# Thread: ooh a tricky question

1. ## ooh a tricky question

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

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

2. You may try using mathematical induction.

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

(A)$\displaystyle \frac{1}{x} + \frac{1}{y} = 1$
(B)$\displaystyle x + y +xy =0$
(C)$\displaystyle xyz = xy + z$
(D)$\displaystyle xyz = x + y + z$
There is something wrong with this question, because $\displaystyle x+y = \displaystyle\Sigma_{n=0}^{\infty}(\cos ^2n \theta + \sin^2n \theta) = \Sigma_{n=0}^{\infty}1$, which is infinite. So at least one of the series for x and y fails to converge. If for example $\displaystyle \theta = \pi/4$ then the three series for x, y, z all diverge, so none of the conditions (A), (B), (C), (D) is true.

5. oh sorry it was just a typing error i corrected the question actually it sin to the power of 2n

6. 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 $\displaystyle x = \dfrac1{1-\cos^2\theta} = \dfrac1{\sin^2\theta}$ and $\displaystyle y = \dfrac1{1-\sin^2\theta} = \dfrac1{\cos^2\theta}$, from which it should be clear that (A) is correct.

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

8. 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?

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

(A)$\displaystyle \frac{1}{x} + \frac{1}{y} = 1$
(B)$\displaystyle x + y +xy =0$
(C)$\displaystyle xyz = xy + z$
(D)$\displaystyle xyz = x + y + z$
me too getting c and d as answer with a(when substituting)

10. 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.

11. 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.

12. yea but how to prove (C) and (D) are correct?? tell me please

13. 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 $\displaystyle x = \dfrac1{\sin^2\theta},\quad y = \dfrac1{\cos^2\theta},\quad z = \dfrac1{1-\sin^2\theta\cos^2\theta}$. Substitute those into the equations (C) and (D), and check that both sides agree.

14. what have you used to get that formula??

15. Originally Posted by grgrsanjay what have you used to get that formula??
Sum of a infinite geometric series, $\displaystyle \displaystyle\sum_{n=0}^\infty x^n = \frac1{1-x}$.

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