sin-4-2-cos-4-2-1-2-

Question Number 17080 by Kunal kumar shukla last updated on 30/Jun/17

\sin^{4} θ / 2 + \cos^{4} θ / 2 ⩾ 1 / 2

Answered by virus last updated on 30/Jun/17

1 - 2 \sin^{2} (θ / 2) \cos^{2} (θ / 2) ⩾ 1 / 2

2 - 4 \sin^{2} (θ / 2) \cos^{2} (θ / 2) ⩾ 1

2 - \sin^{2} θ ⩾ 1

1 - \sin^{2} θ ⩾ 0

\cos^{2} θ ⩾ 0

∴ θ \in (2 n + 1) π / 2

Answered by mrW1 last updated on 02/Jul/17

\sin^{4} \frac{θ}{2} + \cos^{4} \frac{θ}{2}

= \sin^{4} \frac{θ}{2} + {(1 - \sin^{2} \frac{θ}{2})}^{2}

= \sin^{4} \frac{θ}{2} + 1 - {2 \sin}^{2} \frac{θ}{2} + \sin^{4} \frac{θ}{2}

= 2 (\sin^{4} \frac{θ}{2} - \sin^{2} \frac{θ}{2}) + 1

= 2 (\sin^{4} \frac{θ}{2} - \sin^{2} \frac{θ}{2} + \frac{1}{4}) + \frac{1}{2}

= \frac{1}{2} + 2 {(\sin^{2} \frac{θ}{2} - \frac{1}{2})}^{2}

= \frac{1}{2} + 2 {(\frac{1 - \cos θ}{2} - \frac{1}{2})}^{2}

= \frac{1}{2} + \frac{\cos^{2} θ}{2}

= \frac{1}{2} (1 + \cos^{2} θ)

⩾ \frac{1}{2}

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