How to find the general and principal values of ccs\(^{-1}\) x?
Let csc θ = x (| x |≥ 1 i.e., x ≥ 1 or, x ≤ - 1) then θ = csc\(^{-1}\) x .
Here θ has infinitely many values.
Let – \(\frac{π}{2}\) ≤ α ≤ \(\frac{π}{2}\), where α is non-zero(α ≠ 0) positive or negative smallest numerical value of these infinite number of values and satisfies the equation csc θ = x then the angle α is called the principal value of csc\(^{-1}\) x.
Again, if the principal value of csc\(^{-1}\) x is α (– \(\frac{π}{2}\) < α < \(\frac{π}{2}\)) and α ≠ 0 then its general value = nπ + (- 1) n α, where, | x | ≥ 1.
Therefore, tan\(^{-1}\) x = nπ + α, where, (– \(\frac{π}{2}\) < α < \(\frac{π}{2}\)), | x | ≥ 1 and (- ∞ < x < ∞).
Examples to find the general and principal
values of arc csc x:
1. Find the General and Principal Values of csc \(^{-1}\) (√2).
Solution:
Let x = csc\(^{-1}\) (√2)
⇒ csc x = √2
⇒ csc x = csc \(\frac{π}{4}\)
⇒ x = \(\frac{π}{4}\)
⇒ csc\(^{-1}\) (√2) = \(\frac{π}{4}\)
Therefore, principal value of csc\(^{-1}\) (√2) is \(\frac{π}{4}\) and its general value = nπ + (- 1)\(^{n}\) ∙ \(\frac{π}{4}\).
2. Find the General and Principal Values of csc \(^{-1}\) (-√2).
Solution:
Let x = csc\(^{-1}\) (-√2)
⇒ csc x = -√2
⇒ csc x = csc (-\(\frac{π}{4}\))
⇒ x = -\(\frac{π}{4}\)
⇒ csc\(^{-1}\) (-√2) = -\(\frac{π}{4}\)
Therefore, principal value of csc\(^{-1}\) (-√2) is -\(\frac{π}{4}\) and its general value = nπ + (- 1)\(^{n}\) ∙ (-\(\frac{π}{4}\)) = nπ - (- 1)\(^{n}\) ∙ \(\frac{π}{4}\).
● Inverse Trigonometric Functions
11 and 12 Grade Math
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