Which of the following identity is correct:

$$tan (A – B) = (tan A – tan B)/(1 + tan A tan B)$$

$$cos (A + B) = cos (A)cos (B) – sin (A)sin (B)$$

Trigonometric Equations and Identities Solved Examples

Trigonometric Equations and Identities Solved Examples are given on this page. Trigonometric identities have different formulae, which are inequalities involving trigonometric functions of one or more angles. Trigonometric equations and identities are useful for solving trigonometric problems.

Consider a right-angled triangle ABC. Here, ∠CAB = A and ∠BCA = 90°. AB is called the hypotenuse of a triangle. The trigonometric ratios are defined using the adjacent, opposite and hypotenuse sides of the triangle.

Trigonometric Equations And Identities Solved Examples

BC / AB = (Opposite side) / hypotenuse, gives sine of A and is denoted as sin A.

AC / AB = (adjacent side) / hypotenuse gives the cosine of A and is denoted as cos A.

BC / AC = (Opposite side) /(adjacent side), gives the tangent of A and is denoted as tan A.

cosec A = 1/sin A

sec A = 1/cos A

cot A = 1/tan A

Trigonometric Equations and Identities

An equation which consists of 1 or more trigonometric ratios of angles that are unknown is called a trigonometric equation. A trigonometric equation is written as P1 (sin θ, cos θ, tan θ, cot θ, sec θ, cosec θ) = P2 (sin θ, cos θ, tan θ, cot θ, sec θ, cosec θ), where P1 and P2 are rational functions.

A trigonometric equation which is always true and holds well for every angle is called a trigonometric identity.

The steps to solve a trigonometric equation are as follows:

Type – I

The first type of equation involves factorising them or expressing them in quadratic form. They can be solved by converting them into factors and finding the solution for each factor. The solution obtained finally is the association of solutions of all the factors.

Type – II

This type of equation can be written in the form of cos ⁡A + b sin⁡ B = c and then find the solution.

List of Trigonometric Equations and Identities

Trigonometric Identities

a) sin² θ + cos² θ = 1

b) sec² θ = 1 + tan² θ

c) cosec² θ = 1 + cot² θ

General Solution of Some Trigonometric Equations

a) sin θ = 0 ⇒ θ = n π

b) tan θ = 0 ⇒ θ = n π

c) cos θ = 0 ⇒ θ = (2n + 1) π/2

d) sin θ = 1 ⇒ θ = (4n + 1) π/2

e) cos θ = – 1 ⇒ θ = (4n – 1) π/2

f) cos θ = 1 ⇒ θ = 2nπ

g) cos = – 1 ⇒ θ = (2n + 1) π

h) cot θ = 0 ⇒ θ (2n + 1) π/2

General Solution of Some Standard Equations

a) sin θ = sin α ⇒ θ = nπ + (-1)ⁿ θ

b) cos θ = cos α ⇒ θ = 2nπ ± θ

c) tan θ = tan α ⇒ θ = nπ +θ

Angle-Sum and Difference Identities

sin (α + β) = sin (α)cos (β) + cos (α)sin (β)
sin (α – β) = sin (α)cos (β) – cos (α)sin (β)
cos (α + β) = cos (α)cos (β) – sin (α)sin (β)
cos (α – β) = cos (α)cos (β) + sin (α)sin (β)
tan (A + B) = (tan A + tan B)/(1 – tan A tan B)
tan (A – B) = (tan A – tan B)/(1 + tan A tan B)
cot (A + B) = (cot A cot B – 1)/(cot A + cot B)
cot (A – B) = (cot A cot B + 1)/(cot B – cot A)

Negative Angle Identities

sin (-A) = – sin A
cos (-A) = cos A
tan (-A) = – tan A
cosec (-A) = – cosec A
sec (-A) = sec A
cot (-A) = – cot A

Also, Read:

Trigonometric Equations and Its Solutions

Trigonometry Previous Year Questions with Solutions

Trigonometric Equations and Identities Examples

Some trigonometric identity problems are given below. JEE aspirants are advised to practise trigonometric equations and identities solved examples so that they can score better ranks.

Example 1: In Δ ABC, sin (A − B) / sin (A + B) = ?

Solution:

sin (A − B) / sin (A + B) = [sin A cos B − sin B cos A] / [sinC]

= [a / c] [cos B] − [b / c] [cos A]

But cos B = [a² + c² − b²] / 2ac,cos A = b² + c² − a² / 2bc

⇒ [a / c] [cos B] − [b / c] [cos A] = 1 / 2c²

= (a² + c² − b²− b² – c² + a²) / 2c²

= [a²− b²] / [c²]

Example 2: If the angles of a triangle are in the ratio 1: 2: 7, then what is the ratio of its greatest side to the least side?

Solution:

x + 2x + 7x = 180^o

⇒ x = 18^o

Hence, the angles are 18^o, 36^o, 126^o

Greatest side ∝ sin(126^o) Smallest side ∝ sin(18^o) and

Ratio = sin126^o / sin(18^o)

= [√5+1] / [√5−1]

Example 3: The value of sin²5^o+ sin²10^o+ sin²15^o+ . . . + sin²85^o+ sin²90^o is equal to

Solution:

Given expression is sin²5^o+ sin²10^o+ sin²15^o+ . . . + sin²85^o+ sin²90^o.

We know that sin 90^o= 1 or sin²90^o= 1.

Similarly, sin 45^o=1 / √2 or sin²45^o= 1 / 2 and the angles are in AP of 18 terms.

We also know that sin²85^o= [sin (90^o−5^o)]²= cos²5^o.

Therefore, from the complementary rule, we find sin²5^o+ sin²85^o= sin²5^o+cos²5^o= 1

Therefore, sin²5^o+ sin²10^o+ sin²15^o+ . . . + sin²85^o+ sin²90^o =(1 + 1 + 1+ 1 + 1 + 1 + 1 + 1 ) + 1 + 1 / 2 = 9 [1 / 2].

Example 4: Find the value of 1 + cos 56^o + cos58^o − cos66^o.

Solution:

1 + cos 56^o + cos58^o − cos66^o

Apply the conditional identity

cos A + cos B − cos C = − 1 + 4 cos [A / 2] cos [B / 2] sin [C / 2] [∵56^o+ 58^o+ 66^o = 180^o]

We get the value of the required expression equal to 4 cos28^o cos29^o sin33^o.

Example 5: If a cos³α + 3a cos α sin²α = m and a sin³α + 3a cos²α sin α = n, then find (m + n)^2/3 + (m – n)^2/3.

Solution:

Adding and subtracting the given relation, we get

$\begin{array}{l}(m+n)=a{{\cos }^{3}}\alpha +3a\cos \alpha \,{{\sin }^{2}}\alpha +3a{{\cos }^{2}}\alpha .\sin \alpha +a{{\sin }^{3}}\alpha =a{{(\cos \alpha +\sin \alpha )}^{3}}\end{array} $

and similarly

$\begin{array}{l}(m-n)=a\,\,{{(\cos \alpha -\sin \alpha )}^{3}}\end{array} $

Thus,

$\begin{array}{l}{{(m+n)}^{2/3}}+{{(m-n)}^{2/3}}={{a}^{2/3}}{{\{\cos \alpha +\sin \alpha )}^{2}}+{{(\cos \alpha -\sin \alpha )}^{2}}\}\\={{a}^{2/3}}\{2({{\cos }^{2}}\alpha +{{\sin }^{2}}\alpha )\}=2{{a}^{2/3}}\end{array} $

Example 6: If sin A = n sin B, then

$\begin{array}{l}\frac{n-1}{n+1}\tan \,\frac{A+B}{2}=\end{array} $

Solution:

We have

$\begin{array}{l}\sin A=n\sin B\Rightarrow \frac{n}{1}=\frac{\sin A}{\sin B}\\ \Rightarrow \frac{n-1}{n+1}=\frac{\sin A-\sin B}{\sin A+\sin B}=\frac{2\cos \frac{A+B}{2}\sin \frac{A-B}{2}}{2\sin \frac{A+B}{2}\cos \frac{A-B}{2}}\\ =\tan \frac{A-B}{2}\cot \frac{A+B}{2}\\ \Rightarrow \frac{n-1}{n+1}\tan \left( \frac{A+B}{2} \right)\\ =\tan \frac{A-B}{2}.\end{array} $

Example 7: If

$\begin{array}{l}\tan \theta =\frac{\sin \alpha -\cos \alpha }{\sin \alpha +\cos \alpha },\ \text{then}\end{array} $

sin α + cos α and sin α – cos α.

Solution:

We have

$\begin{array}{l}\tan \theta =\frac{\sin \alpha -\cos \alpha }{\sin \alpha +\cos \alpha } \\ \Rightarrow \tan \theta =\frac{\sin \left( \alpha -\frac{\pi }{4} \right)}{\cos \left( \alpha -\frac{\pi }{4} \right)}\Rightarrow \tan \theta =\tan \left( \alpha -\frac{\pi }{4} \right)\\ \Rightarrow \theta =\alpha -\frac{\pi }{4}\Rightarrow \alpha =\theta +\frac{\pi }{4}\\ \sin \alpha +\cos \alpha =\sin \left( \theta +\frac{\pi }{4} \right)+\cos \left( \theta +\frac{\pi }{4} \right)\\ =\sqrt{2}\cos \theta \text \ and \ \sin \alpha -\cos \alpha =\sin \left( \theta +\frac{\pi }{4} \right)-\cos \left( \theta +\frac{\pi }{4} \right)\\ =\frac{1}{\sqrt{2}}\sin \theta +\frac{1}{\sqrt{2}}\cos \theta -\frac{1}{\sqrt{2}}\cos \theta +\frac{1}{\sqrt{2}}\sin \theta \\ =\frac{2}{\sqrt{2}}\sin \theta =\sqrt{2}\sin \theta\end{array} $

Example 8: Evaluate tan π/8.

Solution:

Let x = π/8

2x = π/4

We have tan 2x = 2 tan x /(1-tan² x)

So, tan π/4 = 2 tan π/8 /(1-tan² π/8)

Put y = tan π/8

∴ 1 = 2y / (1-y²)

⇒ 1-y² = 2y

⇒ y²+2y-1 = 0

y = (-2 ± √(4+4))/2

= (-2 ± 2√2)/2

= -1±√2

π/8 lies in the first quadrant. So, tan π/8 is positive.

Hence, tan π/8 = √2 -1.

Trigonometric Equations and Identities Solved Examples

Trigonometric Equations and Identities

List of Trigonometric Equations and Identities

General Solution of Some Trigonometric Equations

General Solution of Some Standard Equations

Also, Read:

Trigonometric Equations and Identities Examples

Trigonometric Equations – Important Topics

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