1Sorry , You missed the main part of the question -- that there are indeed such two angles -- that is what you have to prove ----
P.S -- sin-1x - sin-1y formula comes under many conditions , all of which you have to check ---
Good an' simple --- try this one ---
Given four distinct numbers in the interval (0,1) , show that there exists 2 numbers among them x
and y , such that ------
0 < x \sqrt{1 - y^{2}} - y \sqrt{1 - x^{2}} < \frac{1}{2}
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4 Answers
11Let the nos. be of the form sin-1a, sin-1b, sin-1c, sin-1d....
Since they lie within pi to -pi, difference of some 2 must be less than pi\6....Now applying sin-1x-sin-1y formula , we have our result.
1
341Divide the open interval into \left(0, \frac{1}{3} \right], \left(\frac{1}{3}, \frac{2}{3} \right], \left(\frac{2}{3},1 \right]
Then there exist two numbers \theta, \phi such that \theta- \phi < \frac{1}{3} ...
11Sorry Soumya - indeed it's u (& not me), who actually missed out....in my earlier post, the solution is actually hidden.
Since any real between 0 and 1 can be expressed in terms of sin^{-1}x_i, Let the reals be sin^{-1}x_i =a_i.
Given (Obvious - don't ask me to prove this), that a_1-a_4\le \frac{\pi}{2}
So if I order the reals in an ascending order, we have a_1>a_2>a_3>a_4
Now that gives that the difference of some (a_i-a_{i+1})\le \frac{\pi}{6}
From which our result follows.
Now let's come to ur point....I'll be happy if u've finally understood what we had to prove and what I've proved. I'd surely be happier, if u mention those conditions that ought to have been checked by me before applying sin inverse formual in this problem.
Leaving it to u for completing this thread.