You will hate trigonometry after this

Geometry Level 2

If S=1+sin^2+sin^4+sin^6.........infinity. Then S= _

1 Sin^2 Cos^2 Sec^2

Ashtik Mahapatra by Ashtik Mahapatra , 22, India

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3 solutions

S = 1 + sin ⁡ 2 x + sin ⁡ 4 x + sin ⁡ 6 x + … S = 1 + \sin^2x + \sin^4x + \sin^6x + \ldots

S − 1 = sin ⁡ 2 x + sin ⁡ 4 x + sin ⁡ 6 x + … S - 1 = \sin^2x + \sin^4x + \sin^6x + \ldots

S − 1 = sin ⁡ 2 x ( 1 + sin ⁡ 2 x + sin ⁡ 4 x + sin ⁡ 6 x + … ) S - 1 = \sin^2x( 1 + \sin^2x + \sin^4x + \sin^6x + \ldots)

S − 1 = sin ⁡ 2 x ( S ) S - 1 = \sin^2x(S)

S − sin ⁡ 2 x ( S ) = 1 S - \sin^2x(S) = 1

S ( 1 − sin ⁡ 2 x ) = 1 S(1 - \sin^2x) = 1

S ( cos ⁡ 2 x ) = 1 S(\cos^2x) = 1

S = 1 cos ⁡ 2 x S = \dfrac{1}{\cos^2x}

S = sec ⁡ 2 x S = \boxed{ \sec^2x}

8 Helpful 0 Interesting 0 Brilliant 0 Confused

/Siddhartha Shrivastav nice solution

Ashtik Mahapatra - 7 years, 2 months ago

This is a summation of gemetric series of radix of sin^2 from 0 to inf So the sum = 1/(1-sin^2)= sec^2 easy task direct

Dr.Hesham ElBadawy - 7 years, 1 month ago
Shreyansh Vats
Apr 26, 2014

This problem is relates to the sum of an infinite Geometric Progression. So, simply using the formula a/1-r, we get the answer to be sec^2.

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Soumyadeep Bose
Apr 13, 2014

Value of sin^2 always lies between 0 and 1,so,any higher powers of 2,that is,sin^4,sin^6,etc. will have lesser and lesser values than sin^2,hence the infinite series becomes a GP series with convergence...Thus,S=1/(1-sin^2)=1/(cos^2)=sec^2...Hence S=sec^2

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