Question

Find an angle θ, 0 < θ < `pi/2`, which increases twice as fast as its sine.

Answer 1

As per the given condition,

`("d"theta)/"dt" = 2 "d"/"dt" (sin theta)`

⇒ `("d"theta)/"dt" = 2 cos theta * ("d"theta)/"dt"`

⇒ 1 = 2 cos θ

∴ cos θ = `1/2`

⇒ cos θ = `cos  pi/3`

⇒ θ = `pi/3`

Hence, the required angle is `pi/3`.