Lesson Explainer: Evaluating Trigonometric Functions with Special Angles Mathematics • 10th Grade

In this explainer, we will learn how to evaluate the trigonometric functions with special angles and how to use them to evaluate trigonometric expressions.

We will begin by recalling special angles, together with the sine, cosine, and tangent values of these angles.

Let us consider a unit circle. This enables us to calculate sinπ‘₯, cosπ‘₯, and tanπ‘₯ between 0∘ and 360∘ or 0 and 2πœ‹ radians. All three functions have key values at 0∘, 90∘, 180∘, 270∘, and 360∘. We know that 180=πœ‹βˆ˜radians, so 360=2πœ‹βˆ˜radians, 90=πœ‹2∘radians, and 270=3πœ‹2∘radians. Knowing these conversions enables us to solve trigonometric problems when the angles are given in either degrees or radians.

πœƒ
0∘90∘180∘270∘360∘
sinπœƒ010βˆ’10
cosπœƒ10βˆ’101
tanπœƒ0undefined0undefined0

As the functions are periodic, we can calculate the sine, cosine, and tangent of angles outside of this range by adding multiples of 360∘ for sin and cos, or by adding multiples of 180∘ for tan.

Next, we recall that the special angles are 30∘, 45∘, and 60∘. The sine, cosine, and tangent of these angles are given below.

πœƒ
30∘45∘60∘
sinπœƒ121√2=√22√32
cosπœƒβˆš321√2=√2212
tanπœƒ1√3=√331√3

While we will not consider the derivation or proof of these results in this explainer, it is worth recalling the identity sincostanπ‘₯π‘₯=π‘₯. This enables us to calculate the tangent of any angle if we are given the sine and cosine of that angle.

For example, since sin60=√32∘ and cos60=12∘, then tan60=√32÷12=√3∘.

It will also be necessary for us to recall the definitions of the reciprocal trigonometric functions.

Definition: Reciprocal Trigonometric Functions

The reciprocal trigonometric functions cosecant πœƒ, secant πœƒ, and cotangent πœƒ are the reciprocal of sine πœƒ, cosine πœƒ, and tangent πœƒ such that cscsinseccoscottanπœƒ=1πœƒ,πœƒ=1πœƒ,πœƒ=1πœƒ.

We can use these identities to calculate the cosecant, secant, and cotangent of 30∘, 45∘, and 60∘.

πœƒ
30∘45∘60∘
cscπœƒ2√22√3=2√33
secπœƒ2√3=2√33√22
cotπœƒβˆš311√3=√33

We will now recall the related angles of the trigonometric functions:

One way of recalling whether the sine, cosine, and tangent of any angle between 0∘ and 360∘ are positive or negative is using the CAST diagram. This is a memory device that we use to remember the signs of the trigonometric ratios in each of the four quadrants.

The Quadrant in Which the Terminal Side of the Angle LiesThe Interval in Which the Measure of the Angle BelongsSigns of Trigonometric Functions
csc, sincos, sectan, cot
First0,πœ‹2+++
SecondοŸπœ‹2,πœ‹ο“+βˆ’βˆ’
Thirdο πœ‹,3πœ‹2ο”βˆ’βˆ’+
Fourth3πœ‹2,2πœ‹ο”βˆ’+βˆ’

We note that the angles are measured, from 0∘ to 360∘ or from 0 to 2πœ‹ radians in a counterclockwise direction, where the positive π‘₯-axis is the initial side of the angle. The terminal side is where the angle stops. Any angle between 0∘ and 90∘ lies in the first quadrant. Any angle between 90∘ and 180∘ lies in the second quadrant. Any angle between 180∘ and 270∘ lies in the third quadrant. Any angle between 270∘ and 360∘ lies in the fourth quadrant.

Let us consider an example where we have to evaluate the cosine of an angle using the property of related angles to relate it to a special angle.

Example 1: Using Periodic Identities to Find the Value of a Trigonometric Function Involving Special Angles

Find the value of cos11πœ‹6.

Answer

We begin by recalling that πœ‹=180radians∘.

So, πœ‹6=3011πœ‹6=330.radiansradians∘∘

We therefore need to calculate cos330∘.

Let us recall the property of related angles coscos(360βˆ’πœƒ)=πœƒ.∘

If πœƒ=30∘, then coscoscoscos(360βˆ’30)=30330=30.∘∘∘∘∘

From our knowledge of special angles, we know that cos30=√32∘.

So, cos330=√32.∘

We can therefore conclude that cos11πœ‹6=√32.

Let us now consider a similar example where, this time, we need to evaluate the sine and tangent of given angles.

Example 2: Evaluating Trigonometric Expressions Involving Special Angles

Evaluate 2ο€»πœ‹6ο‡βˆ’8ο€Ό4πœ‹3tansin.

Answer

We begin by recalling that πœ‹=180radians∘.

So, πœ‹6=30.radians∘

Also, πœ‹3=604πœ‹3=240.radiansradians∘∘

We therefore need to calculate 230βˆ’8240tansin∘∘.

From our knowledge of special angles, we know that tan30=√33∘ and sin60=√32∘.

By considering the CAST diagram, as shown below, we see that 240∘ lies in the 3rd quadrant and the sine of any angle here is negative.

Since sinsin(180+πœƒ)=βˆ’πœƒ,∘ then sinsinsinsin(180+60)=βˆ’60(240)=βˆ’60.∘∘∘∘∘

So, sin(240)=βˆ’βˆš32.∘

Substituting the values of tan30∘ and sin240∘ into our expression, we have 230βˆ’8240=2ο€Ώβˆš33ο‹βˆ’8ο€Ώβˆ’βˆš32=2√33+4√3=2√33+12√33=14√33.tansin∘∘

Therefore, 2πœ‹6βˆ’84πœ‹3=14√33tansin.

In the remaining examples in this explainer, we will also need to use reciprocal trigonometric functions.

Example 3: Using Periodic Identities to Evaluate a Trigonometric Expression Involving Special Angles

Find the value of costancsccos135+135+225+225∘∘∘∘.

Answer

In order to answer this question, we need to recall reciprocal trigonometric identities and special angles.

From our knowledge of special angles, sin45=√22∘, cos45=√22∘, and tan45=1∘.

Using the CAST diagram, we know that both the cosine and tangent of any angle in the second quadrant are negative. The sine and cosine of any angle in the third quadrant are also both negative.

The properties of related angles state that coscos(180βˆ’πœƒ)=βˆ’πœƒ.∘

So, coscoscoscos(180βˆ’45)=βˆ’45135=βˆ’45=βˆ’βˆš22.∘∘∘∘∘

The properties of related angles also state that tantan(180βˆ’πœƒ)=βˆ’πœƒ.∘

So, tantan(180βˆ’45)=βˆ’45.∘∘∘

Therefore, tantan135=βˆ’45=βˆ’1.∘∘

They also state that sinsin(180+πœƒ)=βˆ’πœƒ.∘

So, sinsin(180+45)=βˆ’45.∘∘∘

Therefore, sinsin225=βˆ’45=βˆ’βˆš22.∘∘

They also state that coscos(180+πœƒ)=βˆ’πœƒ.∘

So, coscos(180+45)=βˆ’45.∘∘∘

Therefore, coscos225=βˆ’45=βˆ’βˆš22.∘∘

Since cscsinπœƒ=1πœƒ, then cscsin225=1225=βˆ’βˆš2.∘∘

Substituting these values into our expression, we have costancsccos135+135+225+225=βˆ’βˆš22+(βˆ’1)+ο€»βˆ’βˆš2+ο€Ώβˆ’βˆš22=βˆ’βˆš22βˆ’1βˆ’βˆš2βˆ’βˆš22=βˆ’2√2βˆ’1.∘∘∘∘

So, our final answer is βˆ’2√2βˆ’1.

For the next example, we will need to evaluate a second-degree trigonometric expression.

Example 4: Evaluating Trigonometric Expressions Involving Special Angles

Evaluate 3βˆ’4330cos∘.

Answer

Let us recall the property of related angles coscos(360βˆ’πœƒ)=πœƒ.∘

If πœƒ=30∘, then coscoscoscos(360βˆ’30)=30330=30.∘∘∘∘∘

From our knowledge of special angles, we know that cos30=√32∘.

So, cos330=√32.∘

Substituting this value into our expression gives us 3βˆ’4330=3βˆ’4ο€Ώβˆš32=3βˆ’4ο€Ό34=3βˆ’3=0.cos∘

So, our final answer is 0.

In the next example, we will investigate how different angles can satisfy a trigonometric equation involving multiple angles and the power of a trigonometric function.

Example 5: Using Periodic Identities to Evaluate a Trigonometric Expression Involving Special Angles

Which of the following values of π‘₯ does not satisfy the equation sincosπ‘₯+6(3π‘₯)βˆ’12=βˆ’3√2?

  1. 45∘
  2. 135∘
  3. 315∘
  4. 405∘
  5. βˆ’315∘

Answer

In this example, we have been given several values of π‘₯ and are being asked which does not satisfy the given equation. We note that since the given equation is a trigonometric one, it can possibly have an infinite number of solutions due to the properties of related angles.

The easiest way to check which value of π‘₯ does not satisfy the equation is to substitute the given values into the left-hand side one by one and see whether we get the right-hand side.

First, for π‘₯=45∘, we have sincos∘∘45+6(3β‹…45)βˆ’12.

To evaluate this, we first recall that 45∘ is a special angle, and so we can look up the value of the first term directly to find that sin45=√22∘. For the second term, that is, 6(3β‹…45)=6135coscos∘∘, we can use the related angle property: coscos(180βˆ’πœƒ)=βˆ’πœƒ.∘

Taking πœƒ=45∘, we have coscos135=βˆ’45=βˆ’βˆš22.∘∘

Putting this together, we have sincos∘∘45+6(135)βˆ’12=ο€Ώβˆš22+6ο€Ώβˆ’βˆš22ο‹βˆ’12=24βˆ’6√22βˆ’12=12βˆ’3√2βˆ’12=βˆ’3√2.

Thus, since this is equal to the right-hand side, 45∘ satisfies the equation.

Let us repeat this process for the other angles, continuing to use properties of related angles to help us. For π‘₯=135∘, we have sincos∘∘135+6405βˆ’12.

We use the identities sinsin(180βˆ’πœƒ)=πœƒβˆ˜ and coscos(πœƒ+360)=πœƒβˆ˜, with πœƒ=45∘, to get sincossincos∘∘∘∘135+6405βˆ’12=45+645βˆ’12=ο€Ώβˆš22+6ο€Ώβˆš22ο‹βˆ’12=12+6√22βˆ’12=3√2.

As this is not equal to the right-hand side, this shows that 135∘ does not satisfy the equation.

We have already found the answer, but for completeness, let us check the remaining options. For π‘₯=315∘, we have sincos∘∘315+6945βˆ’12.

The first term can be simplified using sinsin(360βˆ’πœƒ)=βˆ’πœƒβˆ˜ (by taking πœƒ=45∘), to get sinsin315=βˆ’45∘∘. The second can be simplified by first using the periodicity of cosine, in other words, coscos(πœƒ+2β‹…360)=πœƒβˆ˜, to get coscos945=225∘∘, and then using coscos(360βˆ’πœƒ)=πœƒβˆ˜ to get coscos225=135∘∘. Recall that we have already found that cos135=βˆ’βˆš22∘. Putting this together, we have sincossincos∘∘∘∘(315)+6(945)βˆ’12=(βˆ’45)+6135βˆ’12=ο€Ώβˆ’βˆš22+6ο€Ώβˆ’βˆš22ο‹βˆ’12=12βˆ’3√2βˆ’12=βˆ’3√2.

Thus, π‘₯=315∘ works. The fourth option is π‘₯=405∘. For this, we have sincos∘∘405+61215βˆ’12.

For both the first and second terms, we can use the periodicity of sine and cosine to simplify these terms. That is, sinsin(πœƒ+360)=πœƒβˆ˜ and coscos(πœƒ+3β‹…360)=πœƒβˆ˜ to get sincossincos∘∘∘∘405+61215βˆ’12=45+6135βˆ’12.

This is the same left-hand side as we had for π‘₯=45∘, so we can conclude that this value satisfies the equation.

Finally, we have π‘₯=βˆ’315∘, which gives us a left-hand side of sincos∘∘(βˆ’315)+6(βˆ’945)βˆ’12.

We can use the periodic identities sinsinπœƒ=(πœƒ+360)∘ and coscosπœƒ=(πœƒ+3β‹…360)∘ to get sincossincos∘∘∘∘(βˆ’315)+6(βˆ’945)βˆ’12=45+6135βˆ’12.

Once more, we have the same left-hand side as we had for π‘₯=45∘ and π‘₯=405∘, so we can conclude that this value satisfies the equation.

In conclusion, option B does not satisfy the equation.

In the previous example, let us note that three of the possible solutions, namely, π‘₯=45∘, π‘₯=405∘, and π‘₯=βˆ’315∘, ended up with the exact same expression on the left-hand side. This was a result of the periodicity of trigonometric functions, which can be seen by considering the positions of these angles on a CAST diagram.

That is to say, the values of cosπ‘₯ and sinπ‘₯ are equal for these values of π‘₯ because they share the same place on the CAST diagram. By extension, the values of cosπ‘₯ and sinπ‘₯ will also be the same.

It is important to be aware, however, that we cannot always use this approach of a CAST diagram. In particular, we should be careful since one of the terms in the equation (i.e., 6(3π‘₯)cos) has a multiple angle. Recall that for trigonometric functions with multiple angles, the periodicity is different. This difference is illustrated in the graphs below.

Specifically, the period of cosπ‘₯ is 360∘ (or 2πœ‹), but the period of cos3π‘₯ is 120∘ (or 2πœ‹3). Nevertheless, since the periods of cosπ‘₯ and cos3π‘₯ overlap, it turns out that angles that are 360∘ apart will still have the same value of cos3π‘₯. We can see this by considering the periodic property of cos3π‘₯, specifically that since it has a period of 120∘, we have coscos3π‘₯=(3(π‘₯+120𝑛)),∘ where 𝑛 is an integer. Since 360∘ is three times 120∘, if we let 𝑛=3π‘˜, where π‘˜ is also an integer, we have coscoscos3π‘₯=(3(π‘₯+120β‹…3π‘˜))=(3(π‘₯+360π‘˜)).∘∘

This shows us that the value of cos3π‘₯ will be the same for any angles that differ by 360∘, so 45∘, 405∘, and βˆ’315∘ will be the same too.

For our final example, let us find the value of an expression involving the product of multiple trigonometric functions.

Example 6: Evaluating Trigonometric Expressions Involving Special Angles

Find the value of 33060βˆ’060+27045sinsincossecsincos∘∘∘∘∘∘.

Answer

In order to answer this question, we need to recall reciprocal trigonometric identities and special angles.

From our knowledge of special angles, we know that sin30=12∘, sin60=√32∘, and cos45=√22∘.

From the sine and cosine graphs below, we see that cos0=1∘ and sin270=βˆ’1∘.

Since seccosπœƒ=1πœƒ, then seccos60=160=1Γ·12=2.∘∘

We can now substitute all of these values into our expression: 33060βˆ’060+27045=3ο€Ό12οˆο€Ώβˆš32ο‹βˆ’(1)(2)+(βˆ’1)ο€Ώβˆš22=3√34βˆ’2βˆ’12=3√34βˆ’104=βˆ’10+3√34.sinsincossecsincos∘∘∘∘∘∘

We will finish this explainer by recapping some of the key points.

Key Points

  • We can evaluate trigonometric functions and expressions using our knowledge of special angles: sinsinsincoscoscostantantan30=12,45=√22,60=√32,30=√32,45=√22,60=12,30=1√3,45=1,60=√3.∘∘∘∘∘∘∘∘∘
  • We can use reciprocal trigonometric identities to solve more complicated problems: cscsinseccoscottanπœƒ=1πœƒ,πœƒ=1πœƒ,πœƒ=1πœƒ.
  • We can also use properties of related angles to evaluate trigonometric expressions:

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