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Trigonometric values, equations and identities

Module by: Sunil Kumar Singh

In this module, we discuss trigonometric values and angles. In particular, we shall learn about two very useful algorithms which help us to find (i) value of trigonometric function when angle is given and (ii) angles when value of trigonometric function is given. In addition, we shall go through various trigonometric equations and identities. We are expected to be already familiar with them. For this reason, solutions of equations and identities are presented here without deduction and are included for reference purpose.

Values of trigonometric function

It is sufficient to know values of trigonometric functions for angles in first quarter. These angles are called acute angles (angle value less than π/2). Here, we develop algorithm, which converts angles in other quadrants in terms of acute angles. Basic idea is that angles can be expressed in terms of combination of acute angle and reference angles like 0, π/2, π and 2π. These angles demark quadrants. Using certain procedure, we can find value of trigonometric function of any angle provided we know the trigonometric value of corresponding acute angle. For the sake of convenience, we shall concentrate on acute angles π/6, π/4 and π/3, whose trigonometric function values are known to us. We follow an algorithm to determine trigonometric values as given here :

1 : Express given angle as sum or difference of acute angle and reference angles 0, π/2, π and 2π.

2 : Write trigonometric function of sum or difference as trigonometric function of acute angle. A trigonometric sum/difference combination of angles involving angles of 0, π and 2π does not change the function. However, a combination involving π/2 changes function from sine to cosine and vice-versa, tangent to cotangent and vice-versa and cosecant to secant and vice-versa.

3 : Apply sign before trigonometric function determined as above in accordance with the sign rule of trigonometric function.

f r + a = + o r − g a f r + a = + o r − g a

where “f” and “g” denote trigonometric functions, “r” denotes reference angles like 0, π/2, π and 2π and “a” denotes acute angle.

Figure 1: Signs of six trigonometric functions in different quadrants.
Trigonometric sign diagram
Trigonometric sign diagram (tis1a.gif)

Let us consider an angle 7π/6. We are required to find sine and cotangent values of this angle. Here, we see that 7π/6 is greater than π. Hence, it is equal to π plus some acute angle, say, x.

π + x = 7 π 6 π + x = 7 π 6 ⇒ x = 7 π 6 - x = π 6 ⇒ x = 7 π 6 - x = π 6 ⇒ sin 7 π 6 = sin π + π 6 ⇒ sin 7 π 6 = sin π + π 6

Since combination involves angle π, the sine of given angle retains the trigonometric function form. However, angle 7π/6 falls in third quadrant, in which sine is negative. Thus,

⇒ sin 7 π 6 = sin π + π 6 = - sin π 6 = - 3 2 ⇒ sin 7 π 6 = sin π + π 6 = - sin π 6 = - 3 2

Similarly,

⇒ cot 7 π 6 = cot π + π 6 = cot π 6 = 1 3 ⇒ cot 7 π 6 = cot π + π 6 = cot π 6 = 1 3

This method is very helpful to determine value of trigonometric function provided we know the value of trigonometric function of corresponding acute angle resulting from combination involving angles 0, π/2, π and 2π. Here, we shall work out few standard identities involving combination of angles with reference angles. We need not remember these identities. Rather, we should rely on the procedure discussed here as all of these can be derived on spot easily.

Reflection in 0

There is no change in function form. Function takes sign in accordance with sign rule.

sin 0 - x = - sin x ; cos 0 - x = cos x ; tan 0 - x = - tan x ; sin 0 - x = - sin x ; cos 0 - x = cos x ; tan 0 - x = - tan x ; cosec 0 - x = - cosec x ; sec 0 - x = sec x ; cot 0 - x = - cot x cosec 0 - x = - cosec x ; sec 0 - x = sec x ; cot 0 - x = - cot x

Reflection in π/2

Reflection in π/2 is also known as co-function identities. Functions are called co-functions when their compliments have same value. As such, sine and cosine are co-functions. In this case, there is change in function form as combination of angle involves π/2. Function takes sign in accordance with sign rule.

sin π 2 - x = cos x ; cos π 2 - x = sin x ; tan π 2 - x = cot x ; sin π 2 - x = cos x ; cos π 2 - x = sin x ; tan π 2 - x = cot x ; cosec π 2 - x = sec x ; sec π 2 - x = cosec x ; cot π 2 - x = tan x cosec π 2 - x = sec x ; sec π 2 - x = cosec x ; cot π 2 - x = tan x

Reflection in π

In this case, there is no change in function form. Function takes sign in accordance with sign rule.

sin π - x = sin x ; cos π - x = - cos x ; tan π - x = - tan x ; sin π - x = sin x ; cos π - x = - cos x ; tan π - x = - tan x ; cosec π - x = cosec x ; sec π - x = - sec x ; cot π - x = - cot x cosec π - x = cosec x ; sec π - x = - sec x ; cot π - x = - cot x

Shift by π/2

Shift refers to horizontal shift of graph. We shall explore this aspect of trigonometric function in detail in a separate module. From transformation point of view, there is change in function form as combination of angle involves π/2. Function takes sign in accordance with sign rule.

sin π 2 + x = cos x ; cos π 2 + x = - sin x ; tan π 2 + x = - cot x ; sin π 2 + x = cos x ; cos π 2 + x = - sin x ; tan π 2 + x = - cot x ; cosec π 2 + x = sec x ; sec π 2 + x = - cosec x ; cot π 2 + x = - tan x cosec π 2 + x = sec x ; sec π 2 + x = - cosec x ; cot π 2 + x = - tan x

Shift by π

In this case, there is no change in function form. Function takes sign in accordance with sign rule.

sin π + x = - sin x ; cos π + x = - cos x ; tan π + x = tan x ; sin π + x = - sin x ; cos π + x = - cos x ; tan π + x = tan x ; cosec π + x = - cosec x ; sec π + x = - sec x ; cot π + x = cot x cosec π + x = - cosec x ; sec π + x = - sec x ; cot π + x = cot x

Shift by 2π

In this case, there is no change in function form. Function takes sign in accordance with sign rule.

sin 2 π + x = sin x ; cos 2 π + x = cos x ; tan 2 π + x = tan x ; sin 2 π + x = sin x ; cos 2 π + x = cos x ; tan 2 π + x = tan x ; cosec 2 π + x = cosec x ; sec 2 π + x = sec x ; cot 2 π + x = cot x cosec 2 π + x = cosec x ; sec 2 π + x = sec x ; cot 2 π + x = cot x

Finding angles

Trigonometric functions are many-one relation. We are required to find angles corresponding to a given trigonometric value. For example, what are angles corresponding to sine value of -√3/2. In other words, we need to find angles whose sine evaluates to this value. Note that these values corresponds to intersection of parallel line y=-√3/2 with the graph of sine curve.

Figure 2: Intersection of sine function with parallel value line.
Graph of sine function
Graph of sine function (tis3a.gif)

For the time being, let us concentrate the interval [0,2π], which corresponds to one cycle of four quadrants. We follow an algorithm as given here to find angles in this interval :

1 : Consider only numerical magnitude of the given value. Find acute angle whose trigonometric function value corresponds to the numerical magnitude of the given value.

2 : Use sign rule and identify quadrants in which trigonometric function has the sign that of given value.

3 : Use value diagram and determine the angles as required.

Figure 3: Angles whose trigonometric function values are same in different quadrants(to be used in conjunction with sign diagram).
Trigonometric value diagram
Trigonometric value diagram (tis2a.gif)

To see the working of the algorithm, let us consider sinx = -√3/2. Considering only the magnitude of numerical value, we have :

sin θ = 3 2 = sin π 3 sin θ = 3 2 = sin π 3

Thus, required acute angle is π/3. Now, sine function is negative in third and fourth quadrants. Looking at the value diagram, the angle in third quadrant is :

⇒ x = π + θ = π + π 3 = 4 π 3 ⇒ x = π + θ = π + π 3 = 4 π 3

Similarly, angle in fourth quadrant is :

⇒ x = 2 π - θ = 2 π - π 3 = 5 π 3 ⇒ x = 2 π - θ = 2 π - π 3 = 5 π 3

Example 1

Problem : Find angles in [0,2π], if

cot x = 1 3 cot x = 1 3

Solution : Considering only the magnitude of numerical value, we have :

⇒ cot θ = 1 3 = cot π 3 ⇒ cot θ = 1 3 = cot π 3

Thus, required acute angle is π/3. Now, cotangent function is positive in first and third quadrants. Looking at the value diagram, the angle in third quadrant is :

⇒ x = π + θ = π + π 3 = 4 π 3 ⇒ x = π + θ = π + π 3 = 4 π 3

Hence angles are π/3 and 4π/3.

Negative angles

When we consider angle as a real number entity, we need to express angles as negative angles as well. The corresponding negative angle (y) is obtained as :

y = x - 2 π y = x - 2 π

Thus, negative angles corresponding to 4π/3 and 5π/3 are :

⇒ y = 4 π 3 - 2 π = - 2 π 3 ⇒ y = 4 π 3 - 2 π = - 2 π 3 ⇒ y = 5 π 3 - 2 π = - π 3 ⇒ y = 5 π 3 - 2 π = - π 3

We can also find negative angle values using a separate negative value diagram (see figure). We draw negative value diagram by demarking quadrants with corresponding angles and writing angle values for negative values. We deduct “2π” from the relation for positive value diagram.

Figure 4: Trigonometric value diagram for negative angles
Trigonometric value diagram
Trigonometric value diagram (tis4a.gif)

Let us consider sinx = -√3/2 again. The acute angle in first quadrant is π/3. Sine is negative in third and fourth quadrants. The angles in these quadrants are :

y = - π + θ = - π + π 3 = - 2 π 3 y = - π + θ = - π + π 3 = - 2 π 3 y = - θ = - π 3 y = - θ = - π 3

Trigonometric equations

Zeroes of sine and cosine functions

Trigonometric equations are formed by equating trigonometric functions to zero. The solutions of these equations are :

1 : sin x = 0 ⇒ x = n π ; n ∈ Z sin x = 0 ⇒ x = n π ; n ∈ Z

2 : cos x = 0 ⇒ x = 2 n + 1 π 2 ; n ∈ Z cos x = 0 ⇒ x = 2 n + 1 π 2 ; n ∈ Z

Definition of other trigonometric functions

We define other trigonometric functions in the light of zeroes of sine and cosine as listed above :

tan x = sin x cos x ; x ≠ 2 n + 1 π 2 ; n ∈ Z tan x = sin x cos x ; x ≠ 2 n + 1 π 2 ; n ∈ Z cot x = cos x sin x ; x ≠ n π ; n ∈ Z cot x = cos x sin x ; x ≠ n π ; n ∈ Z cosec x = 1 sin x ; x ≠ n π ; n ∈ Z cosec x = 1 sin x ; x ≠ n π ; n ∈ Z sec x = 1 cos x ; x ≠ 2 n + 1 π 2 ; n ∈ Z sec x = 1 cos x ; x ≠ 2 n + 1 π 2 ; n ∈ Z

Trigonometric equations

Trigonometric function can be used to any other values as well. Solutions of such equations are given here without deduction for reference purpose. Solutions of three equations involving sine, cosine and tangent functions are listed here :

1. Sine equation

sin x = a = sin y sin x = a = sin y

x = n π + - 1 n y ; n ∈ Z x = n π + - 1 n y ; n ∈ Z

2. Cosine equation

cos x = a = cos y cos x = a = cos y

x = 2 n π ± y ; n ∈ Z x = 2 n π ± y ; n ∈ Z

3. Tangent equation

tan x = a = tan y tan x = a = tan y

x = n π + y ; n ∈ Z x = n π + y ; n ∈ Z

In order to understand the working with trigonometric equation, let us consider an equation :

sin x = - 3 2 sin x = - 3 2

As worked out earlier, -√3/2 is sine value of two angles in the interval [0, π]. Important question here is to know which angle should be used in the solution set. Here,

⇒ sin 4 π 3 = sin 5 π 3 = - 3 2 ⇒ sin 4 π 3 = sin 5 π 3 = - 3 2

We can write general solution using either of two values.

x = n π + - 1 n 4 π 3 ; n ∈ Z x = n π + - 1 n 4 π 3 ; n ∈ Z ⇒ x = n π + - 1 n 5 π 3 ; n ∈ Z ⇒ x = n π + - 1 n 5 π 3 ; n ∈ Z

The solution sets appear to be different, but are same on expansion. Conventionally, however, we use the smaller of two angles which lie in the interval [0, π]. In order to check that two series are indeed same, let us expand series from n=-4 to n=4,

For ⇒ x = n π + - 1 n 4 π 3 ; n ∈ Z ⇒ x = n π + - 1 n 4 π 3 ; n ∈ Z

- 4 π + 4 π 3 = - 8 π 3 , - 3 π - 4 π 3 = - 13 π 3 , - 2 π + 4 π 3 = - 2 π 3 , - π - 4 π 3 = - 7 π 3 , - 4 π + 4 π 3 = - 8 π 3 , - 3 π - 4 π 3 = - 13 π 3 , - 2 π + 4 π 3 = - 2 π 3 , - π - 4 π 3 = - 7 π 3 ,

0 + 4 π / 3 = 4 π 3 , π - 4 π 3 = - π 3 , 2 π + 4 π 3 = 10 π 3 , 3 π - 4 π 3 = 5 π 3 , 4 π + 4 π 3 = 16 π 3 0 + 4 π / 3 = 4 π 3 , π - 4 π 3 = - π 3 , 2 π + 4 π 3 = 10 π 3 , 3 π - 4 π 3 = 5 π 3 , 4 π + 4 π 3 = 16 π 3

Arranging in increasing order :

- 13 π 3 , - 8 π 3 , - 7 π 3 , - 2 π 3 , - π 3 , 4 π 3 , 5 π 3 , 10 π 3 , 16 π 3 - 13 π 3 , - 8 π 3 , - 7 π 3 , - 2 π 3 , - π 3 , 4 π 3 , 5 π 3 , 10 π 3 , 16 π 3

For ⇒ x = n π + - 1 n 5 π 3 ; n ∈ Z ⇒ x = n π + - 1 n 5 π 3 ; n ∈ Z

- 4 π + 5 π 3 = - 7 π 3 , - 3 π - 5 π 3 = - 14 π 3 , - 2 π + 5 π 3 = - π 3 , - π - 5 π 3 = - 8 π 3 , - 4 π + 5 π 3 = - 7 π 3 , - 3 π - 5 π 3 = - 14 π 3 , - 2 π + 5 π 3 = - π 3 , - π - 5 π 3 = - 8 π 3 ,

0 + 5 π 3 = 5 π 3 , π - 5 π 3 = - 2 π 3 , 2 π + 5 π 3 = 11 π 3 , 3 π - 5 π 3 = 4 π 3 , 4 π + 5 π 3 = 17 π 3 0 + 5 π 3 = 5 π 3 , π - 5 π 3 = - 2 π 3 , 2 π + 5 π 3 = 11 π 3 , 3 π - 5 π 3 = 4 π 3 , 4 π + 5 π 3 = 17 π 3

Arranging in increasing order :

- 14 π 3 , - 8 π 3 , - 7 π 3 , - 2 π 3 , - π 3 , 4 π 3 , 5 π 3 , 11 π 3 , 17 π 3 - 14 π 3 , - 8 π 3 , - 7 π 3 , - 2 π 3 , - π 3 , 4 π 3 , 5 π 3 , 11 π 3 , 17 π 3

We see that there are common terms. There are, however, certain terms which do not appear in other series. We can though find those missing terms by evaluating some more values. For example, if we put n = 6 in the second series, then we get the missing term -13π/3. Also, putting n=5,7, we get 10π/3 and 16π/3. Thus, all missing terms in second series are obtained. Similarly, we can compute few more values in first series to find missing terms. We, therefore, conclude that both these series are equal.

Example 2

Problem : Find solution of equation :

2 cos 2 x + 3 sin x = 0 2 cos 2 x + 3 sin x = 0

Solution : Our objective here is to covert equation to linear form. Here, we can not convert sine term to cosine term, but we can convert cos 2 x cos 2 x in terms of sin 2 x sin 2 x .

⇒ 2 1 - sin 2 x + 3 sin x = 0 ⇒ 2 1 - sin 2 x + 3 sin x = 0 ⇒ 2 - 2 sin 2 x + 3 sin x = 0 ⇒ 2 - 2 sin 2 x + 3 sin x = 0 ⇒ 2 sin 2 x − 3 sin x − 2 = 0 ⇒ 2 sin 2 x − 3 sin x − 2 = 0

It is a quadratic equation in sinx. Factoring, we have :

⇒ 2 sin 2 x + sin x − 4 sin x − 2 = 0 ⇒ 2 sin 2 x + sin x − 4 sin x − 2 = 0 ⇒ sin x 2 sin x + 1 − 2 2 sin x + 1 = 0 ⇒ sin x 2 sin x + 1 − 2 2 sin x + 1 = 0 ⇒ 2 sin x + 1 sin x − 2 = 0 ⇒ 2 sin x + 1 sin x − 2 = 0

Either, sinx=-1/2 or sinx = 2. But sinx can not be equal to 2. hence,

⇒ sin x = - 1 2 = sin π + π 6 = sin 7 π 6 ⇒ sin x = - 1 2 = sin π + π 6 = sin 7 π 6 ⇒ x = n π + - 1 n 7 π 6 ; n ∈ Z ⇒ x = n π + - 1 n 7 π 6 ; n ∈ Z

Note : We shall not work with any other examples here as purpose of this module is only to introduce general concepts of angles, identities and equations. These topics are part of separate detailed study.

Trigonometric identities

Reciprocal identities

Reciprocals are defined for values of x for which trigonometric function in the denominator is not zero.

sin x = 1 cosec x ; cos x = 1 sec x ; tan x = 1 cot x ; sin x = 1 cosec x ; cos x = 1 sec x ; tan x = 1 cot x ; cosec x = 1 sin x ; sec x = 1 cos x ; cot x = 1 tan x cosec x = 1 sin x ; sec x = 1 cos x ; cot x = 1 tan x

Negative angle identities

cos - x = cos x ; sin - x = - sin x ; tan - x = - tan x cos - x = cos x ; sin - x = - sin x ; tan - x = - tan x

Pythagorean identities

cos 2 x + sin 2 x = 1 ; 1 + tan 2 x = sec 2 x ; 1 + cot 2 x = cosec 2 x cos 2 x + sin 2 x = 1 ; 1 + tan 2 x = sec 2 x ; 1 + cot 2 x = cosec 2 x

Sum/difference identities

sin x ± y = sin x cos y ± sin y cos x sin x ± y = sin x cos y ± sin y cos x cos x ± y = cos x cos y ∓ sin x sin y cos x ± y = cos x cos y ∓ sin x sin y tan x ± y = tan s x ± tan y / 1 ∓ tan x tan y ; x,y and (x+y) are not odd multiple of π/2 tan x ± y = tan s x ± tan y / 1 ∓ tan x tan y ; x,y and (x+y) are not odd multiple of π/2 cot x ± y = cot x cot y ∓ 1 / cot y ± cot x ; x,y and (x+y) are not odd multiple of π/2 cot x ± y = cot x cot y ∓ 1 / cot y ± cot x ; x,y and (x+y) are not odd multiple of π/2

Double angle identities

sin 2 x = 2 sin x cos x = 2 tan x 1 + tan 2 x sin 2 x = 2 sin x cos x = 2 tan x 1 + tan 2 x cos 2 x = cos 2 x - sin 2 x = 2 cos 2 x - 1 = 1 - 2 sin 2 x = 1 - tan 2 x 1 + tan 2 x cos 2 x = cos 2 x - sin 2 x = 2 cos 2 x - 1 = 1 - 2 sin 2 x = 1 - tan 2 x 1 + tan 2 x tan 2 x = 2 tan x 1 - tan 2 x tan 2 x = 2 tan x 1 - tan 2 x cot 2 x = cot 2 x - 1 2 cot x cot 2 x =