We know that corresponding to every bijection ( one-one onto function) f:A ?B there exists a bijection g:B?A defined by
g(y) = x if only f(x) = y
The function g: B ? A is called the inverse of function f: A? B and is denoted by f?1 . Thus, we have
f(x) = y
So, f?1(y) = x.
We have also learnt that
(f?1 of)(x) = f?1[f(x)]
= f?1(y)
= x, for all x ? A.
And (fo f?1)(y) = f[f?1(y)]
= f(x)
= y, for all x ? B.
We know that trigonometric functions are periodic functions, and hence, in general, all trigonometric functions are not bijections. Consequently, their inverse do not exist. However, if we restrict their domains and co-domains, they can be made bijections and we can obtain their inverses. In the following sections, we shall do all these things to obtain the inverses of trigonometric functions. They are termed as Inverse Trigonometric Functions.
Inverse of the Trigonometric Functions
Here are some Inverse Trigonometric Functions. These are the list of Inverses of Trigonometric Functions:
Inverse of Sine function:-
Consider the function f: R?R given by f(x) = sinx. The graph of the function is shown in the following figure.
Clearly it is a many – one into function as it attains same value at infinitely many points and its range [?1,1] is not same as its co-domain. We know that any function can be made an onto function, if we replace its co-domain by its range. Therefore, f : R?[?1,1] is a many one onto functions.
In order to make f a one- one function, we will have to restrict its domain in such a way that in that domain there is no turn in the function and the function takes every value between ?1 and 1. It is evident from the graph of f(x) = sinx that if we take the domain as [??/2, ?/2], then f(x) becomes one-one. Thus,
f : [??/2, ?/2] ? [?1, 1] given by
f(?) = sin?
is a bijection and hence invertible.
The inverse of the sine function is denoted by sin?1(x). Thus, sin?1 x function with domain [?1, 1] and range [??/2, ?/2] such that
Sin?1x = ?
So, Sin? = x.
Also,
Sin?1(sin?) = ? for all ? ? [??/2, ?/2]
And, Sin(sin?1x) = x for all x ? [?1, 1]
The graph of the function f : [??/2, ?/2] ? [?1, 1] given by f(x) =sinx is shown in fig.(b). in order to obtain the graph of sin-1: [?1, 1] ? [??/2, ?/2] we interchange x and y axes as shown in fig(c).
Students can also avail help with Calculus homework problems involving Inverse Trigonometric function from the online tutors.
Inverse Trigonometric Functions Calculator
Here we shall learn about the Inverse Trigonometric Functions of some more Trigonometric Functions. The Inverse Trigonometric Functions Calculator tool available online provides the Inverse Trigonometric Functions of a given trignometric function.
In the above section, we have discussed about the inverse of sine function and its graph. Similarly we can define the inverses of the other five trigonometric function and their principal branches. The following table gives the domains, ranges and the principal value branches of all inverse trigonometric functions.
Function Domain Range principal value branch
sin-1 x [?1, 1] [??/2, ?/2] ??/2 ? y ? ?/2, where y = sin-1x
cos-1 x [?1, 1] [0, ?] 0 ? y ? ?, where y = cos-1x
tan-1 x R (??/2, ?/2) -?/2 < y < ?/2, where y = tan-1x
cosec-1 x (??, ?1] U [1, ?) [??/2, ?/2] – {0} ??/2 ? y ? ?/2, where y = cosec-1x, y?0
sec-1 x (??, ?1] U [1, ?) [0, ?] – [?/2] 0 ? y ? ?, where y= sec-1x, y ? ?/2
cot-1 x R (0, ?) 0 < y < ?, where y = cot-1x
NOTE:- If no branch of an inverse trigonometric function is mentioned, then it means the principal value branch of that function.
Students can get help with Trigonometry from the online tutors.