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