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We will learn how to find the principal values of inverse trigonometric functions in different types of problems.
The principal value of sinβ1 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, cosβ1 x, tanβ1 x, cscβ1 x, secβ1 x and cotβ1 x are denoted by arc cos x, arc tan x, arc csc x, arc sec x.
1. Find the principal values of sinβ1 (- 1/2)
Solution:
If ΞΈ be the principal value of sinβ1 x then - Ο2 β€ ΞΈ β€ Ο2.
Therefore, If the principal value of sinβ1 (- 1/2) be ΞΈ then sinβ1 (- 1/2) = ΞΈ
β sin ΞΈ = - 1/2 = sin (-Ο6) [Since, - Ο2 β€ ΞΈ β€ Ο2]
Therefore, the principal value of sinβ1 (- 1/2) is (-Ο6).
2. Find the
principal values of the inverse circular function cosβ1 (- β3/2)
Solution:
If the principal value of cosβ1 x is ΞΈ then we know, 0 β€ ΞΈ β€ Ο.
Therefore, If the principal value of cosβ1 (- β3/2) be ΞΈ then cosβ1 (- β3/2) = ΞΈ
β cos ΞΈ = (- β3/2) = cos Ο6 = cos (Ο - Ο6) [Since, 0 β€ ΞΈ β€ Ο]
Therefore, the principal value of cosβ1 (- β3/2) is Ο - Ο6 = 5Ο6.
3. Find the principal values of the inverse trig function tanβ1 (1/β3)
Solution:
If the principal value of tanβ1 x is ΞΈ then we know, - Ο2 < ΞΈ < Ο2.
Therefore, If the principal value of tanβ1 (1/β3) be ΞΈ then tanβ1 (1/β3) = ΞΈ
β tan ΞΈ = 1/β3 = tan Ο6 [Since, - Ο2 < ΞΈ < Ο2]
Therefore, the principal value of tanβ1 (1/β3) is Ο6.
4. Find the principal values of the inverse circular function cotβ1 (- 1)
Solution:
If the principal value of cotβ1 x is Ξ± then we know, - Ο2 β€ ΞΈ β€ Ο2 and ΞΈ β 0.
Therefore, If the principal value of cotβ1 (- 1) be Ξ± then cotβ1 (- 1) = ΞΈ
β cot ΞΈ = (- 1) = cot (-Ο4) [Since, - Ο2 β€ ΞΈ β€ Ο2]
Therefore, the principal value of cotβ1 (- 1) is (-Ο4).
5. Find the principal values of the inverse trig function secβ1 (1)
Solution:If the principal value of secβ1 x is Ξ± then we know, 0 β€ ΞΈ β€ Ο and ΞΈ β Ο2.
Therefore, If the principal value of secβ1 (1) be Ξ± then, secβ1 (1) = ΞΈ
β sec ΞΈ = 1 = sec 0 [Since, 0 β€ ΞΈ β€ Ο]
Therefore, the principal value of secβ1 (1) is 0.
6. Find the principal values of the inverse trig function cscβ1 (- 1).
Solution:
If the principal value of cscβ1 x is Ξ± then we know, - Ο2 β€ ΞΈ β€ Ο2 and ΞΈ β 0.
Therefore, if the principal value of cscβ1 (- 1) be ΞΈ then cscβ1 (- 1) = ΞΈ
β csc ΞΈ = - 1 = csc (-Ο2) [Since, - Ο2 β€ ΞΈ β€ Ο2]
Therefore, the principal value of cscβ1 (- 1) is (-Ο2).
β Inverse Trigonometric Functions
11 and 12 Grade Math
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