RD Chapter 4 Inverse Trigonometric Functions Ex 4.7 Solutions
Question - 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)
Answer - 1 : -
(i) Given sin-1(sinπ/6)
We know that the valueof sin π/6 is ½
By substituting thisvalue 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 principalvalue of sin-1(-π/2, π/2) and sin (π/6) = ½
Therefore sin-1(sinπ/6) = π/6
(ii) Given sin-1(sin7π/6)
But we know that sin7π/6 = – ½
By substituting thisin 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 principalvalue of sin-1(-π/2, π/2) and sin (- π/6) = – ½
Therefore sin-1(sin7π/6) = – π/6
(iii) Given sin-1(sin5π/6)
We know that the valueof sin 5π/6 is ½
By substituting thisvalue 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 principalvalue of sin-1(-π/2, π/2) and sin (π/6) = ½
Therefore sin-1(sin5π/6) = π/6
(iv) Given sin-1(sin13π/7)
Given question can bewritten as sin (2π – π/7)
Sin (2π – π/7) can bewritten as sin (-π/7) [since sin (2π – θ) = sin (-θ)]
By substituting thesevalues in sin-1(sin 13π/7) we get sin-1(sin – π/7)
As sin-1(sinx) = x with x ∈ [-π/2, π/2]
Therefore sin-1(sin13π/7) = – π/7
(v) Given sin-1(sin17π/8)
Given question can bewritten as sin (2π + π/8)
Sin (2π + π/8) can bewritten as sin (π/8)
By substituting thesevalues in sin-1(sin 17π/8) we get sin-1(sin π/8)
As sin-1(sinx) = x with x ∈ [-π/2, π/2]
Therefore sin-1(sin17π/8) = π/8
(vi) Given sin-1{(sin– 17π/8)}
But we know that – sinθ = 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 writtenas – sin (π/8)
– Sin (π/8) = sin(-π/8) [since – sin θ = sin (-θ)]
By substituting thesevalues in sin-1{(sin – 17π/8)} we get,
Sin-1(sin –π/8)
As sin-1(sinx) = x with x ∈ [-π/2, π/2]
Therefore sin-1(sin-π/8) = – π/8
(vii) Given sin-1(sin3)
We know that sin-1(sinx) = x with x ∈ [-π/2, π/2] which isapproximately equal to [-1.57, 1.57]
But here x = 3, whichdoes not lie on the above range,
Therefore we know thatsin (π – x) = sin (x)
Hence sin (π – 3) =sin (3) also π – 3 ∈ [-π/2, π/2]
Sin-1(sin3) = π – 3
(viii) Given sin-1(sin4)
We know that sin-1(sinx) = x with x ∈ [-π/2, π/2] which isapproximately equal to [-1.57, 1.57]
But here x = 4, whichdoes not lie on the above range,
Therefore we know thatsin (π – x) = sin (x)
Hence sin (π – 4) =sin (4) also π – 4 ∈ [-π/2, π/2]
Sin-1(sin4) = π – 4
(ix) Given sin-1(sin12)
We know that sin-1(sinx) = x with x ∈ [-π/2, π/2] which isapproximately equal to [-1.57, 1.57]
But here x = 12, whichdoes not lie on the above range,
Therefore we know thatsin (2nπ – x) = sin (-x)
Hence sin (2nπ – 12) =sin (-12)
Here n = 2 also 12 –4π ∈ [-π/2, π/2]
Sin-1(sin12) = 12 – 4π
(x) Given sin-1(sin2)
We know that sin-1(sinx) = x with x ∈ [-π/2, π/2] which isapproximately equal to [-1.57, 1.57]
But here x = 2, whichdoes not lie on the above range,
Therefore we know thatsin (π – x) = sin (x)
Hence sin (π – 2) =sin (2) also π – 2 ∈ [-π/2, π/2]
Sin-1(sin2) = π – 2
Question - 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)
Answer - 2 : -
(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 thesevalues 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 ofprincipal 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(cos5π/4)
But we know that cos(5π/4) = -1/√2
By substituting thesevalues 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 ofprincipal 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(cos4π/3)
But we know that cos(4π/3) = -1/2
By substituting thesevalues 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 ofprincipal 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(cos13π/6)
But we know that cos(13π/6) = √3/2
By substituting thesevalues 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 ofprincipal 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(cos3)
We know that cos-1(cosθ) = θ if 0 ≤ θ ≤ π
Therefore by applyingthis in given question we get,
Cos-1(cos3) = 3, 3 ∈ [0, π]
(vi) Given cos-1(cos4)
We have cos–1(cosx) = x if x ϵ [0, π] ≈ [0, 3.14]
And here x = 4 whichdoes 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(cos4) = 2π – 4
(vii) Given cos-1(cos5)
We have cos–1(cosx) = x if x ϵ [0, π] ≈ [0, 3.14]
And here x = 5 whichdoes 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(cos5) = 2π – 5
(viii) Given cos-1(cos12)
Cos–1(cosx) = x if x ϵ [0, π] ≈ [0, 3.14]
And here x = 12 whichdoes not lie in the above range.
We know cos (2nπ – x)= cos (x)
Cos (2nπ – 12) = cos(12)
Here n = 2.
Also 4π – 12 belongsin [0, π]
∴ cos–1(cos12) = 4π – 12
Question - 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)
Answer - 3 : -
(i) Given tan-1(tanπ/3)
As tan-1(tanx) = x if x ϵ [-π/2, π/2]
By applying thiscondition in the given question we get,
Tan-1(tanπ/3) = π/3
(ii) Given tan-1(tan6π/7)
We know that tan 6π/7can be written as (π – π/7)
Tan (π – π/7) = – tanπ/7
We know that tan-1(tanx) = x if x ϵ [-π/2, π/2]
Tan-1(tan6π/7) = – π/7
(iii) Given tan-1(tan7π/6)
We know that tan 7π/6= 1/√3
By substituting thisvalue 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 theprincipal value of tan-1 is (-π/2, π/2) and tan (π/6) = 1/√3
Therefore tan-1(tan7π/6) = π/6
(iv) Given tan-1(tan9π/4)
We know that tan 9π/4= 1
By substituting thisvalue 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 theprincipal value of tan-1 is (-π/2, π/2) and tan (π/4) = 1
Therefore tan-1(tan9π/4) = π/4
(v) Given tan-1(tan1)
But we have tan-1(tanx) = x if x ϵ [-π/2, π/2]
By substituting thiscondition in given question
Tan-1(tan 1)= 1
(vi) Given tan-1(tan2)
As tan-1(tanx) = x if x ϵ [-π/2, π/2]
But here x = 2 whichdoes not belongs to 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 thegiven range
Hence tan–1 (tan2) = 2 – π
(vii) Given tan-1(tan4)
As tan-1(tanx) = x if x ϵ [-π/2, π/2]
But here x = 4 whichdoes not belongs to 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 thegiven range
Hence tan–1 (tan2) = 4 – π
(viii) Given tan-1(tan12)
As tan-1(tanx) = x if x ϵ [-π/2, π/2]
But here x = 12 whichdoes not belongs to 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 thegiven range
∴ tan–1 (tan12) = 12 – 4π.
Question - 4 : - Evaluate the following:
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