# 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 sin$$^{-1}$$ 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, cos$$^{-1}$$ x, tan$$^{-1}$$  x, csc$$^{-1}$$  x, sec$$^{-1}$$  x and cot$$^{-1}$$ x are denoted by arc cos x, arc tan x, arc csc x, arc sec x.

1. Find the principal values of sin$$^{-1}$$ (- 1/2)

Solution:

If θ be the principal value of sin$$^{-1}$$ x then - $$\frac{π}{2}$$ ≤ θ ≤ $$\frac{π}{2}$$.

Therefore, If the principal value of sin$$^{-1}$$ (- 1/2) be θ then sin$$^{-1}$$ (- 1/2) = θ

⇒ sin θ = - 1/2 = sin (-$$\frac{π}{6}$$) [Since, - $$\frac{π}{2}$$ ≤ θ ≤ $$\frac{π}{2}$$]

Therefore, the principal value of sin$$^{-1}$$ (- 1/2) is (-$$\frac{π}{6}$$).

2. Find the principal values of the inverse circular function cos$$^{-1}$$ (- √3/2)

Solution:

If the principal value of cos$$^{-1}$$ x is θ then we know, 0 ≤ θ ≤ π.

Therefore, If the principal value of cos$$^{-1}$$  (- √3/2) be θ then cos$$^{-1}$$  (- √3/2) = θ

⇒ cos θ = (- √3/2) = cos $$\frac{π}{6}$$ = cos (π - $$\frac{π}{6}$$) [Since, 0 ≤ θ ≤ π]

Therefore, the principal value of cos$$^{-1}$$  (- √3/2) is π - $$\frac{π}{6}$$ = $$\frac{5π}{6}$$.

3. Find the principal values of the inverse trig function tan$$^{-1}$$ (1/√3)

Solution:

If the principal value of tan$$^{-1}$$ x is θ then we know, - $$\frac{π}{2}$$ < θ < $$\frac{π}{2}$$.

Therefore, If the principal value of tan$$^{-1}$$ (1/√3) be θ then tan$$^{-1}$$ (1/√3) = θ

⇒ tan θ = 1/√3 = tan $$\frac{π}{6}$$ [Since, - $$\frac{π}{2}$$ < θ < $$\frac{π}{2}$$]

Therefore, the principal value of tan$$^{-1}$$ (1/√3) is $$\frac{π}{6}$$.

4. Find the principal values of the inverse circular function cot$$^{-1}$$ (- 1)

Solution:

If the principal value of cot$$^{-1}$$ x is α then we know, - $$\frac{π}{2}$$ ≤ θ ≤ $$\frac{π}{2}$$ and θ ≠ 0.

Therefore, If the principal value of cot$$^{-1}$$ (- 1) be α then cot$$^{-1}$$ (- 1) = θ

⇒ cot θ = (- 1) = cot (-$$\frac{π}{4}$$) [Since, - $$\frac{π}{2}$$ ≤ θ ≤ $$\frac{π}{2}$$]

Therefore, the principal value of cot$$^{-1}$$ (- 1) is (-$$\frac{π}{4}$$).

5. Find the principal values of the inverse trig function sec$$^{-1}$$ (1)

Solution:

If the principal value of sec$$^{-1}$$ x is α then we know, 0 ≤ θ ≤ π and θ ≠ $$\frac{π}{2}$$.

Therefore, If the principal value of sec$$^{-1}$$ (1) be α then, sec$$^{-1}$$ (1) = θ

⇒ sec θ = 1 = sec 0    [Since, 0 ≤ θ ≤ π]

Therefore, the principal value of sec$$^{-1}$$ (1) is 0.

6. Find the principal values of the inverse trig function csc$$^{-1}$$ (- 1).

Solution:

If the principal value of csc$$^{-1}$$ x is α then we know, - $$\frac{π}{2}$$ ≤ θ ≤ $$\frac{π}{2}$$ and θ ≠ 0.

Therefore, if the principal value of csc$$^{-1}$$ (- 1) be θ then csc$$^{-1}$$ (- 1) = θ

⇒ csc θ = - 1 = csc (-$$\frac{π}{2}$$) [Since, - $$\frac{π}{2}$$ ≤ θ ≤ $$\frac{π}{2}$$]

Therefore, the principal value of csc$$^{-1}$$ (- 1) is (-$$\frac{π}{2}$$).

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Inverse Trigonometric Functions