I tried this but could not get much far.Quote:

Originally Posted byJakeD

$\displaystyle x=cosA (0<A<\frac{\pi}{2})$

$\displaystyle y=cosB (0<B<\frac{\pi}{2})$

$\displaystyle cosA + cosB \leq 1 \leq cos^2A + cos^2B$

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Malay

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- Jun 23rd 2006, 09:30 PMmalaygoelQuote:

Originally Posted by**JakeD**

$\displaystyle x=cosA (0<A<\frac{\pi}{2})$

$\displaystyle y=cosB (0<B<\frac{\pi}{2})$

$\displaystyle cosA + cosB \leq 1 \leq cos^2A + cos^2B$

KeepSmiling

Malay - Jun 24th 2006, 02:05 AMJakeDQuote:

Originally Posted by**JakeD**

$\displaystyle \begin{aligned}

0 \le x &\le 1 \\

1-x \le y &\le \sqrt{1-x^2}. \\

\end{aligned}$

The probability, which is a double integral, can then be evaluated as an iterated integral

$\displaystyle \begin{aligned}P &= \underset{D\ \ }{\iint} 1\ dxdy \\

&= \int_0^1 dx\ \int_{1-x}^{\sqrt{1-x^2}} 1\ dy \\

&= \int_0^1 \sqrt{1-x^2} - (1-x)\ dx \\

&= \sin^{-1}(1)/2 - (1 - 1/2) \\

&= \frac{\pi -2}{4}.

$

I used The Integrator to find the antiderivative of $\displaystyle \sqrt{1-x^2} .$ - Jun 24th 2006, 09:52 PMmalaygoelQuote:

Originally Posted by**JakeD**

I want to tell you that this problem is assigned to a student with the instructions to avoid calculus and coordinate geometry for this question(I am sorry, I should have told you this before).

You can use trigonometry and combanotrics,algebra and number theory.

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Malay - Jun 25th 2006, 02:39 AMJakeDQuote:

Originally Posted by**malaygoel**

- Jun 25th 2006, 09:13 PMmalaygoelQuote:

Originally Posted by**JakeD**

Nice

Thanks for your support

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Malay