinsinsinsin∘∘∘∘

Now, we want to apply the cofunction identity sincos(90βˆ’πœƒ)≑π‘₯βˆ˜βˆ’1(90+πœƒ)=βˆ’1(90βˆ’(βˆ’πœƒ))=βˆ’1(βˆ’πœƒ).sinsincos∘∘

Next, we use the fact that the cosine function is even to simplify further: βˆ’1(βˆ’πœƒ)=βˆ’1(πœƒ).coscos

We can then rewrite this in terms of sin(90βˆ’πœƒ)∘ by using the cofunction identity sincos(90βˆ’πœƒ)β‰‘πœƒβˆ˜; this gives us βˆ’1(πœƒ)=βˆ’1(90βˆ’πœƒ).cossin∘

Finally, we are told that sin(90βˆ’πœƒ)=27∘; therefore, βˆ’1(90βˆ’πœƒ)=βˆ’1=βˆ’72.sin∘

Hence, if πœƒ is the smallest positive angle, where sin(90βˆ’πœƒ)=27∘, then csc(270βˆ’πœƒ)=βˆ’72∘.

So far, all of our examples have involved the cofunction identities. Let’s see an example where we need to apply other identities to help us a evaluate a trigonometric expression.

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

𝐴𝐡𝐢 is a right triangle at 𝐡. Find cot𝛼 given that cotπœƒ=43.

Answer

We want to find the value of cot𝛼, from the fact that cotπœƒ=43 and the diagram. To do this, we want to first find an expression for 𝛼 in terms of πœƒ. We can find this from the diagram. First, π‘šβˆ π΅π·π΄=90βˆ’πœƒβˆ˜ because the sum of the angles in triangle 𝐴𝐡𝐷 must be 180∘.

Next, we can see that ∠𝐡𝐷𝐴 and ∠𝐢𝐷𝐴 lie on a straight line, so these angles sum to give us 180∘. This means 90βˆ’πœƒ+𝛼=180𝛼=90+πœƒ.∘∘∘

Therefore, cotcot𝛼=(90+πœƒ).∘

To evaluate this expression, we will use the cofunction identities. To use the cofunction identities, we can start by rewriting our argument: cotcot(90+πœƒ)=(90βˆ’(βˆ’πœƒ)).∘∘

We then write this in terms of the tangent: cottan(90βˆ’(βˆ’πœƒ))=1(90βˆ’(βˆ’πœƒ)).∘∘

One way of evaluating this expression is to use the cofunction identity for the tangent function, which tells us tancot(90βˆ’π‘₯)≑π‘₯.∘

Applying this to our expression, we get 1(90βˆ’(βˆ’πœƒ))=1(βˆ’πœƒ).tancot∘

We then want to write this in terms of cotπœƒ; we can do this by using the fact that cotangent is an odd function: 1(βˆ’πœƒ)=1βˆ’(πœƒ)=βˆ’1=βˆ’34.cotcotοŠͺ

Hence, we have shown that cot𝛼=βˆ’34.

In our final example, we will see how we can evaluate an expression involving multiple different angles.

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

Find the value of sintancotcos1520Γ—7075∘∘∘∘.

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 90βˆ’15=75∘∘∘ and 90βˆ’20=70∘∘∘. 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 sin15∘; we can write this as sinsin15=(90βˆ’75).∘∘∘

We can then use our cofunction identity sincos(90βˆ’πœƒ)β‰‘πœƒβˆ˜ with πœƒ=75∘: sinsincos15=(90βˆ’75)=(75).∘∘∘∘

We can then substitute this into the expression given to us in the question: sintancotcoscostancotcoscottan1520Γ—7075=7520Γ—7075=7020.∘∘∘∘∘∘∘∘∘∘

We can cancel the shared factor of cos75∘, since we know cos75>0∘.

We can do the same to rewrite tan20∘. First, we rewrite the argument: tantan20=(90βˆ’70).∘∘∘

Then, we want to use our cofunction identity: tancot(90βˆ’70)β‰‘πœƒ.∘∘

We set πœƒ=70∘: tantancot20=(90βˆ’70)=70.∘∘∘∘

Finally, we substitute this into our expression: cottancotcot7020=7070=1.∘∘∘∘

Remember, it is important to check cot70β‰ 0∘ 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 tan70>0∘, which means its reciprocal is also positive.

Therefore, we have shown that sintancotcos1520Γ—7075=1.∘∘∘∘

Let’s finish by recapping some basic points.

Key Points

  • We can use the cofunction identities to help us evaluate trigonometric expressions.
  • We can also combine the cofunction identities with all of our other trigonometric identities to help us simplify expressions.
  • We can combine all of the properties and identities of the trigonometric functions to find more identities for the trigonometric functions.

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