Question

Observe the two graphs, Graph 1 and Graph 2 given below and answer the questions that follow.

(i) Which one of the graphs represents y = sin^–1x?   [1]

(ii) Write the domain and range of y = sin^–1x.   [1]

(iii) Prove that `sin^-1  1/sqrt(5) + sin^-1  2/sqrt(5) = π/2`   [2]

(iv) Find the value of `tan^-1 [2sin (2 cos^-1  sqrt(3)/2)]`   [2]

Answer 1

(i) Graph (1) represents y = sin^–1x.

(ii) Domain of sin^–1x is [–1, 1].

Range = `[(-π)/2, π/2]`

(iii) We know

`sin^-1x + sin^-1y = sin^-1 [xsqrt(1 - y^2) + ysqrt(1 - x^2)]`

∴ `sin^-1  1/sqrt(5) + sin^-1  2/sqrt(5)`

= `sin^-1 [1/sqrt(5) sqrt(1 - 4/5) + 2/sqrt(5) sqrt(1 - 1/5)]`

= `sin^-1 [1/sqrt(5) xx 1/sqrt(5) + 2/sqrt(5) xx 2/sqrt(5)]`

= `sin^-1 (1/5 + 4/5)`

= sin^–1 (1)

= `π/2`

∴ `sin^-1  1/sqrt(5) + sin^-1  2/sqrt(5) = π/2`

Hence Proved.

(iv) `tan^-1 [2sin (2cos^-1  sqrt(3)/2)]`

⇒ `tan^-1 [2sin (2 xx π/6)]`

⇒ `tan^-1 [2sin  π/3]`

⇒ `tan^-1 (2 xx sqrt(3)/2)`

⇒ `tan^-1 (sqrt(3))`

⇒ `π/3`