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 (-\(\frac{π}{2}\)).




11 and 12 Grade Math

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