Is the function sin(sin1n) bijective? Which is defined on [-1, 1] to

1.	Is the function sin(sin1n) bijective? Which is defined on [-1, 1] to [-1, 1].
Answer» Let f (n) = sin(sin^-1n) let n₁ n₂∈ [-1, 1] f(n₁) = f(n₂) ⇒ sin(sin^-1n₁) = sin[sin₁(n2)] ⇒ n₁= n₂ hence f is one-one let y = sin (sin^-1 n) = n ∀ y ∈ [-1,1] ∃ [-1, 1] f(n) = n hence f is onto since f is one-one and onto it is bijective.

Is the function sin(sin1n) bijective? Which is defined on [-1, 1] to [-1, 1].

Answer»

Let f (n) = sin(sin^-1n)

let n₁ n₂∈ [-1, 1]

f(n₁) = f(n₂) ⇒ sin(sin^-1n₁)

= sin[sin₁(n2)]

⇒ n₁= n₂

hence f is one-one

let y = sin (sin^-1 n) = n

∀ y ∈ [-1,1] ∃ [-1, 1]

f(n) = n hence f is onto

since f is one-one and onto it is bijective.