Lesson Explainer: Evaluating Trigonometric Functions with Special Angles Mathematics

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 sin๐‘ฅ, cos๐‘ฅ, and tan๐‘ฅ between 0โˆ˜ and 360โˆ˜ or 0 and 2๐œ‹ radians. All three functions have key values at 0โˆ˜, 90โˆ˜, 180โˆ˜, 270โˆ˜, and 360โˆ˜. We know that 180=๐œ‹โˆ˜radians, so 360=2๐œ‹โˆ˜radians, 90=๐œ‹2โˆ˜radians, and 270=3๐œ‹2โˆ˜radians. Knowing these conversions enables us to solve trigonometric problems when the angles are given in either degrees or radians.

๐œƒ
0โˆ˜90โˆ˜180โˆ˜270โˆ˜360โˆ˜
sin๐œƒ010โˆ’10
cos๐œƒ10โˆ’101
tan๐œƒ0undefined0undefined0

As the functions are periodic, we can calculate the sine, cosine, and tangent of angles outside of this range; however, this is not in the scope of this explainer.

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

๐œƒ
30โˆ˜45โˆ˜60โˆ˜
sin๐œƒ121โˆš2=โˆš22โˆš32
cos๐œƒโˆš321โˆš2=โˆš2212
tan๐œƒ1โˆš3=โˆš331โˆš3

While we will not consider the derivation or proof of these results in this explainer, it is worth recalling the identity sincostan๐‘ฅ๐‘ฅ=๐‘ฅ. 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=โˆš32โˆ˜ and cos60=12โˆ˜, then tan60=โˆš32รท12=โˆš3โˆ˜.

Definition: Reciprocal Trigonometric Functions

The reciprocal trigonometric functions cosecant ๐œƒ, secant ๐œƒ, and cotangent ๐œƒ are the reciprocal of sine ๐œƒ, cosine ๐œƒ, and tangent ๐œƒ such that cscsinseccoscottan๐œƒ=1๐œƒ,๐œƒ=1๐œƒ,๐œƒ=1๐œƒ.

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

๐œƒ
30โˆ˜45โˆ˜60โˆ˜
csc๐œƒ2โˆš22โˆš3=2โˆš33
sec๐œƒ2โˆš3=2โˆš33โˆš22
cot๐œƒโˆš311โˆš3=โˆš33

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 0โˆ˜ and 360โˆ˜ 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๏Ÿ0,๐œ‹2๏“+++
Second๏Ÿ๐œ‹2,๐œ‹๏“+โˆ’โˆ’
Third๏ ๐œ‹,3๐œ‹2๏”โˆ’โˆ’+
Fourth๏ 3๐œ‹2,2๐œ‹๏”โˆ’+โˆ’

We note that the angles are measured, from 0โˆ˜ to 360โˆ˜ or from 0 to 2๐œ‹ radians in a counterclockwise direction, where the positive ๐‘ฅ-axis is the initial side of the angle. The terminal side is where the angle stops. Any angle between 0โˆ˜ and 90โˆ˜ lies in the first quadrant. Any angle between 90โˆ˜ and 180โˆ˜ lies in the second quadrant. Any angle between 180โˆ˜ and 270โˆ˜ lies in the third quadrant. Any angle between 270โˆ˜ and 360โˆ˜ lies in the fourth quadrant.

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

Find the value of cos11๐œ‹6.

Answer

We begin by recalling that ๐œ‹=180radiansโˆ˜.

So, ๐œ‹6=3011๐œ‹6=330.radiansradiansโˆ˜โˆ˜

We therefore need to calculate cos330โˆ˜.

Let us recall the property of related angles coscos(360โˆ’๐œƒ)=๐œƒ.โˆ˜

If ๐œƒ=30โˆ˜, then coscoscoscos(360โˆ’30)=30330=30.โˆ˜โˆ˜โˆ˜โˆ˜โˆ˜

From our knowledge of special angles, we know that cos30=โˆš32โˆ˜.

So, cos330=โˆš32.โˆ˜

We can therefore conclude that cos11๐œ‹6=โˆš32.

Example 2: Evaluating Trigonometric Expressions Involving Special Angles

Evaluate 2๏€ป๐œ‹6๏‡โˆ’8๏€ผ4๐œ‹3๏ˆtansin.

Answer

We begin by recalling that ๐œ‹=180radiansโˆ˜.

So, ๐œ‹6=30.radiansโˆ˜

Also, ๐œ‹3=604๐œ‹3=240.radiansradiansโˆ˜โˆ˜

We therefore need to calculate 230โˆ’8240tansinโˆ˜โˆ˜.

From our knowledge of special angles, we know that tan30=โˆš33โˆ˜ and sin60=โˆš32โˆ˜.

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

Since sinsin(180+๐œƒ)=โˆ’๐œƒ,โˆ˜ then sinsinsinsin(180+60)=โˆ’60(240)=โˆ’60.โˆ˜โˆ˜โˆ˜โˆ˜โˆ˜

So, sin(240)=โˆ’โˆš32.โˆ˜

Substituting the values of tan30โˆ˜ and sin240โˆ˜ into our expression, we have 230โˆ’8240=2๏€ฟโˆš33๏‹โˆ’8๏€ฟโˆ’โˆš32๏‹=2โˆš33+4โˆš3=2โˆš33+12โˆš33=14โˆš33.tansinโˆ˜โˆ˜

Therefore, 2๐œ‹6โˆ’84๐œ‹3=14โˆš33tansin.

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+225โˆ˜โˆ˜โˆ˜โˆ˜.

Answer

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

From our knowledge of special angles, sin45=โˆš22โˆ˜, cos45=โˆš22โˆ˜, and tan45=1โˆ˜.

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(180โˆ’๐œƒ)=โˆ’๐œƒ.โˆ˜

So, coscoscoscos(180โˆ’45)=โˆ’45135=โˆ’45=โˆ’โˆš22.โˆ˜โˆ˜โˆ˜โˆ˜โˆ˜

The properties of related angles also state that tantan(180โˆ’๐œƒ)=โˆ’๐œƒ.โˆ˜

So, tantan(180โˆ’45)=โˆ’45.โˆ˜โˆ˜โˆ˜

Therefore, tantan135=โˆ’45=โˆ’1.โˆ˜โˆ˜

They also state that sinsin(180+๐œƒ)=โˆ’๐œƒ.โˆ˜

So, sinsin(180+45)=โˆ’45.โˆ˜โˆ˜โˆ˜

Therefore, sinsin225=โˆ’45=โˆ’โˆš22.โˆ˜โˆ˜

They also state that coscos(180+๐œƒ)=โˆ’๐œƒ.โˆ˜

So, coscos(180+45)=โˆ’45.โˆ˜โˆ˜โˆ˜

Therefore, coscos225=โˆ’45=โˆ’โˆš22.โˆ˜โˆ˜

Since cscsin๐œƒ=1๐œƒ, then cscsin225=1225=โˆ’โˆš2.โˆ˜โˆ˜

Substituting these values into our expression, we have costancsccos135+135+225+225=โˆ’โˆš22+(โˆ’1)+๏€ปโˆ’โˆš2๏‡+๏€ฟโˆ’โˆš22๏‹=โˆ’โˆš22โˆ’1โˆ’โˆš2โˆ’โˆš22=โˆ’2โˆš2โˆ’1.โˆ˜โˆ˜โˆ˜โˆ˜

So, our final answer is โˆ’2โˆš2โˆ’1.

Example 4: Evaluating Trigonometric Expressions Involving Special Angles

Evaluate 3โˆ’4330cos๏Šจโˆ˜.

Answer

Let us recall the property of related angles coscos(360โˆ’๐œƒ)=๐œƒ.โˆ˜

If ๐œƒ=30โˆ˜, then coscoscoscos(360โˆ’30)=30330=30.โˆ˜โˆ˜โˆ˜โˆ˜โˆ˜

From our knowledge of special angles, we know that cos30=โˆš32โˆ˜.

So, cos330=โˆš32.โˆ˜

Substituting this value into our expression gives us 3โˆ’4330=3โˆ’4๏€ฟโˆš32๏‹=3โˆ’4๏€ผ34๏ˆ=3โˆ’3=0.cos๏Šจโˆ˜๏Šจ

So, our final answer is 0.

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

Which of the following values of ๐‘ฅ does not satisfy the equation sincos๏Šจ๐‘ฅ+6(3๐‘ฅ)โˆ’12=โˆ’3โˆš2?

  1. 45โˆ˜
  2. 135โˆ˜
  3. 315โˆ˜
  4. 405โˆ˜
  5. โˆ’315โˆ˜

Answer

In order to answer this question, we will use our knowledge of special angles and properties of related angles of the sine and cosine functions.

Let us begin by considering the five angles we are given and their corresponding position on the CAST diagram.

As 45โˆ˜, 405โˆ˜, and โˆ’315โˆ˜ have the same position on the diagram, we know that the sine and cosine of these answers will be equal.

For example, we know that sin45=โˆš22โˆ˜.

So, sinandsin405=โˆš22โˆ’315=โˆš22.โˆ˜โˆ˜

Likewise, cos45=โˆš22โˆ˜.

So, cosandcos405=โˆš22โˆ’315=โˆš22.โˆ˜โˆ˜

We therefore need only consider the three options A, B, and C.

Firstly, let us consider ๐‘ฅ=45โˆ˜. Substituting this into the left-hand side of our equation, we have sincossincos๏Šจโˆ˜โˆ˜๏Šจโˆ˜โˆ˜(45)+6(3ร—45)โˆ’12=(45)+6(135)โˆ’12.

We recall the property of related angles coscos(180โˆ’๐œƒ)=โˆ’๐œƒ.โˆ˜

If ๐œƒ=45โˆ˜, then coscoscoscos(180โˆ’45)=โˆ’45135=โˆ’45.โˆ˜โˆ˜โˆ˜โˆ˜โˆ˜

Substituting sin45=โˆš22โˆ˜ and cos135=โˆ’โˆš22โˆ˜ into this expression gives us ๏€ฟโˆš22๏‹+6๏€ฟโˆ’โˆš22๏‹โˆ’12=12โˆ’3โˆš2โˆ’12=โˆ’3โˆš2.๏Šจ

Secondly, let us consider ๐‘ฅ=135โˆ˜. Substituting this into the left-hand side of our equation, we have sincossincos๏Šจโˆ˜โˆ˜๏Šจโˆ˜โˆ˜(135)+6(3ร—135)โˆ’12=(135)+6(405)โˆ’12.

We recall the property of related angles sinsin(180โˆ’๐œƒ)=๐œƒ.โˆ˜

If ๐œƒ=45โˆ˜, then sinsinsinsin(180โˆ’45)=45135=45.โˆ˜โˆ˜โˆ˜โˆ˜โˆ˜

Substituting sin135=โˆš22โˆ˜ and cos405=โˆš22โˆ˜ into this expression gives us ๏€ฟโˆš22๏‹+6๏€ฟโˆš22๏‹โˆ’12=12+3โˆš2โˆ’12=3โˆš2.๏Šจ

Finally, let us consider ๐‘ฅ=315โˆ˜. Substituting this into the left-hand side of our equation, we have sincossincos๏Šจโˆ˜โˆ˜๏Šจโˆ˜โˆ˜(315)+6(3ร—315)โˆ’12=(315)+6(945)โˆ’12.

We recall the property of related angles sinsin(360โˆ’๐œƒ)=โˆ’๐œƒ.โˆ˜

If ๐œƒ=45โˆ˜, then sinsinsinsin(360โˆ’45)=โˆ’45315=โˆ’45.โˆ˜โˆ˜โˆ˜โˆ˜โˆ˜

Substituting sin315=โˆ’โˆš22โˆ˜ and cos945=โˆ’โˆš22โˆ˜ into this expression gives us ๏€ฟโˆ’โˆš22๏‹+6๏€ฟโˆ’โˆš22๏‹โˆ’12=12โˆ’3โˆš2โˆ’12=โˆ’3โˆš2.๏Šจ

We can therefore conclude that ๐‘ฅ=135โˆ˜ does not satisfy the equation sincos๏Šจ๐‘ฅ+6(3๐‘ฅ)โˆ’12=โˆ’3โˆš2, whereas all other four options do.

Example 6: Evaluating Trigonometric Expressions Involving Special Angles

Find the value of 33060โˆ’060+27045sinsincossecsincosโˆ˜โˆ˜โˆ˜โˆ˜โˆ˜๏Šจโˆ˜.

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=12โˆ˜, sin60=โˆš32โˆ˜, and cos45=โˆš22โˆ˜.

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

Since seccos๐œƒ=1๐œƒ, then seccos60=160=1รท12=2.โˆ˜โˆ˜

We can now substitute all of these values into our expression: 33060โˆ’060+27045=3๏€ผ12๏ˆ๏€ฟโˆš32๏‹โˆ’(1)(2)+(โˆ’1)๏€ฟโˆš22๏‹=3โˆš34โˆ’2โˆ’12=3โˆš34โˆ’104=โˆ’10+3โˆš34.sinsincossecsincosโˆ˜โˆ˜โˆ˜โˆ˜โˆ˜๏Šจโˆ˜๏Šจ

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=โˆš22,60=โˆš32,30=โˆš32,45=โˆš22,60=12,30=1โˆš3,45=1,60=โˆš3.โˆ˜โˆ˜โˆ˜โˆ˜โˆ˜โˆ˜โˆ˜โˆ˜โˆ˜
  • We can use reciprocal trigonometric identities to solve more complicated problems: cscsinseccoscottan๐œƒ=1๐œƒ,๐œƒ=1๐œƒ,๐œƒ=1๐œƒ.
  • We can also use properties of related angles to evaluate trigonometric expressions:

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