13http://targetiit.com/iit_jee_forum/posts/help_1694.html
this is being asked for the 3rd time in tiit....
1)If ar+1=√1/2(ar+1) then prove that cos[√1-a02/a1a2....∞]=a0
2)If A+B+C+D=2∩ prove that sinA/2.sinB/2.sinC/2.sinD/2≤1/4
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12 Answers
3how to use latex pls help me out there.i think my browser is not supporting latex
133sir i have worked for more than 1 and half hr still didnt think abt substitution.
Sir can u prove the next inequality.
341You havent mentioned this but gping by the result we have to prove it appears we can take that |a_0|<1
It is evident that if you let a_0 = \ cos \theta, then a_1 = \sqrt{\frac{1}{2} (1+\cos \theta)} = \cos \frac{\theta}{2}
You can now prove that a_n = \cos \frac{\theta}{2^{n-1}}
\frac{\sqrt{1-a_0^2}}{a_1a_2...a_n} = \frac{\sin \theta}{\cos \frac{\theta}{2} \cos \frac{\theta}{4} ...\cos \frac{\theta}{2^{n-1}}} = 2^{n-1}\sin \frac{\theta}{2^{n-1}} \rightarrow \theta \ \text{as} \ n \rightarrow \infty
and the result follows
341The second one is best done by using Jensen's Inequality.
Consider the function f(x) = log (sin x).
This function is convex.
Hence we can write
\dfrac{\left(\log \sin \frac{A}{2} + \log \sin \frac{B}{2}+\log \sin \frac{C}{2}+\log \sin \frac{D}{2}\right)}{4} \le \log \sin \left(\dfrac{\frac{A}{2} + \frac{B}{2}+\frac{C}{2}+\frac{D}{2}}{4}\right)
But \frac{A+B+C+D}{8} = \frac{2 \pi}{8} = \frac{\pi}{4} and I am sure you can finish it now
341If you want a non-calculus function, we can do it this way.
Essentially you will be proving convexity but still: We will prove that
\sin \theta_1 \sin \theta_2 \le \sin^2 \frac{\theta_1 + \theta_2}{2}
This is true as otherwise
\sin \theta_1 \sin \theta_2 > \sin^2 \frac{\theta_1+\theta_2}{2} \\ \\ \Rightarrow 2 \sin \theta_1 \sin \theta_2 > 1 - \cos (\theta_1 + \theta_2) \\ \\ \Rightarrow \cos (\theta_1 - \theta_2)>1
That means by replacing the set (A,B,C,D) with
\left(\frac{A+B}{2}, \frac{A+B}{2},C,D \right) we are not changing the sum, but increasing the value of the given function
Hence we have
\sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \sin \frac{D}{2} \le \sin^2 \frac{A+B}{4} \sin^2 \frac{C+D}{4} \le \sin^4 \frac{A+B+C+D}{8}
as above
11Sir can't we do the 2nd via trigno?
tryin cos and sin formulas
i tried but in vain!