Principal Values of Inverse Trigonometric Functions

We will learn how to find the principal values of inverse trigonometric functions in different types of problems.

The principal value of sinβˆ’1 x for x > 0, is the length of the arc of a unit circle centred at the origin which subtends an angle at the centre whose sine is x. For this reason sin^-1 x is also denoted by arc sin x. Similarly, cosβˆ’1 x, tanβˆ’1  x, cscβˆ’1  x, secβˆ’1  x and cotβˆ’1 x are denoted by arc cos x, arc tan x, arc csc x, arc sec x.

1. Find the principal values of sinβˆ’1 (- 1/2)      

Solution: 

If ΞΈ be the principal value of sinβˆ’1 x then - Ο€2 ≀ ΞΈ ≀ Ο€2.

Therefore, If the principal value of sinβˆ’1 (- 1/2) be ΞΈ then sinβˆ’1 (- 1/2) = ΞΈ

β‡’ sin ΞΈ = - 1/2 = sin (-Ο€6) [Since, - Ο€2 ≀ ΞΈ ≀ Ο€2]

Therefore, the principal value of sinβˆ’1 (- 1/2) is (-Ο€6).



2. Find the principal values of the inverse circular function cosβˆ’1 (- √3/2)

Solution:

 If the principal value of cosβˆ’1 x is ΞΈ then we know, 0 ≀ ΞΈ ≀ Ο€.

Therefore, If the principal value of cosβˆ’1  (- √3/2) be ΞΈ then cosβˆ’1  (- √3/2) = ΞΈ

β‡’ cos ΞΈ = (- √3/2) = cos Ο€6 = cos (Ο€ - Ο€6) [Since, 0 ≀ ΞΈ ≀ Ο€]

Therefore, the principal value of cosβˆ’1  (- √3/2) is Ο€ - Ο€6 = 5Ο€6.

 

3. Find the principal values of the inverse trig function tanβˆ’1 (1/√3)

Solution:

If the principal value of tanβˆ’1 x is ΞΈ then we know, - Ο€2 < ΞΈ < Ο€2.

Therefore, If the principal value of tanβˆ’1 (1/√3) be ΞΈ then tanβˆ’1 (1/√3) = ΞΈ

β‡’ tan ΞΈ = 1/√3 = tan Ο€6 [Since, - Ο€2 < ΞΈ < Ο€2]

Therefore, the principal value of tanβˆ’1 (1/√3) is Ο€6.

 

4. Find the principal values of the inverse circular function cotβˆ’1 (- 1)

Solution:

If the principal value of cotβˆ’1 x is Ξ± then we know, - Ο€2 ≀ ΞΈ ≀ Ο€2 and ΞΈ β‰  0.

Therefore, If the principal value of cotβˆ’1 (- 1) be Ξ± then cotβˆ’1 (- 1) = ΞΈ

β‡’ cot ΞΈ = (- 1) = cot (-Ο€4) [Since, - Ο€2 ≀ ΞΈ ≀ Ο€2]  

Therefore, the principal value of cotβˆ’1 (- 1) is (-Ο€4).    

 

5. Find the principal values of the inverse trig function secβˆ’1 (1)

Solution:

If the principal value of secβˆ’1 x is Ξ± then we know, 0 ≀ ΞΈ ≀ Ο€ and ΞΈ β‰  Ο€2.

Therefore, If the principal value of secβˆ’1 (1) be Ξ± then, secβˆ’1 (1) = ΞΈ

β‡’ sec ΞΈ = 1 = sec 0    [Since, 0 ≀ ΞΈ ≀ Ο€]

Therefore, the principal value of secβˆ’1 (1) is 0.

 

6. Find the principal values of the inverse trig function cscβˆ’1 (- 1).

Solution:

If the principal value of cscβˆ’1 x is Ξ± then we know, - Ο€2 ≀ ΞΈ ≀ Ο€2 and ΞΈ β‰  0.

Therefore, if the principal value of cscβˆ’1 (- 1) be ΞΈ then cscβˆ’1 (- 1) = ΞΈ

β‡’ csc ΞΈ = - 1 = csc (-Ο€2) [Since, - Ο€2 ≀ ΞΈ ≀ Ο€2]

Therefore, the principal value of cscβˆ’1 (- 1) is (-Ο€2).

● Inverse Trigonometric Functions




11 and 12 Grade Math

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