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 to solve trigonometric problems.

Consider a right-angled triangle ABC. Here ∠CAB = A and ∠BCA = 90°. AB is called 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 denoted as sin A.

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

BC / AC = (Opposite side) /(adjacent side), gives the tangent of A and 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 good for every angle is called a trigonometric identity.

Steps to solve a trigonometric equation are as follows:

Type – I

The first type of equations involves factorising them or expressing in quadratic form. They can be solved by converting them into factors and find 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 identities problems are given below. JEE aspirants are advised to learn 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 = 180o

⇒ x = 18o

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

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

Ratio = sin126o / sin(18o)

= [√5+1] / [√5−1]

Example 3: The value of sin25o + sin210o + sin215o + . . . + sin285o + sin290o is equal to

Solution:

Given expression is sin25o + sin210o + sin215o + . . . + sin285o + sin290o.

We know that sin 90o = 1 or sin2 90o = 1.

Similarly, sin 45o=1 / √2 or sin245o = 1 / 2 and the angles are in A.P. of 18 terms.

We also know that sin285o = [sin (90o−5o)]2 = cos25o.

Therefore, from the complementary rule, we find sin25o + sin285o = sin25o +cos25o = 1

Therefore, sin25o + sin210o + sin215o + . . . + sin285o + sin290o =(1 + 1 + 1+ 1 + 1 + 1 + 1 + 1 ) + 1 + 1 / 2 = 9 [1 / 2].

Example 4: Find the value of 1 + cos 56o + cos58o − cos66o.

Solution:

1 + cos 56o + cos58o − cos66o

Apply the conditional identity

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

We get the value of the required expression equal to 4 cos28o cos29o sin33o.

Example 5: If

\(\begin{array}{l}a\,{{\cos }^{3}}\alpha +3a\,\cos \alpha \,{{\sin }^{2}}\alpha =m\end{array} \)
and
\(\begin{array}{l}a\,{{\sin }^{3}}\alpha +3a\,{{\cos }^{2}}\alpha \sin \alpha =n, \text \ then \ find \ {{(m+n)}^{2/3}}+{{(m-n)}^{2/3}}.\end{array} \)

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

\(\begin{array}{l}\sin A=n\sin B,\end{array} \)
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

\(\begin{array}{l}\tan \theta =\frac{\sin \alpha -\cos \alpha }{\sin \alpha +\cos \alpha }, \text \ then \ find \ \sin \alpha +\cos \alpha\end{array} \)
and
\(\begin{array}{l}\sin \alpha -\cos \alpha\end{array} \)
.

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 – Important Topics

Trigonometric Equations - Important Topics
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