RD Sharma Solutions For Class 12 Maths Exercise 4.7 Chapter 4 Inverse Trigonometric Functions

RD Sharma Solutions Class 12 Maths Exercise 4.7 Chapter 4 Inverse Trigonometric functions is provided here. Students can use the PDF of solutions to get their doubts cleared instantly and solve problems as per the latest exam pattern. It helps students to self analyse their knowledge about 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 which are given here. This exercise deals with the properties of Inverse Trigonometric Functions.

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Exercise 4.7 Page No: 4.42

1. Evaluate each of the following:

(i) sin-1(sin Ï€/6)

(ii) sin-1(sin 7Ï€/6)

(iii) sin-1(sin 5Ï€/6)

(iv) sin-1(sin 13Ï€/7)

(v) sin-1(sin 17Ï€/8)

(vi) sin-1{(sin – 17Ï€/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 Ï€/6)

We know that the value of sin Ï€/6 is Â½

By substituting this value in sin-1(sin Ï€/6)

We get, sin-1 (1/2)

Now let y = sin-1 (1/2)

Sin (Ï€/6) = Â½

The range of principal value of sin-1(-Ï€/2, Ï€/2) and sin (Ï€/6) = Â½

Therefore sin-1(sin Ï€/6) = Ï€/6

(ii) Given sin-1(sin 7Ï€/6)

But we know that sin 7Ï€/6 = – Â½

By substituting this in sin-1(sin 7Ï€/6) we get,

Sin-1 (-1/2)

Now let y = sin-1 (-1/2)

– Sin y = Â½

– Sin (Ï€/6) = Â½

– Sin (Ï€/6) = sin (- Ï€/6)

The range of principal value of sin-1(-Ï€/2, Ï€/2) and sin (- Ï€/6) = – Â½

Therefore sin-1(sin 7Ï€/6) = – Ï€/6

(iii) Given sin-1(sin 5Ï€/6)

We know that the value of sin 5Ï€/6 is Â½

By substituting this value in sin-1(sin 5Ï€/6)

We get, sin-1 (1/2)

Now let y = sin-1 (1/2)

Sin (Ï€/6) = Â½

The range of principal value of sin-1(-Ï€/2, Ï€/2) and sin (Ï€/6) = Â½

Therefore sin-1(sin 5Ï€/6) = Ï€/6

(iv) Given sin-1(sin 13Ï€/7)

Given question can be written as sin (2Ï€ â€“ Ï€/7)

Sin (2Ï€ â€“ Ï€/7) can be written as sin (-Ï€/7) [since sin (2Ï€ â€“ Î¸) = sin (-Î¸)]

By substituting these values in sin-1(sin 13Ï€/7) we get sin-1(sin – Ï€/7)

As sin-1(sin x) = x with x âˆˆ [-Ï€/2, Ï€/2]

Therefore sin-1(sin 13Ï€/7) = – Ï€/7

(v) Given sin-1(sin 17Ï€/8)

Given question can be written as sin (2Ï€ + Ï€/8)

Sin (2Ï€ + Ï€/8) can be written as sin (Ï€/8)

By substituting these values in sin-1(sin 17Ï€/8) we get sin-1(sin Ï€/8)

As sin-1(sin x) = x with x âˆˆ [-Ï€/2, Ï€/2]

Therefore sin-1(sin 17Ï€/8) = Ï€/8

(vi) Given sin-1{(sin – 17Ï€/8)}

But we know that â€“ sin Î¸ = sin (-Î¸)

Therefore (sin -17Ï€/8) = – sin 17Ï€/8

– Sin 17Ï€/8 = – sin (2Ï€ + Ï€/8) [since sin (2Ï€ â€“ Î¸) = -sin (Î¸)]

It can also be written as â€“ sin (Ï€/8)

â€“ Sin (Ï€/8) = sin (-Ï€/8) [since â€“ sin Î¸ = sin (-Î¸)]

By substituting these values in sin-1{(sin – 17Ï€/8)} we get,

Sin-1(sin – Ï€/8)

As sin-1(sin x) = x with x âˆˆ [-Ï€/2, Ï€/2]

Therefore sin-1(sin -Ï€/8) = – Ï€/8

(vii) Given sin-1(sin 3)

We know that sin-1(sin x) = x with x âˆˆ [-Ï€/2, Ï€/2] which is approximately equal to [-1.57, 1.57]

But here x = 3, which does not lie on the above range,

Therefore we know that sin (Ï€ â€“ x) = sin (x)

Hence sin (Ï€ â€“ 3) = sin (3) also Ï€ â€“ 3 âˆˆ [-Ï€/2, Ï€/2]

Sin-1(sin 3) = Ï€ â€“ 3

(viii) Given sin-1(sin 4)

We know that sin-1(sin x) = x with x âˆˆ [-Ï€/2, Ï€/2] which is approximately equal to [-1.57, 1.57]

But here x = 4, which does not lie on the above range,

Therefore we know that sin (Ï€ â€“ x) = sin (x)

Hence sin (Ï€ â€“ 4) = sin (4) also Ï€ â€“ 4 âˆˆ [-Ï€/2, Ï€/2]

Sin-1(sin 4) = Ï€ â€“ 4

(ix) Given sin-1(sin 12)

We know that sin-1(sin x) = x with x âˆˆ [-Ï€/2, Ï€/2] which is approximately equal to [-1.57, 1.57]

But here x = 12, which does not lie on the above range,

Therefore we know that sin (2nÏ€ â€“ x) = sin (-x)

Hence sin (2nÏ€ â€“ 12) = sin (-12)

Here n = 2 also 12 â€“ 4Ï€ âˆˆ [-Ï€/2, Ï€/2]

Sin-1(sin 12) = 12 â€“ 4Ï€

(x) Given sin-1(sin 2)

We know that sin-1(sin x) = x with x âˆˆ [-Ï€/2, Ï€/2] which is approximately equal to [-1.57, 1.57]

But here x = 2, which does not lie on the above range,

Therefore we know that sin (Ï€ â€“ x) = sin (x)

Hence sin (Ï€ â€“ 2) = sin (2) also Ï€ â€“ 2 âˆˆ [-Ï€/2, Ï€/2]

Sin-1(sin 2) = Ï€ â€“ 2

2. Evaluate each of the following:

(i) cos-1{cos (-Ï€/4)}

(ii) cos-1(cos 5Ï€/4)

(iii) cos-1(cos 4Ï€/3)

(iv) cos-1(cos 13Ï€/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 (-Ï€/4)}

We know that cos (-Ï€/4) = cos (Ï€/4) [since cos (-Î¸) = cos Î¸

Also know that cos (Ï€/4) = 1/âˆš2

By substituting these values in cos-1{cos (-Ï€/4)} we get,

Cos-1(1/âˆš2)

Now let y = cos-1(1/âˆš2)

Therefore cos y = 1/âˆš2

Hence range of principal value of cos-1 is [0, Ï€] and cos (Ï€/4) = 1/âˆš2

Therefore cos-1{cos (-Ï€/4)} = Ï€/4

(ii) Given cos-1(cos 5Ï€/4)

But we know that cos (5Ï€/4) = -1/âˆš2

By substituting these values in cos-1{cos (5Ï€/4)} we get,

Cos-1(-1/âˆš2)

Now let y = cos-1(-1/âˆš2)

Therefore cos y = – 1/âˆš2

– Cos (Ï€/4) = 1/âˆš2

Cos (Ï€ – Ï€/4) = – 1/âˆš2

Cos (3 Ï€/4) = – 1/âˆš2

Hence range of principal value of cos-1 is [0, Ï€] and cos (3Ï€/4) = -1/âˆš2

Therefore cos-1{cos (5Ï€/4)} = 3Ï€/4

(iii) Given cos-1(cos 4Ï€/3)

But we know that cos (4Ï€/3) = -1/2

By substituting these values in cos-1{cos (4Ï€/3)} we get,

Cos-1(-1/2)

Now let y = cos-1(-1/2)

Therefore cos y = – 1/2

– Cos (Ï€/3) = 1/2

Cos (Ï€ – Ï€/3) = – 1/2

Cos (2Ï€/3) = – 1/2

Hence range of principal value of cos-1 is [0, Ï€] and cos (2Ï€/3) = -1/2

Therefore cos-1{cos (4Ï€/3)} = 2Ï€/3

(iv) Given cos-1(cos 13Ï€/6)

But we know that cos (13Ï€/6) = âˆš3/2

By substituting these values in cos-1{cos (13Ï€/6)} we get,

Cos-1(âˆš3/2)

Now let y = cos-1(âˆš3/2)

Therefore cos y = âˆš3/2

Cos (Ï€/6) = âˆš3/2

Hence range of principal value of cos-1 is [0, Ï€] and cos (Ï€/6) = âˆš3/2

Therefore cos-1{cos (13Ï€/6)} = Ï€/6

(v) Given cos-1(cos 3)

We know that cos-1(cos Î¸) = Î¸ if 0 â‰¤ Î¸ â‰¤ Ï€

Therefore by applying this in given question we get,

Cos-1(cos 3) = 3, 3 âˆˆ [0, Ï€]

(vi) Given cos-1(cos 4)

We have cosâ€“1(cos x) = x if xÂ ÏµÂ [0, Ï€] â‰ˆ [0, 3.14]

And here x = 4 which does not lie in the above range.

We know that cos (2Ï€ â€“ x) = cos(x)

Thus, cos (2Ï€ â€“ 4) = cos (4) so 2Ï€â€“4 belongs in [0, Ï€]

Hence cosâ€“1(cos 4) = 2Ï€ â€“ 4

(vii) Given cos-1(cos 5)

We have cosâ€“1(cos x) = x if xÂ ÏµÂ [0, Ï€] â‰ˆ [0, 3.14]

And here x = 5 which does not lie in the above range.

We know that cos (2Ï€ â€“ x) = cos(x)

Thus, cos (2Ï€ â€“ 5) = cos (5) so 2Ï€â€“5 belongs in [0, Ï€]

Hence cosâ€“1(cos 5) = 2Ï€ â€“ 5

(viii) Given cos-1(cos 12)

Cosâ€“1(cos x) = x if xÂ ÏµÂ [0, Ï€] â‰ˆ [0, 3.14]

And here x = 12 which does not lie in the above range.

We know cos (2nÏ€ â€“ x) = cos (x)

Cos (2nÏ€ â€“ 12) = cos (12)

Here n = 2.

Also 4Ï€ â€“ 12 belongs in [0, Ï€]

âˆ´Â cosâ€“1(cos 12) = 4Ï€ â€“ 12

3. Evaluate each of the following:

(i) tan-1(tan Ï€/3)

(ii) tan-1(tan 6Ï€/7)

(iii) tan-1(tan 7Ï€/6)

(iv) tan-1(tan 9Ï€/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 Ï€/3)

As tan-1(tan x) = x if x Ïµ [-Ï€/2, Ï€/2]

By applying this condition in the given question we get,

Tan-1(tan Ï€/3) = Ï€/3

(ii) Given tan-1(tan 6Ï€/7)

We know that tan 6Ï€/7 can be written as (Ï€ â€“ Ï€/7)

Tan (Ï€ â€“ Ï€/7) = – tan Ï€/7

We know that tan-1(tan x) = x if x Ïµ [-Ï€/2, Ï€/2]

Tan-1(tan 6Ï€/7) = – Ï€/7

(iii) Given tan-1(tan 7Ï€/6)

We know that tan 7Ï€/6 = 1/âˆš3

By substituting this value in tan-1(tan 7Ï€/6) we get,

Tan-1 (1/âˆš3)

Now let tan-1 (1/âˆš3) = y

Tan y = 1/âˆš3

Tan (Ï€/6) = 1/âˆš3

The range of the principal value of tan-1 is (-Ï€/2, Ï€/2) and tan (Ï€/6) = 1/âˆš3

Therefore tan-1(tan 7Ï€/6) = Ï€/6

(iv) Given tan-1(tan 9Ï€/4)

We know that tan 9Ï€/4 = 1

By substituting this value in tan-1(tan 9Ï€/4) we get,

Tan-1 (1)

Now let tan-1 (1) = y

Tan y = 1

Tan (Ï€/4) = 1

The range of the principal value of tan-1 is (-Ï€/2, Ï€/2) and tan (Ï€/4) = 1

Therefore tan-1(tan 9Ï€/4) = Ï€/4

(v) Given tan-1(tan 1)

But we have tan-1(tan x) = x if x Ïµ [-Ï€/2, Ï€/2]

By substituting this condition in given question

Tan-1(tan 1) = 1

(vi) Given tan-1(tan 2)

As tan-1(tan x) = x if x Ïµ [-Ï€/2, Ï€/2]

But here x = 2 which does not belongs to above range

We also have tan (Ï€ â€“ Î¸) = â€“tan (Î¸)

Therefore tan (Î¸ â€“ Ï€) = tan (Î¸)

Tan (2 â€“ Ï€) = tan (2)

Now 2 â€“ Ï€ is in the given range

Hence tanâ€“1Â (tan 2) = 2 â€“ Ï€

(vii) Given tan-1(tan 4)

As tan-1(tan x) = x if x Ïµ [-Ï€/2, Ï€/2]

But here x = 4 which does not belongs to above range

We also have tan (Ï€ â€“ Î¸) = â€“tan (Î¸)

Therefore tan (Î¸ â€“ Ï€) = tan (Î¸)

Tan (4 â€“ Ï€) = tan (4)

Now 4 â€“ Ï€ is in the given range

Hence tanâ€“1Â (tan 2) = 4 â€“ Ï€

(viii) Given tan-1(tan 12)

As tan-1(tan x) = x if x Ïµ [-Ï€/2, Ï€/2]

But here x = 12 which does not belongs to above range

We know that tan (2nÏ€ â€“ Î¸) = â€“tan (Î¸)

Tan (Î¸ â€“ 2nÏ€) = tan (Î¸)

Here n = 2

Tan (12 â€“ 4Ï€) = tan (12)

Now 12 â€“ 4Ï€ is in the given range

âˆ´Â tanâ€“1Â (tan 12) = 12 â€“ 4Ï€.

Access other exercises of RD Sharma Solutions For Class 12 Chapter 4 – Inverse Trigonometric Functions

Exercise 4.1 Solutions

Exercise 4.2 Solutions

Exercise 4.3 Solutions

Exercise 4.4 Solutions

Exercise 4.5 Solutions

Exercise 4.6 Solutions

Exercise 4.8 Solutions

Exercise 4.9 Solutions

Exercise 4.10 Solutions

Exercise 4.11 Solutions

Exercise 4.12 Solutions

Exercise 4.13 Solutions

Exercise 4.14 Solutions