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Range of x satisfying  sin^{-1} x>1

Option: 1

(- 1 ,1]


Option: 2

(-sin 1 ,1]


Option: 3

(\sin 1 ,1]


Option: 4

[\sin 1 ,1]


Answers (1)

best_answer

 

 

Inverse Trigonometric Function -

Inverse Trigonometric Function

\\\mathrm{ Trigonometric\;function \quad\quad\quad\quad\quad\quad \quad Inverse\;trigo\;function}\\\mathrm{{\color{Teal} Domain: Measure \;of\; an\; angle}\quad\quad\quad\quad{\color{Red} Domain: Ratio}}\\\mathrm{{\color{Red} Range: Ratio }\quad\quad\quad\quad\quad\quad\quad\quad \quad\quad\quad {\color{Teal} Range: Measure\; of\; an\; angle}}

For example, if f(x) = sin x, then we would write f −1 (x) = sin−1 x. Be aware that sin−1 x does not mean 1/sinx . The following examples illustrate the inverse trigonometric functions:

  1. sin (π/6) = ½, then π/6 = sin-1 (½ ) 

  2. cos(π) = −1, then π = cos−1 (−1) 

  3. tan (π/4)  = 1, then (π/4) = tan−1 (1)

As we know that trigonometric functions are periodic and hence many-one in their domain.

For inverse of trigonometric functions to be defined, the actual domain of trigonometric function must be restricted to make one-one function 

The domain of the sine function is R and range is [-1, 1]. If we restrict its domain to [-π/2, π/2] then it become one-one with onto and having range [-1, 1].

Actually, sine function can be restricted to any of the intervals [--3π/2, -π/2], [-π/2, π/2], [π/2, 3π/2] and so on. It becomes one-one and its range is [-1, 1]. We can, therefore, define the inverse of sine function in each of these intervals. 

 

Thus, sin-1 is a function whose domain is [-1, 1] and the range could be any of the intervals [--3π/2, -π/2], [-π/2, π/2] or  [π/2, 3π/2] and so on. Corresponding to each such interval, we get a branch of the function sin -1 .  The branch with range  [-π/2, π/2] is called the principal value branch, whereas other intervals as range give different branches of sin-1.

When we refer to the function sin-1, we take it as the function whose domain is [-1, 1] and range is [-π/2, π/2].

In a similar way, we define other trigonometric functions by restricting their domain.

 

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sin^{-1} x>1\\ 1<sin^{-1} x \leq \frac{\pi}{2}\\ \sin 1<x \leq \sin \frac{\pi}{2} \\ x \epsilon (sin 1 \sin,1]

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sudhir.kumar

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