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Geometry Level 5

Let P = tan ⁡ θ 2 + 2 tan ⁡ θ + 4 tan ⁡ 2 θ + 8 cot ⁡ 4 θ P = \tan \dfrac \theta2 + 2\tan \theta + 4\tan2\theta + 8\cot4\theta and given that 1 1 − cos ⁡ θ + i sin ⁡ θ \dfrac1{1- \cos\theta + i \sin\theta} can be expressed in the form of a + b i a + bi , where a a and b b are real numbers with i = − 1 i = \sqrt{-1} . Compute a ⋅ b P \dfrac{a\cdot b}{ P } .


The answer is -0.25.

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Raj Rajput by RAJ RAJPUT , 25, India

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3 solutions

Raj Rajput
Jan 30, 2016

11 Helpful 0 Interesting 0 Brilliant 0 Confused

That is a wonderful answer,, but i m sorry because i don t understand the concept,, may you give more explanation about the concept,, thanks,,

Dea Driastuti - 5 years, 4 months ago

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Which point is difficult for you ??

RAJ RAJPUT - 5 years, 4 months ago
Ariel Gershon
Feb 4, 2016

Let x = tan ⁡ θ 2 x = \tan{\dfrac{\theta}{2}} . We can use the double-angle formula to get tan ⁡ ( θ ) = 2 x 1 − x 2 \tan{(\theta)} = \dfrac{2x}{1-x^2} . Applying the formula again and simplifying, we get: tan ⁡ ( 2 θ ) = 4 x ( 1 − x 2 ) 1 − 6 x 2 + x 4 \tan{(2\theta)} = \dfrac{4x(1-x^2)}{1-6x^2+x^4} . If we apply the formula one more time and take the reciprocal, we get: cot ⁡ ( 4 θ ) = 1 tan ⁡ ( 4 θ ) = ( 1 − 6 x 2 + x 4 ) 2 − 16 x 2 ( 1 − x 2 ) 2 8 x ( 1 − x 2 ) ( 1 − 6 x 4 + x 4 ) \cot{(4\theta)} = \dfrac{1}{\tan{(4\theta)}} = \dfrac{(1-6x^2+x^4)^2-16x^2(1-x^2)^2}{8x(1-x^2)(1-6x^4+x^4)}

Therefore, P = x + 4 x 1 − x 2 + 16 x ( 1 − x 2 ) 1 − 6 x 2 + x 4 + ( 1 − 6 x 2 + x 4 ) 2 − 16 x 2 ( 1 − x 2 ) 2 x ( 1 − x 2 ) ( 1 − 6 x 4 + x 4 ) P = x + \dfrac{4x}{1-x^2} + \dfrac{16x(1-x^2)}{1-6x^2+x^4} + \dfrac{(1-6x^2+x^4)^2-16x^2(1-x^2)^2}{x(1-x^2)(1-6x^4+x^4)} After some simplifying, the result is P = 1 x = cot ⁡ θ 2 P = \dfrac{1}{x} = \cot{\dfrac{\theta}{2}} .

Now onto the second part of the problem:

1 1 − cos ⁡ ( θ ) + i sin ⁡ ( θ ) = 1 − cos ⁡ ( θ ) − i sin ⁡ ( θ ) ( 1 − cos ⁡ ( θ ) ) 2 + sin ⁡ 2 ( θ ) = 1 − cos ⁡ ( θ ) − i sin ⁡ ( θ ) 2 − 2 cos ⁡ ( θ ) = 1 2 − sin ⁡ ( θ ) 2 − 2 cos ⁡ ( θ ) i \dfrac{1}{1-\cos(\theta)+ i \sin(\theta)} = \dfrac{1-\cos(\theta) - i \sin(\theta)}{(1-\cos(\theta))^2 + \sin^2(\theta)} = \dfrac{1-\cos(\theta) - i \sin(\theta)}{2 - 2\cos(\theta)} = \dfrac{1}{2} - \dfrac{\sin(\theta)}{2 - 2\cos(\theta)} i

Thus a = 1 2 a = \dfrac{1}{2} and b = − sin ⁡ ( θ ) 2 − 2 cos ⁡ ( θ ) b = - \dfrac{\sin(\theta)}{2 - 2\cos(\theta)} . Substitute sin ⁡ ( θ ) = 2 sin ⁡ θ 2 cos ⁡ θ 2 \sin(\theta) = 2\sin\dfrac{\theta}{2} \cos\dfrac{\theta}{2} and cos ⁡ ( θ ) = 1 − 2 sin ⁡ 2 θ 2 \cos(\theta) = 1 - 2\sin^2\dfrac{\theta}{2} :

b = − 2 sin ⁡ θ 2 cos ⁡ θ 2 4 sin ⁡ 2 θ 2 = − 1 2 cot ⁡ θ 2 b = - \dfrac{2\sin\dfrac{\theta}{2} \cos\dfrac{\theta}{2}}{4\sin^2\dfrac{\theta}{2}} = - \dfrac{1}{2}\cot\dfrac{\theta}{2}

Hence, a ∗ b P = 1 2 ∗ ( − 1 2 cot ⁡ θ 2 ) ÷ cot ⁡ θ 2 = − 1 4 \dfrac{a*b}{P} = \dfrac{1}{2} * \left(-\dfrac{1}{2}\cot\dfrac{\theta}{2}\right) ÷ \cot{\dfrac{\theta}{2}} = \boxed{\dfrac{-1}{4}}

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Seriously you have done this or it was just to show another method regarding this problem ??

RAJ RAJPUT - 5 years, 4 months ago

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Yeah, this is how I actually solved it. Your method is a lot simpler but I didn't think of it.

Ariel Gershon - 5 years, 4 months ago

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Your method is a bit complicated but provides a positive thinking "STICKING TO BASICS FIRST" may even also solve the problem. Good work man :) :)

RAJ RAJPUT - 5 years, 4 months ago

Put theta =60° and get d ans

0 Helpful 0 Interesting 0 Brilliant 0 Confused

Really........

Spandan Senapati - 4 years, 3 months ago

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