Write the principal value of branch (range) of f(x)=sin inverse x
Answers
F(x)=sin^-1x
Domain=[-1,+1]
Range= [-pi/2 , +pi/2]
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The principal value of range of f(x) = sin⁻¹x , is [-π/2 , π/2].
Concepts :
- Every inverse function is a bijective function. means, inverse function is one - one as well as onto function.
- Trigonometric function is many - one function because a parallel line of x-axis cuts the graph of any trigonometric identity more than one point.
- To get an inverse function of any trigonometric identity, we have to take an specific range for which function works as a one - one function. this range is known as principal value of range of inverse of a trigonometric identity.
Let's come to the point.
Here, y = f(x) = sin⁻¹x
⇒ x = siny
you observe that domain of siny is the range of f(x) = sin⁻¹x and the range of siny is the domain of f(x) = sin⁻¹x.
Domain of f(x) = Range of siny = [-1 , 1]
Now, -1 ≤ siny ≤ 1
siny = -1 = sin(-π/2) ⇒y = nπ + (-1)ⁿ(-π/2)
siny = 1 = sin(π/2) ⇒y = nπ + (-1)ⁿ(π/2)
To get the principal value of range of f(x), take n = 0
y = -π/2 for siny = -1
y = π/2 for siny = 1
∴ -1 ≤ siny ≤ 1
⇒ -π/2 ≤ siny ≤ π/2
Therefore the principal value of range of f(x) = sin⁻¹x, is [-π/2 , π/2].
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