RD Sharma Solutions Class 12 Maths Exercise 4.7 Chapter 4 Inverse Trigonometric functions are provided here. Students can use the PDF of solutions to get their doubts cleared instantly and solve problems as per the latest exam pattern. They also help students to self analyse their understanding of the concepts covered under each exercise of Chapter 4.
To perform better in the board exam, students can make use of RD Sharma Solutions Class 12 Maths Chapter 4 Inverse Trigonometric functions Exercise 4.7 from the links given here. This exercise deals with the properties of Inverse Trigonometric Functions.
RD Sharma Solutions for Class 12 Chapter 4 – Inverse Trigonometric Functions Exercise 4.7
Access answers to Maths RD Sharma Solutions for Class 12 Chapter 4 – Inverse Trigonometric Functions Exercise 4.7
Exercise 4.7 Page No: 4.42
1. Evaluate each of the following:
(i) sin-1(sin π/6)
(ii) sin-1(sin 7π/6)
(iii) sin-1(sin 5π/6)
(iv) sin-1(sin 13π/7)
(v) sin-1(sin 17π/8)
(vi) sin-1{(sin – 17π/8)}
(vii) sin-1(sin 3)
(viii) sin-1(sin 4)
(ix) sin-1(sin 12)
(x) sin-1(sin 2)
Solution:
(i) Given sin-1(sin π/6)
We know that the value of sin π/6 is ½
By substituting this value in sin-1(sin π/6)
We get, sin-1 (1/2)
Now let y = sin-1 (1/2)
Sin (π/6) = ½
The range of principal value of sin-1(-π/2, π/2) and sin (π/6) = ½
Therefore, sin-1(sin π/6) = π/6
(ii) Given sin-1(sin 7π/6)
But we know that sin 7π/6 = – ½
By substituting this in sin-1(sin 7π/6) we get,
Sin-1 (-1/2)
Now let y = sin-1 (-1/2)
– Sin y = ½
– Sin (π/6) = ½
– Sin (π/6) = sin (- π/6)
The range of principal value of sin-1(-π/2, π/2) and sin (- π/6) = – ½
Therefore, sin-1(sin 7π/6) = – π/6
(iii) Given sin-1(sin 5π/6)
We know that the value of sin 5π/6 is ½
By substituting this value in sin-1(sin 5π/6)
We get, sin-1 (1/2)
Now let y = sin-1 (1/2)
Sin (π/6) = ½
The range of principal value of sin-1(-π/2, π/2) and sin (π/6) = ½
Therefore, sin-1(sin 5π/6) = π/6
(iv) Given sin-1(sin 13π/7)
The given question can be written as sin (2π – π/7)
Sin (2π – π/7) can be written as sin (-π/7) [since sin (2π – θ) = sin (-θ)]
By substituting these values in sin-1(sin 13π/7) we get sin-1(sin – π/7)
As sin-1(sin x) = x with x ∈ [-π/2, π/2]
Therefore, sin-1(sin 13π/7) = – π/7
(v) Given sin-1(sin 17π/8)
The given question can be written as sin (2π + π/8)
Sin (2π + π/8) can be written as sin (π/8)
By substituting these values in sin-1(sin 17π/8) we get sin-1(sin π/8)
As sin-1(sin x) = x with x ∈ [-π/2, π/2]
Therefore, sin-1(sin 17π/8) = π/8
(vi) Given sin-1{(sin – 17π/8)}
But we know that – sin θ = sin (-θ)
Therefore, (sin -17π/8) = – sin 17π/8
– Sin 17π/8 = – sin (2π + π/8) [since sin (2π – θ) = -sin (θ)]
It can also be written as – sin (π/8)
– Sin (π/8) = sin (-π/8) [since – sin θ = sin (-θ)]
By substituting these values in sin-1{(sin – 17π/8)} we get,
Sin-1(sin – π/8)
As sin-1(sin x) = x with x ∈ [-π/2, π/2]
Therefore, sin-1(sin -π/8) = – π/8
(vii) Given sin-1(sin 3)
We know that sin-1(sin x) = x with x ∈ [-π/2, π/2], which is approximately equal to [-1.57, 1.57]
But here, x = 3, which does not lie in the above range.
Therefore, we know that sin (π – x) = sin (x)
Hence, sin (π – 3) = sin (3) also π – 3 ∈ [-π/2, π/2]
Sin-1(sin 3) = π – 3
(viii) Given sin-1(sin 4)
We know that sin-1(sin x) = x with x ∈ [-π/2, π/2], which is approximately equal to [-1.57, 1.57]
But here, x = 4, which does not lie in the above range.
Therefore, we know that sin (π – x) = sin (x)
Hence, sin (π – 4) = sin (4) also π – 4 ∈ [-π/2, π/2]
Sin-1(sin 4) = π – 4
(ix) Given sin-1(sin 12)
We know that sin-1(sin x) = x with x ∈ [-π/2, π/2], which is approximately equal to [-1.57, 1.57]
But here, x = 12, which does not lie in the above range.
Therefore, we know that sin (2nπ – x) = sin (-x)
Hence, sin (2nπ – 12) = sin (-12)
Here, n = 2 also 12 – 4π ∈ [-π/2, π/2]
Sin-1(sin 12) = 12 – 4π
(x) Given sin-1(sin 2)
We know that sin-1(sin x) = x with x ∈ [-π/2, π/2], which is approximately equal to [-1.57, 1.57]
But here, x = 2, which does not lie in the above range.
Therefore, we know that sin (π – x) = sin (x)
Hence, sin (π – 2) = sin (2) also π – 2 ∈ [-π/2, π/2]
Sin-1(sin 2) = π – 2
2. Evaluate each of the following:
(i) cos-1{cos (-π/4)}
(ii) cos-1(cos 5π/4)
(iii) cos-1(cos 4π/3)
(iv) cos-1(cos 13π/6)
(v) cos-1(cos 3)
(vi) cos-1(cos 4)
(vii) cos-1(cos 5)
(viii) cos-1(cos 12)
Solution:
(i) Given cos-1{cos (-π/4)}
We know that cos (-π/4) = cos (π/4) [since cos (-θ) = cos θ
Also know that cos (π/4) = 1/√2
By substituting these values in cos-1{cos (-π/4)} we get,
Cos-1(1/√2)
Now let y = cos-1(1/√2)
Therefore, cos y = 1/√2
Hence, range of principal value of cos-1 is [0, π] and cos (π/4) = 1/√2
Therefore, cos-1{cos (-π/4)} = π/4
(ii) Given cos-1(cos 5π/4)
But we know that cos (5π/4) = -1/√2
By substituting these values in cos-1{cos (5π/4)} we get,
Cos-1(-1/√2)
Now let y = cos-1(-1/√2)
Therefore, cos y = – 1/√2
– Cos (π/4) = 1/√2
Cos (π – π/4) = – 1/√2
Cos (3 π/4) = – 1/√2
Hence, range of principal value of cos-1 is [0, π] and cos (3π/4) = -1/√2
Therefore, cos-1{cos (5π/4)} = 3π/4
(iii) Given cos-1(cos 4π/3)
But we know that cos (4π/3) = -1/2
By substituting these values in cos-1{cos (4π/3)} we get,
Cos-1(-1/2)
Now let y = cos-1(-1/2)
Therefore, cos y = – 1/2
– Cos (π/3) = 1/2
Cos (π – π/3) = – 1/2
Cos (2π/3) = – 1/2
Hence, range of principal value of cos-1 is [0, π] and cos (2π/3) = -1/2
Therefore, cos-1{cos (4π/3)} = 2π/3
(iv) Given cos-1(cos 13π/6)
But we know that cos (13π/6) = √3/2
By substituting these values in cos-1{cos (13π/6)} we get,
Cos-1(√3/2)
Now let y = cos-1(√3/2)
Therefore, cos y = √3/2
Cos (π/6) = √3/2
Hence, range of principal value of cos-1 is [0, π] and cos (π/6) = √3/2
Therefore, cos-1{cos (13π/6)} = π/6
(v) Given cos-1(cos 3)
We know that cos-1(cos θ) = θ if 0 ≤ θ ≤ π
Therefore, by applying this in the given question, we get,
Cos-1(cos 3) = 3, 3 ∈ [0, π]
(vi) Given cos-1(cos 4)
We have cos–1(cos x) = x if x ϵ [0, π] ≈ [0, 3.14]
And here, x = 4, which does not lie in the above range.
We know that cos (2π – x) = cos(x)
Thus, cos (2π – 4) = cos (4) so 2π–4 belongs in [0, π]
Hence, cos–1(cos 4) = 2π – 4
(vii) Given cos-1(cos 5)
We have cos–1(cos x) = x if x ϵ [0, π] ≈ [0, 3.14]
And here, x = 5, which does not lie in the above range.
We know that cos (2π – x) = cos(x)
Thus, cos (2π – 5) = cos (5) so 2π–5 belongs in [0, π]
Hence, cos–1(cos 5) = 2π – 5
(viii) Given cos-1(cos 12)
Cos–1(cos x) = x if x ϵ [0, π] ≈ [0, 3.14]
And here, x = 12, which does not lie in the above range.
We know cos (2nπ – x) = cos (x)
Cos (2nπ – 12) = cos (12)
Here, n = 2
Also, 4π – 12 belongs in [0, π]
∴ cos–1(cos 12) = 4π – 12
3. Evaluate each of the following:
(i) tan-1(tan π/3)
(ii) tan-1(tan 6π/7)
(iii) tan-1(tan 7π/6)
(iv) tan-1(tan 9π/4)
(v) tan-1(tan 1)
(vi) tan-1(tan 2)
(vii) tan-1(tan 4)
(viii) tan-1(tan 12)
Solution:
(i) Given tan-1(tan π/3)
As tan-1(tan x) = x if x ϵ [-π/2, π/2]
By applying this condition to the given question, we get,
Tan-1(tan π/3) = π/3
(ii) Given tan-1(tan 6π/7)
We know that tan 6π/7 can be written as (π – π/7)
Tan (π – π/7) = – tan π/7
We know that tan-1(tan x) = x if x ϵ [-π/2, π/2]
Tan-1(tan 6π/7) = – π/7
(iii) Given tan-1(tan 7π/6)
We know that tan 7π/6 = 1/√3
By substituting this value in tan-1(tan 7π/6) we get,
Tan-1 (1/√3)
Now let tan-1 (1/√3) = y
Tan y = 1/√3
Tan (π/6) = 1/√3
The range of the principal value of tan-1 is (-π/2, π/2) and tan (π/6) = 1/√3
Therefore, tan-1(tan 7π/6) = π/6
(iv) Given tan-1(tan 9π/4)
We know that tan 9π/4 = 1
By substituting this value in tan-1(tan 9π/4) we get,
Tan-1 (1)
Now let tan-1 (1) = y
Tan y = 1
Tan (π/4) = 1
The range of the principal value of tan-1 is (-π/2, π/2) and tan (π/4) = 1
Therefore, tan-1(tan 9π/4) = π/4
(v) Given tan-1(tan 1)
But we have tan-1(tan x) = x if x ϵ [-π/2, π/2]
By substituting this condition in the given question,
Tan-1(tan 1) = 1
(vi) Given tan-1(tan 2)
As tan-1(tan x) = x if x ϵ [-π/2, π/2]
But here, x = 2, which does not belong to the above range.
We also have tan (π – θ) = –tan (θ)
Therefore, tan (θ – π) = tan (θ)
Tan (2 – π) = tan (2)
Now, 2 – π is in the given range
Hence, tan–1 (tan 2) = 2 – π
(vii) Given tan-1(tan 4)
As tan-1(tan x) = x if x ϵ [-π/2, π/2]
But here, x = 4, which does not belong to the above range.
We also have tan (π – θ) = –tan (θ)
Therefore, tan (θ – π) = tan (θ)
Tan (4 – π) = tan (4)
Now, 4 – π is in the given range.
Hence tan–1 (tan 2) = 4 – π
(viii) Given tan-1(tan 12)
As tan-1(tan x) = x if x ϵ [-π/2, π/2]
But here, x = 12, which does not belong to he above range.
We know that tan (2nπ – θ) = –tan (θ)
Tan (θ – 2nπ) = tan (θ)
Here, n = 2
Tan (12 – 4π) = tan (12)
Now, 12 – 4π is in the given range.
∴ tan–1 (tan 12) = 12 – 4π.
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