Lesson Explainer: Evaluating Trigonometric Functions with Special Angles | Nagwa Lesson Explainer: Evaluating Trigonometric Functions with Special Angles | Nagwa

Lesson Explainer: Evaluating Trigonometric Functions with Special Angles Mathematics • First Year of Secondary School

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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 by adding multiples of 360โˆ˜ for sin and cos, or by adding multiples of 180โˆ˜ for tan.

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˜.

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 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.

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 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.

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 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.

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

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.

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 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 this example, we have been given several values of ๐‘ฅ 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 ๐‘ฅ 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 ๐‘ฅ=45โˆ˜, we have sincos๏Šจโˆ˜โˆ˜45+6(3โ‹…45)โˆ’12.

To evaluate this, we first recall that 45โˆ˜ is a special angle, and so we can look up the value of the first term directly to find that sin45=โˆš22โˆ˜. For the second term, that is, 6(3โ‹…45)=6135coscosโˆ˜โˆ˜, we can use the related angle property: coscos(180โˆ’๐œƒ)=โˆ’๐œƒ.โˆ˜

Taking ๐œƒ=45โˆ˜, we have coscos135=โˆ’45=โˆ’โˆš22.โˆ˜โˆ˜

Putting this together, we have sincos๏Šจโˆ˜โˆ˜๏Šจ45+6(135)โˆ’12=๏€ฟโˆš22๏‹+6๏€ฟโˆ’โˆš22๏‹โˆ’12=24โˆ’6โˆš22โˆ’12=12โˆ’3โˆš2โˆ’12=โˆ’3โˆš2.

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

Let us repeat this process for the other angles, continuing to use properties of related angles to help us. For ๐‘ฅ=135โˆ˜, we have sincos๏Šจโˆ˜โˆ˜135+6405โˆ’12.

We use the identities sinsin(180โˆ’๐œƒ)=๐œƒโˆ˜ and coscos(๐œƒ+360)=๐œƒโˆ˜, with ๐œƒ=45โˆ˜, to get sincossincos๏Šจโˆ˜โˆ˜๏Šจโˆ˜โˆ˜๏Šจ135+6405โˆ’12=45+645โˆ’12=๏€ฟโˆš22๏‹+6๏€ฟโˆš22๏‹โˆ’12=12+6โˆš22โˆ’12=3โˆš2.

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

We have already found the answer, but for completeness, let us check the remaining options. For ๐‘ฅ=315โˆ˜, we have sincos๏Šจโˆ˜โˆ˜315+6945โˆ’12.

The first term can be simplified using sinsin(360โˆ’๐œƒ)=โˆ’๐œƒโˆ˜ (by taking ๐œƒ=45โˆ˜), to get sinsin315=โˆ’45โˆ˜โˆ˜. The second can be simplified by first using the periodicity of cosine, in other words, coscos(๐œƒ+2โ‹…360)=๐œƒโˆ˜, to get coscos945=225โˆ˜โˆ˜, and then using coscos(360โˆ’๐œƒ)=๐œƒโˆ˜ to get coscos225=135โˆ˜โˆ˜. Recall that we have already found that cos135=โˆ’โˆš22โˆ˜. Putting this together, we have sincossincos๏Šจโˆ˜โˆ˜โˆ˜๏Šจโˆ˜๏Šจ(315)+6(945)โˆ’12=(โˆ’45)+6135โˆ’12=๏€ฟโˆ’โˆš22๏‹+6๏€ฟโˆ’โˆš22๏‹โˆ’12=12โˆ’3โˆš2โˆ’12=โˆ’3โˆš2.

Thus, ๐‘ฅ=315โˆ˜ works. The fourth option is ๐‘ฅ=405โˆ˜. For this, we have sincos๏Šจโˆ˜โˆ˜405+61215โˆ’12.

For both the first and second terms, we can use the periodicity of sine and cosine to simplify these terms. That is, sinsin(๐œƒ+360)=๐œƒโˆ˜ and coscos(๐œƒ+3โ‹…360)=๐œƒโˆ˜ to get sincossincos๏Šจโˆ˜โˆ˜๏Šจโˆ˜โˆ˜405+61215โˆ’12=45+6135โˆ’12.

This is the same left-hand side as we had for ๐‘ฅ=45โˆ˜, so we can conclude that this value satisfies the equation.

Finally, we have ๐‘ฅ=โˆ’315โˆ˜, which gives us a left-hand side of sincos๏Šจโˆ˜โˆ˜(โˆ’315)+6(โˆ’945)โˆ’12.

We can use the periodic identities sinsin๐œƒ=(๐œƒ+360)โˆ˜ and coscos๐œƒ=(๐œƒ+3โ‹…360)โˆ˜ to get sincossincos๏Šจโˆ˜โˆ˜๏Šจโˆ˜โˆ˜(โˆ’315)+6(โˆ’945)โˆ’12=45+6135โˆ’12.

Once more, we have the same left-hand side as we had for ๐‘ฅ=45โˆ˜ and ๐‘ฅ=405โˆ˜, 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, ๐‘ฅ=45โˆ˜, ๐‘ฅ=405โˆ˜, and ๐‘ฅ=โˆ’315โˆ˜, 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 cos๐‘ฅ and sin๐‘ฅ are equal for these values of ๐‘ฅ because they share the same place on the CAST diagram. By extension, the values of cos๏Šจ๐‘ฅ and sin๏Šจ๐‘ฅ 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(3๐‘ฅ)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 cos๐‘ฅ is 360โˆ˜ (or 2๐œ‹), but the period of cos3๐‘ฅ is 120โˆ˜ (or 2๐œ‹3). Nevertheless, since the periods of cos๐‘ฅ and cos3๐‘ฅ overlap, it turns out that angles that are 360โˆ˜ apart will still have the same value of cos3๐‘ฅ. We can see this by considering the periodic property of cos3๐‘ฅ, specifically that since it has a period of 120โˆ˜, we have coscos3๐‘ฅ=(3(๐‘ฅ+120๐‘›)),โˆ˜ where ๐‘› is an integer. Since 360โˆ˜ is three times 120โˆ˜, if we let ๐‘›=3๐‘˜, where ๐‘˜ is also an integer, we have coscoscos3๐‘ฅ=(3(๐‘ฅ+120โ‹…3๐‘˜))=(3(๐‘ฅ+360๐‘˜)).โˆ˜โˆ˜

This shows us that the value of cos3๐‘ฅ will be the same for any angles that differ by 360โˆ˜, so 45โˆ˜, 405โˆ˜, and โˆ’315โˆ˜ 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 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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