CASE STUDY CHALLENGE
Inverse Trigonometric Functions
If a function f : X → Y defined as f(x) = y is one-one and onto, then we can define a unique function g : Y → X such that g(y) = x, where x ∈ X and y = f(x), y ∈ Y. Function g is called the inverse of function f.
The domain of sine function is R and function sine : R → R is neither one-one nor onto. The following graph shows the sine function:
Let sine function be defined from set A to [-1, 1] such that inverse of sine function exists, i.e., sin-1 x is defined from [-1, 1] to A.
On the basis of the above information, answer the following questions:
Answer the following:
| (i) | If A is the interval other than principal value branch, give an example of one such interval. | [1] |
| (ii) | If sin-1(x) is defined from [-1, 1] to its principal value branch, find the value of sin-1(-1/2) – sin-1(1). | [1] |
| (iii) |
(a) Draw the graph of sin-1 x from [-1, 1] to its principal value branch. — OR —
(b) Find the domain and range of f(x) = 2 sin-1(1 – x). |
[2] |