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Byju's Answer

Standard XII

Mathematics

Commutative Law of Binary Operation

The inverse o...

Question

The inverse of a function f is only possible if and only if f is surjective.
But the inverse of trigonometric functions like sine , cosine is possible even if they are not surjective. Why so?

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Solution

$Inverse of these function is defined for a particular range and not on whole domain of sinx or cosx . And that is why we always write the domain and range of these inverse function because inverse exist in these range only .$

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Similar questions

Let

f : A \to B

be a function defined by

y = f (x)

where f is a bijective function, means f is injective (one-one) as well as surjective (onto), then there exist a unique mapping

g : B \to A

such that

f (x) = y

if and only if

g (y) = x \forall x ϵ A, y ϵ B

Then function g is said to be inverse of f and vice versa so we write

g = f^{- 1} : B \to A [{f (x), x} : {x, f (x)} ϵ f^{- 1}]

when branch of an inverse function is not given (define) then we consider its principal value branch.

Which of the following is not correct?

If we consider only the principle values of the inverse trigonometric functions, then the value of $t a n (c o s^{- 1} \frac{1}{5 \sqrt{2}} - s i n^{- 1} \frac{4}{\sqrt{17}})$

Let

f : A \to B

be a function defined by

y = f (x)

where f is a bijective function, means f is injective (one-one) as well as surjective (onto), then there exist a unique mapping

g : B \to A

such that

f (x) = y

if and only if

g (y) = x \forall x ϵ A, y ϵ B

Then function g is said to be inverse of f and vice versa so we write

g = f^{- 1} : B \to A [{f (x), x} : {x, f (x)} ϵ f^{- 1}]

when branch of an inverse function is not given (define) then we consider its principal value branch.

For

0 \leq x \leq 1

the range of

{tan}^{- 1} (\frac{1 + x}{1 - x})

is?

Let

f : A \to B

be a function defined by

y = f (x)

where f is a bijective function, means f is injective (one-one) as well as surjective (onto), then there exist a unique mapping

g : B \to A

such that

f (x) = y

if and only if

g (y) = x \forall x ϵ A, y ϵ B

Then function g is said to be inverse of f and vice versa so we write

g = f^{- 1} : B \to A [{f (x), x} : {x, f (x)} ϵ f^{- 1}]

when branch of an inverse function is not given (define) then we consider its principal value branch.

x < 0

then

{tan}^{- 1} x + {tan}^{- 1} \frac{1}{x}

equals?

Let

f : A \to B

be a function defined by

y = f (x)

where f is a bijective function, means f is injective (one-one) as well as surjective (onto), then there exist a unique mapping

g : B \to A

such that

f (x) = y

if and only if

g (y) = x \forall x ϵ A, y ϵ B

Then function g is said to be inverse of f and vice versa so we write

g = f^{- 1} : B \to A [{f (x), x} : {x, f (x)} ϵ f^{- 1}]

when branch of an inverse function is not given (define) then we consider its principal value branch.

- 1 < x < 0

,then

{tan}^{- 1} x

equals?

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The inverse of a function f is only possible if and only if f is surjective. But the inverse of trigonometric functions like sine , cosine is possible even if they are not surjective. Why so?

Inverse of these function is defined for a particular range and not on whole domain of sinx or cosx.And that is why we always write the domain and range of theseinverse function because inverse exist in these range only.

The inverse of a function f is only possible if and only if f is surjective.
But the inverse of trigonometric functions like sine , cosine is possible even if they are not surjective. Why so?

$Inverse of these function is defined for a particular range and not on whole domain of sinx or cosx . And that is why we always write the domain and range of these inverse function because inverse exist in these range only .$