Trigonometric problem

Geometry Level 3

Solve for angle A A (in degrees) given that 0 < 5 A < 90 0 < 5A < 90 and

1 cos ⁡ 0 × cos ⁡ A + 1 cos ⁡ A × cos ⁡ 2 A + 1 cos ⁡ 2 A × cos ⁡ 3 A + 1 cos ⁡ 3 A × cos ⁡ 4 A + 1 cos ⁡ 4 A × cos ⁡ 5 A = 2 − 3 sin ⁡ A \frac{ 1 } {\cos 0 \times \cos A } + \frac{ 1 } { \cos A \times \cos 2A } + \frac{ 1 } { \cos 2A \times \cos 3A } + \frac{1}{ \cos 3A \times \cos 4A } + \frac{1}{ \cos 4A \times \cos 5A } = \frac{ 2 - \sqrt{3} } { \sin A }


The answer is 3.

Sahil Bansal by Sahil Bansal , 23, India

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1 solution

Sahil Bansal
Nov 14, 2016

Solution:

First take S i n A SinA on the other side.Now solve it using the equation

t a n [ ( n − 1 ) A ] tan[(n-1)A] + S i n A C o s [ ( n − 1 ) A ] × C o s ( n A ) \frac{SinA}{Cos[(n-1)A] \times Cos(nA)} = t a n ( n A ) tan(nA) (True for any n n )

Hence the equation reduces to

t a n 5 A tan5A = 2 − 3 2 - \sqrt 3

5 A 5A = 15 15

A A = 3 3

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