Plz explain the points Inverse Trigonometric Trigonometric functions are not one-one - Maths - Inverse Trigonometric Functions

Question

Plz explain the points

Inverse Trigonometric

Trigonometric functions are not one-one and onto in their usual
natural domain Hence, they are not invertible

They can be made one-one and onto ( i e, invertible ) by
restricting their domain In this case, the

∘ Range of the inverse of trigonometric function is the
proper subset of the domain of that trigonometric function
∘ The branch of the inverse trigonometric function with the
restricted range is called the Principal Value Brach

There are 6 inverse trigonometric functions They can be
described as

Shivam Mishra · Accepted Answer

Dear Student,
Please find below the solution to the asked query:

Inverse of any function exists if and only if it is one-one and onto
Let is understand this with an example
Consider f:R→R fx=sinxNow sin is periodic function with a period of π
 Hence it will not be one-one
Range is sinx is -1,1 which is not equal to codomain, hence it will be not be onto
But if we restrict domain and codomain of sinx, then we can make it one-one and ontoand hence then inverse will exist
e g
 f:-π2,π2→-1,1 fx=sinx is one-one as well as onto
Hence inverse will existg:-1,1→-π2,π2 gx=sin-1xSimilarly we make all other trigonometric ratios one-one and onto before finding inverse

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