IIT JEE , AIEEE Forum - Mathematics , Trignometry - Triugonometry question-please explain

rahulraj1

If A,B,C and D are angles of a quadilateral and sin A/2.sinB/2.sinC/2.sinD/2=1/4,prove that A=B=C=D=/2

rahulraj1

Anyone please reply me.

risin

You can prove it conversely.

ssbrightstar1

please reply this question

hsbhatt

You will need to invoke an inequality known as Jensen's Inequality.

$\frac{A}{2} + \frac{B}{2} + \frac{C}{2} + \frac{D}{2} = \pi$

Since sin(x) is a concave function in the interval [0, 180^o], from Jensen's Inequality

$\frac{\sin \frac{A}{2}+ \sin \frac{B}{2}+\sin \frac{C}{2}+\sin \frac{D}{2}} {4} \le \sin \frac{\pi}{4} = \frac{1}{\sqrt 2}$

Hence from AM-GM Inequality

$\left (\sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \sin \frac{D}{2} \right )^{\frac{1}{4}} \le \frac{1}{\sqrt 2}$

$\Rightarrow \left (\sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \sin \frac{D}{2} \right ) \le \frac{1}{4}$

We are given that $\sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \sin \frac{D}{2} = \frac{1}{4}$

The equality condition in this case occurs when $A = B = C = D = \frac{\pi}{2}$

rahulraj1

Very-very thanx sir.

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