# ooh a tricky question

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• September 3rd 2010, 08:27 AM
grgrsanjay
ooh a tricky question
i don't know weather its grouped correctly(sorry)
for $0 < \theta < \frac{\pi}{2}$if
$x = \displaystyle\Sigma_{n=0}^{\infty}(\cos ^{2n} \theta)$
$y = \displaystyle\Sigma_{n=0}^{\infty}(\sin^{2n} \theta)$
$z = \displaystyle\Sigma_{n=0}^{\infty}(\sin^{2n} \theta \cos^{2n} \theta)$
then which of the following are true(multiple choice)

(A) $\frac{1}{x} + \frac{1}{y} = 1$
(B) $x + y +xy =0$
(C) $xyz = xy + z$
(D) $xyz = x + y + z$
• September 3rd 2010, 09:33 AM
wonderboy1953
You may try using mathematical induction.
• September 3rd 2010, 05:23 PM
grgrsanjay
THE ACTUAL TIME GIVEN FOR PROBLEM IS 3 MINUTES using mathematical induction it takes 10 minutes at least

i think we should expand it and take the G.P sum of n terms and do something might work

this is a tricky sum for me :(:(
• September 3rd 2010, 11:58 PM
Opalg
Quote:

Originally Posted by grgrsanjay
i don't know weather its grouped correctly(sorry)
for $0 < \theta < \frac{\pi}{2}$if
$x = \displaystyle\Sigma_{n=0}^{\infty}(\cos ^2n \theta)$
$y = \displaystyle\Sigma_{n=0}^{\infty}(\sin^2n \theta)$
$z = \displaystyle\Sigma_{n=0}^{\infty}(\sin^2n \theta \cos^2n \theta)$
then which of the following are true(multiple choice)

(A) $\frac{1}{x} + \frac{1}{y} = 1$
(B) $x + y +xy =0$
(C) $xyz = xy + z$
(D) $xyz = x + y + z$

There is something wrong with this question, because $x+y = \displaystyle\Sigma_{n=0}^{\infty}(\cos ^2n \theta + \sin^2n \theta) = \Sigma_{n=0}^{\infty}1$, which is infinite. So at least one of the series for x and y fails to converge. If for example $\theta = \pi/4$ then the three series for x, y, z all diverge, so none of the conditions (A), (B), (C), (D) is true.
• September 4th 2010, 04:11 AM
grgrsanjay
oh sorry it was just a typing error i corrected the question actually it sin to the power of 2n
• September 4th 2010, 05:38 AM
Opalg
Quote:

Originally Posted by grgrsanjay
oh sorry it was just a typing error i corrected the question actually it sin to the power of 2n

In that case, these are geometric series, and the sums are given by $x = \dfrac1{1-\cos^2\theta} = \dfrac1{\sin^2\theta}$ and $y = \dfrac1{1-\sin^2\theta} = \dfrac1{\cos^2\theta}$, from which it should be clear that (A) is correct.
• September 4th 2010, 06:24 AM
grgrsanjay
yea i got (A) correct so any other options are correct??? but when substitute a value (C) and (D) also seem correct to me
• September 4th 2010, 06:27 AM
undefined
Quote:

Originally Posted by grgrsanjay
yea i got (A) correct so any other options are correct??? but when substitute a value (C) and (D) also seem correct to me

Is this a multiple choice question or isn't it?
• September 4th 2010, 09:10 AM
ggn
Quote:

Originally Posted by grgrsanjay
which of the following are true(multiple choice)says it is multiple

(A) $\frac{1}{x} + \frac{1}{y} = 1$
(B) $x + y +xy =0$
(C) $xyz = xy + z$
(D) $xyz = x + y + z$

me too getting c and d as answer with a(when substituting)
• September 4th 2010, 09:24 AM
undefined
Quote:

Originally Posted by ggn
says it is multiple

Yes thank you I can read, my point is that in a valid multiple choice question it's not possible to have more than one correct option, therefore if A is correct then B,C,D are automatically incorrect. If the OP wanted to call into question the validity of the problem, the OP could have specified that.
• September 4th 2010, 09:43 AM
Opalg
Quote:

Originally Posted by grgrsanjay
yea i got (A) correct so any other options are correct??? but when substitute a value (C) and (D) also seem correct to me

That's right! I only looked at (A), because when I saw that was correct I assumed that it would be the only correct choice. But (C) and (D) are also correct.
• September 5th 2010, 04:32 AM
grgrsanjay
yea but how to prove (C) and (D) are correct?? tell me please
• September 5th 2010, 05:54 AM
Opalg
Quote:

Originally Posted by grgrsanjay
yea but how to prove (C) and (D) are correct?? tell me please

The formulas for x, y and z are $x = \dfrac1{\sin^2\theta},\quad y = \dfrac1{\cos^2\theta},\quad z = \dfrac1{1-\sin^2\theta\cos^2\theta}$. Substitute those into the equations (C) and (D), and check that both sides agree.
• September 5th 2010, 06:01 AM
grgrsanjay
what have you used to get that formula??
• September 5th 2010, 06:07 AM
Opalg
Quote:

Originally Posted by grgrsanjay
what have you used to get that formula??

Sum of a infinite geometric series, $\displaystyle\sum_{n=0}^\infty x^n = \frac1{1-x}$.
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