Class 11 Mathematics Trigonometry Notes | Inverse Trigonometric Functions & Equations

.rll-youtube-player .play{--wpr-bg-fcebace7-a573-43e6-a411-1513d17a0068: url('https://nepalenotes.com/wp-content/plugins/wp-rocket/assets/img/youtube.png');}

Trigonometry – Original PDF

Class 11 Mathematics Trigonometry Notes

Typed notes from the supplied 63-page handwritten PDF. The chapter covers inverse trigonometric functions, identities, solved examples, general solutions of trigonometric equations and Exercise 7.2.

Contents

Inverse Trigonometric Functions
Important Inverse Trigonometric Identities
Solved Examples
General Solution of Trigonometric Equations
Exercise 7.2
More Trigonometric Equations

Inverse Trigonometric Functions

Example 1

(a) Show that θ = sin^-1(sin θ)

Let sin θ = x.
Therefore, θ = sin^-1x = sin^-1(sin θ).

(b) Show that θ = cos^-1(cos θ)

Let cos θ = x.
Therefore, θ = cos^-1x = cos^-1(cos θ).

(c) Show that θ = tan^-1(tan θ)

Let tan θ = x.
Therefore, θ = tan^-1x = tan^-1(tan θ).

Example 2

The notes similarly establish:

θ = sin(sin^-1θ)
θ = cos(cos^-1θ)

Example 3

For a given angle x:

sin^-1x = cosec^-1(1/x)

cos^-1x = sec^-1(1/x)

tan^-1x = cot^-1(1/x)

Important Inverse Trigonometric Identities

Example 4: Express sin^-1x in other inverse circular functions

(a)

sin^-1x = cos^-1√(1-x²)

(b)

sin^-1x = tan^-1(x/√(1-x²))

(c)

sin^-1x = sec^-1(1/√(1-x²))

(d)

sin^-1x = cot^-1(√(1-x²)/x)

(e)

sin^-1x = cosec^-1(1/x)

Example 5

For any numerical value of x:

sin^-1x + cos^-1x = π/2

tan^-1x + cot^-1x = π/2

cosec^-1x + sec^-1x = π/2

Example 6

For a given numerical value of t:

sin^-1(-t) = -sin^-1t

cos^-1(-t) = π – cos^-1t

tan^-1(-t) = -tan^-1t

cot^-1(-t) = π – cot^-1t

Solved Examples

Example 7

Prove that:

2sin^-1x = sin^{-1
The source lets sin^-1x = θ, then uses sin 2θ = 2sinθ cosθ.}

Example 8

Using the tangent addition formula, the notes prove:

tan^-1x + tan^-1y = tan^-1((x+y)/(1-xy))

subject to the principal-value conditions used in the handwritten solution.

Example 9

sin^-1x + sin^-1y = sin^-1

Example 10

The notes show:

cos(sin^-1x + cos^-1y) = y√(1-x²) – x√(1-y²)

Example 11

Prove:

2tan^-1(1/3) + tan^-1(1/7) = π/4

The source combines the inverse tangents using the tangent addition formula and reaches tan^-1(1) = π/4.

Example 12

The source proves an identity involving:

tan(2tan^-1x) = 2tan(tan^-1x + tan^-1x)

and simplifies the right-hand side using tan(A+B).

Example 13

The handwritten notes solve an expression of the form:

sin^-1(2a/(1+a²)) – cos^{-1
by putting a = tan A and b = tan B, and converting the fractions to sin 2A and cos 2B.}

Example 14

The source proves a relation involving cot^-13 and cosec^-1√5 by constructing right triangles and evaluating cos(A-B).

Example 16

Find the value of:

tan(1/2 cos^-1(√5/3))

The source lets cos^-1(√5/3) = 2A and uses the half-angle identity:

tan A = √((1-cos2A)/(1+cos2A))

Example 16 (next example in source)

Find:

sec²(tan^-12) + cosec²(cot^-13)

The source simplifies:

= 1 + tan²(tan^-12) + 1 + cot²(cot^-13)
= 1 + 4 + 1 + 9 = 15

Example 17

Solve:

sin^-1(x/2) = cos^-1x

The source converts cos^-1x to a square-root form, squares both sides and obtains:

x = ±2/√5

Example 18

The source proves an identity involving:

sin^-1(4/5) + sin^-1(5/13) + sin^-1(16/65)

using the inverse-sine addition formula repeatedly and obtains:

π/2

Example 19

If:

tan^-1a + tan^-1b + tan^-1c = π

then the source proves:

a + b + c = abc

Example 20

If:

tan^-1x + tan^-1y + tan^-1z = π/2

the source proves:

xy + yz + zx = 1

Example 21

If:

cos^-1x + cos^-1y + cos^-1z = π

then the source proves:

x² + y² + z² + 2xyz = 1

Example 22

If:

sin^-1x + sin^-1y + sin^-1z = π

the source derives a relation involving x, y and z by expanding sin(A+B+C).

Example 23

Solve:

tan^-1(2x/(1-x²)) + cot^-1((1-x²)/(2x)) = π/3

The source converts the cotangent inverse and uses the double-angle identity, obtaining:

x = tan 15°

Example 24

Prove:

2tan^-1x = sin^{-1-1-1
The source uses the double-angle formulas for sine, cosine and tangent.}

Example 25

The notes prove the half-angle inverse relations:

cos^-1x = 2sin^-1√((1-x)/2)

cos^-1x = 2cos^-1√((1+x)/2)

Example 26

Find the value of:

cos(tan^-1(sin(cot^-1x)))

The source reduces the nested expression using:

sin(cot^-1x) = 1/√(1+x²)

and then simplifies the cosine of an inverse tangent.

Example 27

The source establishes an identity involving:

tan^-1( (√(1+x²)+√(1-x²)) / (√(1+x²)-√(1-x²)) )

by setting x² = cos 2θ and using half-angle identities.

General Solution of Trigonometric Equations

Basic General Solutions

sin x = 0 ⇒ x = nπ, n ∈ Z

cos x = 0 ⇒ x = (2n+1)π/2, n ∈ Z

tan x = 0 ⇒ x = nπ, n ∈ Z

cot x = 0 ⇒ x = (2n+1)π/2, n ∈ Z

General solution of sin x = k

Let sin θ = k. Then the source derives:

x = nπ + (-1)ⁿθ, n ∈ Z

General solution of cos x = k

Let cos θ = k. Then:

x = 2nπ ± θ, n ∈ Z

General solution of tan x = k

Let tan θ = k. Then:

x = nπ + θ, n ∈ Z

General solution of sin 2x = sin 2θ

The source obtains:

x = nπ ± θ

Exercise 7.2

1. Find the general value

(a) sin x = 1/2

x = nπ + (-1)ⁿπ/6, n ∈ Z

(b) cos x = √3/2

x = 2nπ ± π/6, n ∈ Z

(c) tan x = 1/√3

x = nπ + π/6, n ∈ Z

(d) sec x = √2

cos x = 1/√2
x = 2nπ ± π/4, n ∈ Z

(e) cosec x = 1

sin x = 1
x = nπ + (-1)ⁿπ/2, n ∈ Z

(f) cot x = -√3

tan x = -1/√3
x = nπ – π/6, n ∈ Z

(g) tan x = 7/4

x = nπ + tan^-1(7/4), n ∈ Z

(h) tan x + √3 = 0

tan x = -√3
x = nπ – π/3, n ∈ Z

(i) 2cos x + 1 = 0

cos x = -1/2

(j) 2sin x – 1 = 0

sin x = 1/2
x = nπ + (-1)ⁿπ/6

(k) tan 2x = -√3

2x = nπ – π/3
x = nπ/2 – π/6

2. Find the general values

(a) cos 4x = cos 2x

The source factors the difference and obtains solution families including:

x = nπ/3, nπ

(b) tan 3x = tan 2x

x = nπ

(c) sin 5x – sin x = 0

Using sin A – sin B:

2cos 3x sin 2x = 0

The source then obtains corresponding solution families.

(d) sin 2x = cos 2x

tan 2x = 1
x = (4n+1)π/8

(e) sin 4x + sin 2x = 0

2sin 3x cos x = 0

(f) sec 3x = sec 2x

The source converts secants to cosines and factors the result to obtain solution families including:

x = 2nπ/5, 2nπ

3. Find the general values

(a) 2cos²x – 1 = 0

cos²x = 1/2
cos x = 1/√2
x = 2nπ ± π/4

(b) 4sin²x – 1 = 0

sin²x = 1/4

(c) tan²x – 3 = 0

tan x = √3
x = nπ + π/3

(d) cosec 2x – 2 = 0

sin 2x = 1/2

More Trigonometric Equations

5(a). sin 2θ + cos θ = 0

2sinθ cosθ + cosθ = 0
cosθ(2sinθ + 1) = 0

Thus the source considers:

cosθ = 0
or
sinθ = -1/2

5(c). cos x + cos 2x + cos 3x = 0

2cos 2x cos x + cos 2x = 0
cos 2x(2cos x + 1) = 0

5(d). cos x + sin x = cos 2x + sin 2x

The source transforms the equation using sum-to-product identities and obtains two families of solutions.

5(e). sin 9θ – sin θ = 0

2cos 5θ sin 4θ = 0

5(f). sin θ + sin 2θ + sin 3θ + sin 4θ = 0

The source groups terms using sum-to-product identities and reduces the equation to factors involving cosθ and sine terms.

5(g). cosec θ = cot θ + 1

1/sinθ = cosθ/sinθ + 1
1 – cosθ = sinθ

The source converts this with half-angle identities and reaches:

tan(θ/2) = 1

6(a). √3 sinθ – cosθ = √2

Dividing by 2:

(√3/2)sinθ – (1/2)cosθ = 1/√2

The source rewrites the left side as sin(θ – π/6) and finds the general solution.

6(b). sinθ + cosθ = √2

(1/√2)sinθ + (1/√2)cosθ = 1
cos(θ – π/4) = 1

θ = 2nπ + π/4

6(c). sinθ – √3 cosθ = 2

The source divides by 2 and rewrites the expression using sin(θ – π/3).

Additional equation: cos x + √3 sin x = √2

The source divides by 2:

(1/2)cosx + (√3/2)sinx = 1/√2

and converts it to:

cos(x – π/3) = 1/√2

7(a). 2cos²x – 5cosx + 2 = 0, 0 ≤ x ≤ 2π

The source factors:

(cosx – 2)(2cosx – 1) = 0

Since cosx = 2 is impossible:

cosx = 1/2

Within the stated interval, the source obtains the admissible values.

7(b). sin²θ – cosθ = 1/4, 0 ≤ θ ≤ 2π

The source substitutes sin²θ = 1 – cos²θ, forms a quadratic in cosθ and solves the admissible branch.

7(c). 2sin²x + 3cosx = 0, 0 ≤ x ≤ 2π

Using sin²x = 1 – cos²x:

2(1-cos²x) + 3cosx = 0

The source factors the resulting quadratic and selects the possible cosine value.

7(d). √3 cosx + sinx = 1, -2π ≤ x ≤ 2π

The source divides by 2 and writes:

cos(x – π/6) = 1/2

then lists the values lying in the stated interval.

Example 13: tanθ + tan2θ + tan3θ = 0

The source uses:

tan3θ = (tan2θ + tanθ)/(1 – tan2θ tanθ)

and derives the solution families shown in the handwritten notes.

Final example

Solve:

tan3θ + tan2θ = 2tan2θ

The source reduces this to:

tan3θ – tan2θ = 0
sin(3θ – 2θ) = 0
sinθ = 0

θ = nπ, n ∈ Z

Note: This typed version follows the supplied handwritten PDF closely. Some expressions on a few pages are faint or partially obscured, so those portions are summarized conservatively instead of being silently corrected or invented.

Discussion

Share a helpful question, idea, or explanation with other students.

Class 11 Mathematics Trigonometry Notes

Inverse Trigonometric Functions

Example 1

(a) Show that θ = sin-1(sin θ)

(b) Show that θ = cos-1(cos θ)

(c) Show that θ = tan-1(tan θ)

Example 2

Example 3

Important Inverse Trigonometric Identities

Example 4: Express sin-1x in other inverse circular functions

(a)

(b)

(c)

(d)

(e)

Example 5

Example 6

Solved Examples

Example 7

Example 8

Example 9

Example 10

Example 11

Example 12

Example 13

Example 14

Example 16

Example 16 (next example in source)

Example 17

Example 18

Example 19

Example 20

Example 21

Example 22

Example 23

Example 24

Example 25

Example 26

Example 27

General Solution of Trigonometric Equations

Basic General Solutions

General solution of sin x = k

General solution of cos x = k

General solution of tan x = k

General solution of sin 2x = sin 2θ

Exercise 7.2

1. Find the general value

(a) sin x = 1/2

(b) cos x = √3/2

(c) tan x = 1/√3

(d) sec x = √2

(e) cosec x = 1

(f) cot x = -√3

(g) tan x = 7/4

(h) tan x + √3 = 0

(i) 2cos x + 1 = 0

(j) 2sin x – 1 = 0

(k) tan 2x = -√3

2. Find the general values

(a) cos 4x = cos 2x

(b) tan 3x = tan 2x

(c) sin 5x – sin x = 0

(d) sin 2x = cos 2x

(e) sin 4x + sin 2x = 0

(f) sec 3x = sec 2x

3. Find the general values

(a) 2cos²x – 1 = 0

(b) 4sin²x – 1 = 0

(c) tan²x – 3 = 0

(d) cosec 2x – 2 = 0

More Trigonometric Equations

5(a). sin 2θ + cos θ = 0

5(c). cos x + cos 2x + cos 3x = 0

5(d). cos x + sin x = cos 2x + sin 2x

5(e). sin 9θ – sin θ = 0

5(f). sin θ + sin 2θ + sin 3θ + sin 4θ = 0

5(g). cosec θ = cot θ + 1

6(a). √3 sinθ – cosθ = √2

6(b). sinθ + cosθ = √2

6(c). sinθ – √3 cosθ = 2

(a) Show that θ = sin^-1(sin θ)

(b) Show that θ = cos^-1(cos θ)

(c) Show that θ = tan^-1(tan θ)

Example 4: Express sin^-1x in other inverse circular functions