Trigonometry! 47

Geometry Level 4

For 0 < ϕ < π 2 0<\phi<\frac {\pi}{2} , if x = ∑ n = 0 ∞ cos ⁡ 2 n ϕ x=\displaystyle \sum_{n=0}^{\infty} \cos^{2n} \phi y = ∑ n = 0 ∞ sin ⁡ 2 n ϕ y=\displaystyle \sum_{n=0}^{\infty} \sin^{2n} \phi z = ∑ n = 0 ∞ cos ⁡ 2 n ϕ sin ⁡ 2 n ϕ z=\displaystyle \sum_{n=0}^{\infty} \cos^{2n} \phi \sin^{2n} \phi then which of the following holds true?

  1. x y z = x z + y xyz=xz+y

  2. x y z = x y + z xyz=xy+z

  3. x y z = x + y + z xyz=x+y+z

  4. x y z = y z + x xyz=yz+x

Note that two or three options are correct. Enter the sum of serial numbers of the correct options.

This problem is part of the set Trigonometry .


The answer is 5.

Omkar Kulkarni by Omkar Kulkarni , 22, India

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1 solution

Prakhar Gupta
Feb 3, 2015

Using Infinite GP formula we have:- x = 1 1 − cos ⁡ 2 x = 1 sin ⁡ 2 x x = \dfrac{1}{1-\cos^{2}x} = \dfrac{1}{\sin^{2}x} y = 1 1 − sin ⁡ 2 x = 1 cos ⁡ 2 x y = \dfrac{1}{1-\sin^{2}x} = \dfrac{1}{\cos^{2}x} z = 1 1 − sin ⁡ 2 x cos ⁡ 2 x z = \dfrac{1}{1-\sin^{2}x \cos^{2}x} We observe LHS:- x y z = 1 sin ⁡ 2 x cos ⁡ 2 x ( 1 − sin ⁡ 2 x cos ⁡ 2 x ) xyz = \dfrac{1}{\sin^{2}x \cos^{2}x(1-\sin^{2}x \cos^{2}x)} x y z = sin ⁡ 2 x cos ⁡ 2 x + ( 1 − sin ⁡ 2 x cos ⁡ 2 x ) sin ⁡ 2 x cos ⁡ 2 x ( 1 − sin ⁡ 2 x cos ⁡ 2 x ) xyz = \dfrac{\sin^{2}x \cos^{2}x +(1-\sin^{2}x \cos^{2}x)}{\sin^{2}x \cos^{2}x(1-\sin^{2}x \cos^{2}x)} x y z = 1 1 − sin ⁡ 2 x cos ⁡ 2 x + 1 sin ⁡ 2 x cos ⁡ 2 x xyz = \dfrac{1}{1-\sin^{2}x \cos^{2}x} + \dfrac{1}{\sin^{2}x \cos^{2}x} x y z = z + x y xyz = z + xy Also :- x y z = z + 1 sin ⁡ 2 x cos ⁡ 2 x xyz = z +\dfrac{1}{\sin^{2}x \cos^{2}x} x y z = z + sin ⁡ 2 x + cos ⁡ 2 x sin ⁡ 2 x cos ⁡ 2 x xyz = z + \dfrac{\sin^{2}x + \cos^{2}x}{\sin^{2}x \cos^{2}x} x y z = z + 1 sin ⁡ 2 x + 1 cos ⁡ 2 x xyz = z + \dfrac{1}{\sin^{2}x} + \dfrac{1}{\cos^{2}x} x y z = z + x + y xyz = z + x + y

7 Helpful 0 Interesting 0 Brilliant 0 Confused

what if we divide the numerator and denominator of "z" by first (cosx)^2and then by (sinx)^2.we get cosecx=secx which implies that x=45?

avn bha - 6 years ago

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I'm sorry, I didn't get you. Can you please elaborate.

Prakhar Gupta - 6 years ago

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