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Geometry Level 3

s = 1 − cos ⁡ a + cos ⁡ 2 a − cos ⁡ 3 a + ⋯ r = 1 − sin ⁡ a + sin ⁡ 2 a − sin ⁡ 3 a + ⋯ s = 1- \cos a + \cos^2 a - \cos^3 a + \cdots \\ r = 1-\sin a + \sin^2 a - \sin^3 a + \cdots

We are given the two equations above. State the expression below in terms of r r and/or s s .

1 − sin ⁡ a cos ⁡ a + sin ⁡ 2 a cos ⁡ 2 a − sin ⁡ 3 a cos ⁡ 3 a + ⋯ 1 - \sin a \cos a + \sin^2 a \cos^2 a - \sin^3 a \cos^3 a + \cdots

s*r s r/(1+s r) s r/(1-(s+r) + 2 s*r) 1/(1+s*r)

Raven Herd by Raven Herd , 21, India

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1 solution

Raven Herd
Dec 23, 2015

Since |sin a| <=1 and |cos a| <=1 thus s and r are infinite geometric progressions. 1/(1+sin a) = r and 1/(1+cos a)= s sin a = 1/r -1 and cos a = 1/s - 1 the progression in question equals 1/(1 + (sin a) *(cos a)) Plug in the values of sin a and cos a we get option A as the answer and we are done.

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