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

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

Answered by virus last updated on 30/Jun/17

$$1-2\sin {}^2\left(\theta /2\right)\cos {}^2\left(\theta /2\right)⩾1/2$$$$2-4\sin {}^2\left(\theta /2\right)\cos {}^2\left(\theta /2\right)⩾1$$$$2-\sin {}^2\theta ⩾1$$$$1-\sin {}^2\theta ⩾0$$$$\cos {}^2\theta ⩾0$$$$\therefore \theta \in (2n+1)\pi /2$$

Answered by mrW1 last updated on 02/Jul/17

$${\sin }^4\frac{\theta }{2}+{\cos }^4\frac{\theta }{2}$$$$={\sin }^4\frac{\theta }{2}+{(1-{\sin }^2\frac{\theta }{2})}^2$$$$={\sin }^4\frac{\theta }{2}+1-{2\sin }^2\frac{\theta }{2}+{\sin }^4\frac{\theta }{2}$$$$=2({\sin }^4\frac{\theta }{2}-{\sin }^2\frac{\theta }{2})+1$$$$=2({\sin }^4\frac{\theta }{2}-{\sin }^2\frac{\theta }{2}+\frac{1}{4})+\frac{1}{2}$$$$=\frac{1}{2}+2{({\sin }^2\frac{\theta }{2}-\frac{1}{2})}^2$$$$=\frac{1}{2}+2{(\frac{1-\cos \theta }{2}-\frac{1}{2})}^2$$$$=\frac{1}{2}+\frac{{\cos }^2\theta }{2}$$$$=\frac{1}{2}(1+{\cos }^2\theta )$$$$⩾\frac{1}{2}$$

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