Trigonometry IIT JEE Study Material - Trigonometric Equations, Ratios, Expressions

Trigonometry is one of the important branches of mathematics that studies triangles and their measurements. In this article, you will learn trigonometric ratios, graphs of trigonometric functions, identities, maximum and minimum values, main formulas and much more.

Trigonometric Ratios

Trigonometric ratios 1

\begin{array}{l} \sin θ = \frac{A B}{A C} = \frac{o p p}{h y p} \end{array}

\begin{array}{l} \cos θ = \frac{B C}{A C} = \frac{a d j}{h y p} \end{array}

\begin{array}{l} \tan θ = \frac{A B}{B C} = \frac{o p p}{a d j} \end{array}

\begin{array}{l} \cot θ = \frac{B C}{A B} \end{array}

\begin{array}{l} \sec θ = \frac{A C}{B C} \end{array}

\begin{array}{l} \cos e c θ = \frac{A C}{A B} \end{array}

Trigonometric Circular Function

Trigonometric circular function

\begin{array}{l} < A O P = \frac{a r c A P}{r} = \frac{l}{r} \end{array}

cos θ = x

sin θ = y

\begin{array}{l} \tan θ = \frac{x}{y} \end{array}

Graphs of T Ratios

Sine

y = sin x

Graph of sine function

Domain R

Range (-1,1)

Cosine

y = cos x

Graph of cosine function

Tangent

y = tan x

Graph of tangent function

Co – tangent

y = cot x

Graph of co tangent function

Secant

\begin{array}{l} y = \sec x, \end{array}

Graph of secant function

\begin{array}{l} D o m a i n : R - {(2 x + 1) \frac{π}{2}} \end{array}

and

\begin{array}{l} R a n g e : (- \infty - 1) \cup (1, \infty) \end{array}

Cosecant

y = cosec x

Graph of cosecant function

\begin{array}{l} D o m a i n : R - {n π} \end{array}

and

\begin{array}{l} R a n g e : (- \infty - 1) \cup (1, \infty) \end{array}

Also Read:

Trigonometric Equations and Its Solutions

Trigonometry JEE Main Previous Year Questions

Inverse Trig JEE Main Previous Year Questions

Trignometric Identities

\begin{array}{l} \sin^{2} θ + \cos^{2} θ = 1 \end{array}

\begin{array}{l} 1 + \tan^{2} θ = \sec^{2} θ \end{array}

\begin{array}{l} 1 + \cot^{2} θ = \cos e c^{2} θ \end{array}

\begin{array}{l} \sin^{4} θ + \cos^{4} = 1 - 2 \sin^{2} θ \cos^{2} θ \end{array}

\begin{array}{l} \sin^{6} θ + \cos^{6} = 1 - 3 \sin^{2} θ \cos^{2} θ \end{array}

Remember:

\begin{array}{l} \sec θ - \tan θ = \frac{1}{\sec θ + \tan θ} \end{array}

and

\begin{array}{l} c o s e c θ - \cot θ = \frac{1}{c o s e c θ + \cot θ} \end{array}

T-Ratio at Some Standard Angles

T ratio of standard angles

T-Ratio at Some Specific Angles

T-ratio

	$\begin{array}{l} 7 \frac{1^{\circ}}{2} \end{array}$	15^o	$\begin{array}{l} 22 \frac{1^{\circ}}{2} \end{array}$	18^o
$\begin{array}{l} \sin θ \end{array}$	$\begin{array}{l} \frac{\sqrt{4 - \sqrt{2} - \sqrt{6}}}{2 \sqrt{2}} \end{array}$	$\begin{array}{l} \frac{\sqrt{3} - 1}{2 \sqrt{2}} \end{array}$	$\begin{array}{l} \frac{1}{2} \sqrt{2 - \sqrt{2}} \end{array}$	$\begin{array}{l} \frac{\sqrt{5 - 1}}{4} \end{array}$
$\begin{array}{l} \cos θ \end{array}$	$\begin{array}{l} \frac{\sqrt{4 + \sqrt{2} + \sqrt{6}}}{2 \sqrt{2}} \end{array}$	$\begin{array}{l} \frac{\sqrt{3} + 1}{2 \sqrt{2}} \end{array}$	$\begin{array}{l} \frac{1}{2} \sqrt{2 + \sqrt{2}} \end{array}$	$\begin{array}{l} \frac{1}{4} \sqrt{10 + 2 \sqrt{5}} \end{array}$
$\begin{array}{l} \tan θ \end{array}$	$\begin{array}{l} (\sqrt{3} - \sqrt{2}) (\sqrt{2} - 1) \end{array}$	$\begin{array}{l} 2 - \sqrt{3} \end{array}$	$\begin{array}{l} \sqrt{2 - 1} \end{array}$	$\begin{array}{l} \frac{\sqrt{25 + 10 \sqrt{15}}}{5} \end{array}$

Maximum and Minimum Value of Standard Trigonometric Expressions

Maximum value of
$\begin{array}{l} a \cos θ \pm b \sin θ = \sqrt{a^{2} + b^{2}} \end{array}$
Minimum value of
$\begin{array}{l} a \cos θ \pm b \sin θ = - \sqrt{a^{2} + b^{2}} \end{array}$
Maximum value of
$\begin{array}{l} a \cos θ \pm b \sin θ + c = c + \sqrt{a^{2} + b^{2}} \end{array}$
Minimum value of
$\begin{array}{l} a \cos θ \pm b \sin θ + c = c - \sqrt{a^{2} + b^{2}} \end{array}$

Problems on Trigonometry

Example 1: If

\begin{array}{l} \sec^{4} x - {c o s e c}^{4} x - 2 \sec^{2} x + 2 \cos e c^{2} x = \frac{15}{4} \end{array}

. Find tan x.

Solution:

\begin{array}{l} (\sec^{4} x - 2 \sec^{2} x) - (\cos e c^{4} x - 2 \cos e c^{2} x) = \frac{15}{4} \end{array}

or

\begin{array}{l} (\sec^{4} x - 2 \sec^{2} x + 1) - (\cos e c^{4} x - 2 \cos e c^{2} x + 1) = \frac{15}{4} \end{array}

\begin{array}{l} {(\sec^{2} x - 1)}^{2} - {(\cos e c^{2} x - 1)}^{2} = \frac{15}{4} \end{array}

\begin{array}{l} {(\tan^{2} x)}^{2} - {(\cot^{2} x)}^{2} = \frac{15}{4} \end{array}

\begin{array}{l} \tan^{4} x - \frac{1}{\tan^{4} x} = \frac{15}{4} \end{array}

\begin{array}{l} \tan^{4} x - \frac{1}{{t a n}^{4} x} = 4 - \frac{1}{4} \end{array}

On comparing,

\begin{array}{l} \tan^{4} x = 4 \end{array}

\begin{array}{l} \tan^{2} x = 2 \end{array}

\begin{array}{l} t a n x = \pm \sqrt{2} \end{array}

Example 2:

If sin θ, cos θ, and tan θ are in G.P., then

\begin{array}{l} \cos^{9} θ + \cos^{6} θ + 3 \cos^{3} θ \cos^{2} θ - 1 = ? \end{array}

Solution:

\begin{array}{l} \sin θ . t a n θ = c o s^{2} θ \end{array}

\begin{array}{l} \sin^{2} θ = \cos^{3} θ \end{array}

Now,

\begin{array}{l} {(\cos^{3} θ)}^{3} + {(\cos^{3} θ)}^{2} + 3 \cos^{3} θ \cos^{2} θ - 1 \end{array}

\begin{array}{l} = \sin^{6} θ + \sin^{4} θ + 3 \sin^{2} θ \cos^{2} θ - 1 \end{array}

\begin{array}{l} = \sin^{6} θ + \sin^{4} θ - (1 - 3 \sin^{2} θ \cos^{2} θ) \end{array}

\begin{array}{l} = \sin^{6} θ + \sin^{4} θ - (\sin^{6} θ + \cos^{6} θ) \end{array}

\begin{array}{l} = \sin^{4} θ - \cos^{6} θ \end{array}

\begin{array}{l} = \sin^{4} θ - {(\cos^{3} θ)}^{2} \end{array}

\begin{array}{l} = \sin^{4} θ - \sin^{4} θ = 0 \end{array}

Example 3:

If sec x – tan x = P, then sec x = ?

Solution:

Given: sec x – tan x = P ….(1)

We know,

\begin{array}{l} \sec x - \tan x = \frac{1}{\sec x + \tan x} \end{array}

or

\begin{array}{l} \sec x + t a n x = \frac{1}{P} \dots . (2) \end{array}

Adding (1) and (2)

\begin{array}{l} 2 \sec x = P + \frac{1}{P} \end{array}

or

\begin{array}{l} \sec x = \frac{P^{2} + 1}{2 P} \end{array}

Example 4:

Prove that

\begin{array}{l} \sin (- 420) . \cos (390) + \cos (- 660) \sin (330) = - 1 \end{array}

Solution:

\begin{array}{l} - \sin (420) \cos (390) + \cos (660) \sin 330 \end{array}

\begin{array}{l} = - \sin (2 π + 60) \cos (2 π + 30) + \cos (4 π - 60) \sin (2 π - 30) \end{array}

\begin{array}{l} = - \sin 60 \cos 30 + \cos 60 (- \sin 30) \end{array}

\begin{array}{l} = - \frac{\sqrt{3}}{2} . \frac{\sqrt{3}}{2} - \frac{1}{2} . \frac{1}{2} \end{array}

\begin{array}{l} = - \frac{3}{4} - \frac{1}{4} = - \frac{4}{4} = - 1. \end{array}

Hence proved.

Trigonometric Ratios of Compound Angles

The algebraic sum of two or more angles is generally called compound angles, and angles are known as constituent angles.

Some Important Results

$\begin{array}{l} \sin (A \pm B) = \sin A \cos B \pm \cos A s i n B \end{array}$
$\begin{array}{l} \cos (A \pm B) = \cos A \cos B \mp \sin A s i n B \end{array}$
$\begin{array}{l} \tan (A \pm B) = \frac{\tan A \pm \tan B}{1 \tan A \tan B} \end{array}$
$\begin{array}{l} \cot (A \pm B) = \frac{\cot A \cot B \mp 1}{\cot B \pm \cot A} \end{array}$
$\begin{array}{l} \sin (A + B) \sin (A - B) = \sin^{2} A - \sin^{2} B = \cos^{2} B - \cos^{2} A \end{array}$
$\begin{array}{l} \cos (A + B) \cos (A - B) = \cos^{2} A - \sin^{2} B = \cos^{2} B - \sin^{2} A \end{array}$
$\begin{array}{l} \sin (A + B + C) = \sin A \cos B \cos C + \sin B \cos A \cos C + \sin C \cos A \cos B - \sin A \sin B \sin C \end{array}$
$\begin{array}{l} \cos (A + B + C) = \cos A \cos B \cos C - \cos A \sin B \sin C - \cos B \sin A \sin C - \cos C \sin A \sin B \end{array}$
$\begin{array}{l} \tan (A + B + C) = \frac{\tan A + \tan B + \tan C - \tan A \tan B \tan C}{1 - \tan A \tan B - \tan B \tan C - \tan C \tan A} \end{array}$

Remember

If A+B+C = 0, then

\begin{array}{l} \tan A + \tan B + \tan C = \tan A \tan B \tan C \end{array}

Let us solve some problems to understand the concept in a better way.

Example 1

\begin{array}{l} If \sin α = \frac{3}{5} & \cos β = \frac{9}{41} α, β \in I quadrant . \end{array}

Find sin(α – β).

Solution

\begin{array}{l} \sin α = \frac{3}{5} \end{array}

\begin{array}{l} \sin^{2} α + \cos^{2} α = 1 \end{array}

\begin{array}{l} \cos^{2} α = 1 - \frac{9}{25} = \frac{16}{25} \end{array}

\begin{array}{l} \cos α = \frac{4}{5}, \frac{- 4}{5} \end{array}

as in first quadrant.

\begin{array}{l} \cos β = \frac{9}{41} \end{array}

\begin{array}{l} \cos^{2} β + \sin^{2} β = 1 \end{array}

\begin{array}{l} \sin^{2} β = 1 - \frac{81}{1681} = \frac{1600}{1681} \end{array}

\begin{array}{l} \sin β = \frac{40}{41} \end{array}

\begin{array}{l} \sin (α - β) = \sin α \cos β - \cos α \sin β \end{array}

\begin{array}{l} = \frac{3}{5} \times \frac{9}{41} - \frac{4}{5} \times \frac{40}{41} = \frac{27 - 160}{205} \end{array}

\begin{array}{l} = \frac{- 133}{205} . \end{array}

Example 2

\begin{array}{l} Find value of \frac{\sin (B - C)}{\cos B \cos C} + \frac{\sin (C - A)}{\cos C \cos A} + \frac{\sin (A - B)}{\cos A \cos B} . \end{array}

Solution

\begin{array}{l} \frac{\sin B \cos C - \cos B \sin C}{\cos B \cos C} + \frac{\sin C \cos A - \cos C \sin A}{\cos C \cos A} + \frac{\sin A \cos B - \cos A \sin B}{\cos A \cos B} \end{array}

\begin{array}{l} t a n B - t a n C + t a n C - t a n A + t a n A - t a n B = 0 \end{array}

Transformation Formulae

$\begin{array}{l} 2 \sin A \cos B = \sin (A + B) + \sin (A - B) \end{array}$
$\begin{array}{l} 2 \sin B \cos A = \sin (A + B) - \sin (A - B) \end{array}$
$\begin{array}{l} 2 \cos A \cos B = \cos (A + B) + \cos (A - B) \end{array}$
$\begin{array}{l} 2 \sin A \sin B = \cos (A - B) - \cos (A + B) \end{array}$
$\begin{array}{l} \sin C + \sin D = 2 \sin (\frac{C + D}{2}) \cos (\frac{C - D}{2}) \end{array}$
$\begin{array}{l} \sin C - \sin D = 2 \cos (\frac{C + D}{2}) \sin (\frac{C - D}{2}) \end{array}$
$\begin{array}{l} \cos C - \cos D = 2 \cos (\frac{C + D}{2}) \cos (\frac{C - D}{2}) \end{array}$
$\begin{array}{l} \cos C - \cos D = 2 \sin (\frac{C + D}{2}) \sin (\frac{D - C}{2}) \end{array}$

T-Ratio of Multiple Angles

$\begin{array}{l} \sin 2 A = 2 \sin A \cos A = \frac{2 \tan A}{1 + \tan^{2} A} \end{array}$
$\begin{array}{l} \cos 2 A = 2 \cos^{2} A - 1 = 1 - 2 \sin^{2} A = \frac{1 - \tan^{2} A}{1 + \tan^{2} A} = \cos^{2} A - \sin^{2} A \end{array}$
$\begin{array}{l} \tan 2 A = \frac{2 \tan A}{1 - \tan^{2} A} \end{array}$
$\begin{array}{l} \sin 3 A = 3 \sin A - 4 \sin^{3} A \end{array}$
$\begin{array}{l} \cos 3 A = 4 \cos^{3} A - 3 \cos A \end{array}$
$\begin{array}{l} \tan 3 A = \frac{3 \tan A - \tan^{3} A}{1 - 3 \tan^{2} A} \end{array}$

Remember

\begin{array}{l} 1 + \cos 2 A = 2 \cos^{2} A \end{array}

\begin{array}{l} 1 - \cos 2 A = 2 \sin^{2} A \end{array}

\begin{array}{l} \frac{1 - \cos 2 A}{1 + \cos 2 A} = \tan^{2} A \end{array}

Values of T-Ratios at Some Useful Angles

$\begin{array}{l} \sin 75^{\circ} = \frac{\sqrt{3} + 1}{2 \sqrt{2}} = \cos 15^{\circ} \end{array}$
$\begin{array}{l} \cos 75^{\circ} = \frac{\sqrt{3} - 1}{2 \sqrt{2}} = \sin 15^{\circ} \end{array}$
$\begin{array}{l} \tan 75^{\circ} = 2 + \sqrt{3} = \cot 15^{\circ} \end{array}$
$\begin{array}{l} \cot 75^{\circ} = 2 - \sqrt{3} = \tan 15^{\circ} \end{array}$
$\begin{array}{l} \sin 9^{\circ} = \frac{\sqrt{3 + \sqrt{5}} - \sqrt{5 - \sqrt{5}}}{4} = \cos 81^{\circ} \end{array}$
$\begin{array}{l} \cos 9^{\circ} = \frac{\sqrt{3 + \sqrt{5}} - \sqrt{5 - \sqrt{5}}}{4} = \sin 81^{\circ} \end{array}$

Important Results

\begin{array}{l} \sin θ \sin (60 - θ) \sin (60 + θ) = \frac{1}{4} \sin 3 θ \end{array}

\begin{array}{l} \cos θ \cos (60 - θ) \cos (60 + θ) = \frac{1}{4} \cos 3 θ \end{array}

\begin{array}{l} \tan θ \tan (60 - θ) \tan (60 + θ) = \tan 3 θ \end{array}

Let’s solve a few more examples

Example 1

\begin{array}{l} Find value of \frac{\sin A \sin 2 A + \sin 3 A \sin 6 A + \sin 4 A \sin 13 A}{2 \sin A \cos 2 A + s i n 3 A c o s 6 A + s i n 4 A c o s 13 A} . \end{array}

Solution

\begin{array}{l} \frac{\cos (- A) - \cos (3 A) + \cos 3 A - \cos 9 A + \cos 9 A - \cos 17 A}{\sin 3 A + \sin (- A) + \sin 9 A + s i n (- 3 A) + \sin (17 A) + \sin (- 9 A)} \end{array}

\begin{array}{l} = \frac{\cos A - \cos 17 A}{\sin 17 A - \sin A} \end{array}

\begin{array}{l} = \frac{2 \sin 9 A \sin 8 A}{2 \cos 9 A \sin 8 A} \end{array}

= tan 9A

Example 2

\begin{array}{l} \sin^{2} 5^{\circ} + \sin^{2} 10^{\circ} + \sin^{2} 15^{\circ} + \dots \dots \dots . + \sin^{2} 90^{\circ} = ? \end{array}

Solution

\begin{array}{l} \sin^{2} 5^{\circ} + \sin^{2} 10^{\circ} + \dots \dots \dots . + \sin^{2} 85^{\circ} + \sin^{2} 90^{\circ} \end{array}

\begin{array}{l} = \sin^{2} 5^{\circ} + \sin^{2} 10^{\circ} + \dots \dots \dots . + \cos^{2} 5^{\circ} + \sin^{2} 90^{\circ} \end{array}

\begin{array}{l} = (\sin^{2} 5^{\circ} + \cos^{2} 5^{\circ}) + \dots \dots \dots . + \sin^{2} 45^{\circ} + \sin^{2} 90^{\circ} \end{array}

= [1 + 1 + … + 1] + (1/√2)² + 1

\begin{array}{l} = 8 \times 1 + 1 + \frac{1}{2} \end{array}

\begin{array}{l} = 8 + 1 + \frac{1}{2} = 9 \frac{1}{2} \end{array}

Example 3

Prove that

\begin{array}{l} \sin α + s i n (α + \frac{2 π}{3}) + \sin (α + \frac{4 π}{3}) = 0 \end{array}

Solution

\begin{array}{l} 2 \sin (\frac{2 α}{2} + \frac{4 π}{6}) \cos (- \frac{4 π}{6}) + \sin (α + \frac{2 π}{3}) \end{array}

\begin{array}{l} = 2 \sin (α + \frac{2 π}{3}) \cos (\frac{2 π}{3}) + \sin (α + \frac{2 π}{3}) \end{array}

\begin{array}{l} = 2 \times (- \frac{1}{2}) \sin (α + \frac{2 π}{3}) + \sin (α + \frac{2 π}{3}) \end{array}

\begin{array}{l} = - \sin (α + \frac{2 π}{3}) + \sin (α + \frac{2 π}{3}) \end{array}

= 0

Example 4

\begin{array}{l} If a \sin θ = b \sin (θ + \frac{2 π}{3}) = c \sin (θ + \frac{4 π}{3}), then \end{array}

prove that ab + bc + ca = 0.

Solution

\begin{array}{l} a \sin θ = b \sin (θ + \frac{2 π}{3}) = c \sin (θ + \frac{4 π}{3}) = k \end{array}

\begin{array}{l} \sin θ = \frac{k}{a} \end{array}

\begin{array}{l} \sin (θ + \frac{2 π}{3}) = \frac{k}{b} \end{array}

\begin{array}{l} \sin (θ + \frac{4 π}{3}) = \frac{k}{c} \end{array}

We know

\begin{array}{l} \sin θ + \sin (θ + \frac{2 π}{3}) + \sin (θ + \frac{4 π}{3}) = 0 \end{array}

\begin{array}{l} \frac{k}{a} + \frac{k}{b} + \frac{k}{c} = 0 \end{array}

\begin{array}{l} \frac{k (b c + a c + a b)}{a b c} = 0 \end{array}

\begin{array}{l} a b + b c + c a = 0 \end{array}

Summation of Series in Trigonometry

1.

\begin{array}{l} \sin α + \sin (α + β) + \sin (α + 2 β) + \dots \dots \dots . + \sin (α + (n - 1) β) = \frac{\sin [α + \frac{(n - 1)}{2} β] \sin \frac{n β}{2}}{\sin \frac{β}{2}} \end{array}

When the angles of sine are in AP, the sum of the sine series is given by the above formula.

However, if α = β in the above case,

Then

\begin{array}{l} \sin α + \sin 2 α + \sin 3 α + \dots \dots \dots . + \sin α = \frac{\sin (\frac{n + 1}{2}) α \sin \frac{n α}{2}}{\sin \frac{α}{2}} \end{array}

2. When cosine angles are in AP,

\begin{array}{l} c o s α + c o s 2 α + \cos α + \dots \dots \dots . + \cos n α = \frac{\cos (\frac{n + 1}{2}) α \sin \frac{n α}{2}}{\sin \frac{α}{2}} \end{array}

Remember

If A, B and C are angles of Δ,

$\begin{array}{l} \sin (B + C) = \sin A \end{array}$
$\begin{array}{l} \cos (B + C) = - \cos A \end{array}$
$\begin{array}{l} \sin (\frac{B + C}{2}) = \cos \frac{A}{2} \end{array}$
$\begin{array}{l} \cos (\frac{B + C}{2}) = \sin \frac{A}{2} \end{array}$
$\begin{array}{l} \sin 2 A + \sin 2 B + \sin 2 C = 4 \sin A \sin B \sin C \end{array}$
$\begin{array}{l} \cos 2 A + \cos 2 B + \cos 2 C = 1 - 4 \cos A \cos B \cos C \end{array}$
$\begin{array}{l} \tan A + \tan B + \tan C = \tan A \tan B \tan C \end{array}$

Trigonometric Equations

A solution of a trigonometric equation is the value of the unknown angle that satisfies the equation.

Following are general solutions of trigonometric equations in the standard form:

Equation	General Solution
$\begin{array}{l} \sin θ = 0 \end{array}$	$\begin{array}{l} θ = n π n \in z \end{array}$
$\begin{array}{l} \cos θ = 0 \end{array}$	$\begin{array}{l} θ = (2 n + 1) \frac{π}{2} n \in z \end{array}$
$\begin{array}{l} \tan θ = 0 \end{array}$	$\begin{array}{l} θ = n π, n \in z \end{array}$
$\begin{array}{l} \sin θ = \sin α \end{array}$	$\begin{array}{l} θ = n π + {(- 1)}^{n} α, n \in z \end{array}$
$\begin{array}{l} \cos θ = \cos α \end{array}$	$\begin{array}{l} θ = 2 n π \pm α, n \in z \end{array}$
$\begin{array}{l} \tan θ = \tan α \end{array}$	$\begin{array}{l} θ = n π + α, n \in z \end{array}$

\begin{array}{l} \begin{array}{c} \sin^{2} θ = \sin^{2} α \\ \cos^{2} θ = \cos^{2} α \\ \tan^{2} θ = \tan^{2} α \end{array}} θ = n π \pm α, n \in z \end{array}

Example 1

Solve

\begin{array}{l} 4 \sin x \sin 2 x \sin 4 x = \sin 3 x \end{array}

Solution

\begin{array}{l} 4 \sin x \sin (3 x - x) \sin (3 x + x) = \sin 3 x \end{array}

,

\begin{array}{l} 4 \sin x (\sin^{2} 3 x - \sin^{2} x) = \sin 3 x \end{array}

,

\begin{array}{l} 4 \sin x . \sin^{2} 3 x - 4 \sin^{3} x = 3 \sin x - 4 \sin^{3} x \end{array}

,

\begin{array}{l} 4 \sin x \sin^{2} 3 x - 3 \sin x = 0 \end{array}

,

\begin{array}{l} \sin x (4 \sin^{2} 3 x - 3) = 0 \end{array}

,

\begin{array}{l} \sin x = 0 a n d \sin^{2} 3 x = \frac{3}{4} \end{array}

,

\begin{array}{l} n \in z \end{array}

,

\begin{array}{l} \sin^{2} 3 x = \frac{3}{4} \end{array}

,

\begin{array}{l} \sin^{2} 3 x = {(\frac{\sqrt{3}}{4})}^{2} = \sin^{2} \frac{π}{3} \end{array}

,

\begin{array}{l} 3 x = m π \pm \frac{π}{3} m \in z \end{array}

,

\begin{array}{l} x = \frac{m π}{3} \pm \frac{π}{9} \end{array}

Example 2

Solve

\begin{array}{l} \sin 3 θ + \cos 2 θ = 0 \end{array}

Solution

\begin{array}{l} \cos 2 θ = - \sin 3 θ \end{array}

.

Taking +ve sign

\begin{array}{l} θ = 2 n π + \frac{π}{2} + 3 θ \end{array}

,

\begin{array}{l} - θ = 2 n π + \frac{π}{2} \end{array}

,

\begin{array}{l} θ = - 2 n π - \frac{π}{2} \end{array}

or

Taking -ve sign

\begin{array}{l} θ = 2 n π + \frac{π}{2} - 3 θ; n \in z \end{array}

,

\begin{array}{l} 4 θ = 2 n π + \frac{π}{2} \end{array}

,

\begin{array}{l} θ = \frac{n π}{2} + \frac{π}{8} \end{array}

Example 3

Solve

\begin{array}{l} \sqrt{3} \sec 2 θ = 2 \end{array}

Solution

\begin{array}{l} \frac{\sqrt{3}}{\cos 2 θ} = 2 \end{array}

,

\begin{array}{l} \cos 2 θ = \frac{\sqrt{3}}{2} = \cos \frac{π}{6} \end{array}

,

\begin{array}{l} 2 θ = 2 n π \pm \frac{π}{6} n \in z \end{array}

,

\begin{array}{l} θ = n π \pm \frac{π}{12} n \in z \end{array}

Properties and Solutions of Triangle

Solutions of Triangle

Solutions of triangle

Here A, B and C are the angles, and a, b and c are the lengths of sides of the triangle ABC.

Sine Rule

The sine rule relates the length of the sides of a triangle to the sines of its angles.

\begin{array}{l} \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2 R \end{array}

Where 2R is the diameter of the triangle’s circumcircle.

Cosine Rule

\begin{array}{l} \cos A = \frac{b^{2} + c^{2} - a^{2}}{2 b c} \end{array}

and

\begin{array}{l} \cos B = \frac{a^{2} + c^{2} - b^{2}}{2 a c} \end{array}

and

\begin{array}{l} \cos C = \frac{a^{2} + b^{2} - c^{2}}{2 a b} \end{array}

Area of Δ

\begin{array}{l} = \frac{1}{2} b c \sin A \end{array}

\begin{array}{l} = \frac{1}{2} c a \sin B \end{array}

\begin{array}{l} = \frac{1}{2} a b \sin C \end{array}

\begin{array}{l} = \sqrt{S (S - a) (S - b) (S - c)} \end{array}

\begin{array}{l} = \frac{1}{2} \times b a s e \times h e i g h t \end{array}

Projection Formula

a=b cos c+c cos B

b=c cos A+a cos C

c=a cos B+b cos A

Napier’s Analogy

This is also called the Law of Tangents.

\begin{array}{l} \tan (\frac{B - C}{3}) = (\frac{b - c}{b + c}) \cot \frac{A}{2} \end{array}

,

\begin{array}{l} \tan (\frac{C - A}{2}) = (\frac{c - a}{c + a}) \cot \frac{B}{2} \end{array}

,

\begin{array}{l} \tan (\frac{A - B}{2}) = (\frac{a - b}{a + b}) \cot \frac{C}{2} \end{array}

Example

In a triangle ABC,

\begin{array}{l} \frac{a + b}{13} = \frac{a + c}{12} = \frac{b + c}{11} \end{array}

Then prove that

\begin{array}{l} \frac{\cos A}{7} = \frac{\cos B}{19} = \frac{\cos C}{25} \end{array}

Solution

\begin{array}{l} \frac{b + c}{11} = \frac{c + a}{12} = \frac{a + b}{13} = k \end{array}

,

\begin{array}{l} b + c = 11 k, c + a = 12 k, a + b = 13 k \end{array}

,

\begin{array}{l} 2 (a + b + c) = 36 k \end{array}

,

\begin{array}{l} a + b + c = 18 k \end{array}

b+c=11k

a=7k

Similarly, b=6k

c=5k

\begin{array}{l} \cos A = \frac{b^{2} c^{2} - a^{2}}{2 b c} = \frac{36 k^{2} + 25 k^{2} - 49 k^{2}}{60 k_{2}} = \frac{1}{5} \end{array}

Similarly,

\begin{array}{l} \cos B = \frac{19}{35} a n d \cos C = \frac{5}{7} \end{array}

,

\begin{array}{l} \cos A : \cos B : \cos C = \frac{1}{5} : \frac{19}{35} : \frac{8}{7} \end{array}

=7:19:25

Circum Circle of a Triangle

The circle passing through the vertices of a triangle is called the circum circle of a triangle.

The radius is the circum radius.

\begin{array}{l} R = \frac{a}{2 s i n A} = \frac{b}{2 \sin B} = \frac{c}{2 \sin C} \end{array}

Also,

\begin{array}{l} R = \frac{a b c}{4 Δ} \end{array}

;

\begin{array}{l} Δ i s t h e a r e a o f Δ A B C \end{array}

Inradius of a Triangle

\begin{array}{l} r = \frac{Δ}{s} w h e r e Δ = a r e a o f Δ \end{array}

,

\begin{array}{l} s = \frac{a + b + c}{2} \end{array}

,

\begin{array}{l} r = (s - a) \tan \frac{A}{2} = (s - b) \tan \frac{B}{2} = (s - c) \tan \frac{C}{2} \end{array}

,

\begin{array}{l} r = 4 R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \end{array}

Example

In a triangle ABC, angle C = 60 degrees.

Then prove that

\begin{array}{l} \frac{1}{a + c} + \frac{1}{b + c} = \frac{3}{a + b + c} \end{array}

Solution

\begin{array}{l} \cos c = \frac{1}{2} = \frac{a^{2} + b^{2} - c^{2}}{2 b a} \end{array}

,

\begin{array}{l} a b = a^{2} + b^{2} - c^{2} \to (i) \end{array}

,

\begin{array}{l} \frac{1}{a + c} + \frac{1}{b + c} = \frac{3}{a + b + c} \end{array}

,

\begin{array}{l} \frac{a + b + c + c}{(a + c) (b + c)} = \frac{3}{a + b + c} \end{array}

,

\begin{array}{l} (a + b + 2 c) (a + b + c) = 3 (a + c) (b + c) \end{array}

,

\begin{array}{l} a^{2} + b^{2} + 2 a b + 2 c^{2} + 3 a c + 3 b c = 3 a b + 3 a c + 3 b c + 3 c^{2} \end{array}

,

\begin{array}{l} a^{2} b^{2} - a b = c^{2} \to (i i) \end{array}

From (i) and (ii), we can say

\begin{array}{l} \frac{1}{a + c} + \frac{1}{b + c} = \frac{3}{a + b + c} \end{array}

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Frequently Asked Questions

Q1

What is trigonometry?

Trigonometry is a branch of mathematics that deals with the relationship between the sides of a triangle (right triangle) with its angles.

Q2

Name the six basic trigonometric functions.

The basic trigonometric functions are sin, cos, tan, sec, cosec and cot.

Q3

Give the Pythagorean trigonometric identities.

sin²a + cos²a = 1
1 + tan²a = sec²a
1 + cot² a = cosec²a

Q4

Give an application of trigonometry.

The height of an object or the distance between two distinct objects can be calculated using trigonometric ratios.

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