If-P-n-cos-n-sin-n-0-pi-2-n-2-then-minimum-of-P-n-will-be-1-1-2-1-2-3-2-4-1-2-

Question Number 18402 by Tinkutara last updated on 20/Jul/17

If P_{n} = \cos^{n} θ + \sin^{n} θ, θ \in [0, \frac{π}{2}], n \in

(- \infty, 2), then minimum of P_{n} will be

(1) 1

(2) \frac{1}{2}

(3) \sqrt{2}

(4) \frac{1}{\sqrt{2}}

Answered by Tinkutara last updated on 28/Jul/17

{(\sin θ)}^{- n} ⩾ \dots ⩾ {(\sin θ)}^{- 1} ⩾ \sin θ ⩾ \sin^{2} θ

{(\cos θ)}^{- n} ⩾ \dots ⩾ {(\cos θ)}^{- 1} ⩾ \cos θ ⩾ \cos^{2} θ

{(\sin θ)}^{- n} + {(\cos θ)}^{- n} > \sin^{2} θ + \cos^{2} θ = 1

∴ Minimum of P_{n} = 1