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I was trying to solve a problem taken from an Physics Olympiad when I came across a curious and complex mathematical expression. I can not prove with what I know so far about mathematics, does could anyone tell me if the answer is correct?

My answer: $d\tan(\arcsin ((\sin(180-t)*H)/ \sqrt(H² + 4d^2 - 4dH\cos(180-t))))$

The correct answer:$(dH\sin(t))/(2d+H\cos(t))$

The mathematical problem: $d\tan(\arcsin ((\sin(180-t)*H)/ \sqrt(H² + 4d^2 - 4dH\cos(180-t)))) = (dH\sin(t))/(2d+H\cos(t))$

Is it true?

The original problem: http://pir2.forumeiros.com/t23212-angulo-em-relacao-ao-solo

My solution:

enter image description here

 = 180° - ϴ

(A’B)² = (AB)² + (AA’)² - 2. (AB). (AA’).cos(180° - ϴ)
A’B = v(H² + 4d² - 4.H.d.cos(180° - ϴ))

sen(Â’)/AB = sen(Â)/A’B
sen(Â’) = sen(180° - ϴ).H/ v(H² + 4d² - 4.H.d.cos(180° - ϴ))
Â’ = arc sen (sen(180° - ϴ).H/ v(H² + 4d² - 4.H.d.cos(180° - ϴ)))

enter image description here

tg(Â’) = CE/A’C
tg(Â’) = L/d
L = d.tg(Â’)
L = d.tg(arc sen (sen(180° - ϴ).H/ v(H² + 4d² - 4.H.d.cos(180° - ϴ))))
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Cross-posted from physics.stackexchange.com/q/61660/2451 –  Qmechanic Apr 20 '13 at 16:57

1 Answer 1

up vote 2 down vote accepted

Let $$ a = \frac{\sin(180-t)*H}{\sqrt{H^2 + 4d^2 - 4dH\cos(180-t)}} = \frac{\sin(t)*H}{\sqrt{H^2 + 4d^2 + 4dH\cos(t)}}$$ $$1-a^2=\frac{H^2 + 4d^2 + 4dH\cos(t) - \sin^2(t)*H^2}{H^2 + 4d^2 + 4dH\cos(t)} = \frac{H^2(1-\sin^2(t)) + 4d^2 + 4dH\cos(t)}{H^2 + 4d^2 + 4dH\cos(t)} = \frac{H^2cos^2(t) + 4d^2 + 4dH\cos(t)}{H^2 + 4d^2 + 4dH\cos(t)} = \frac{(H\cos(t)+2d)^2}{H^2 + 4d^2 + 4dH\cos(t)}$$ $$\sqrt{1-a^2} = \frac{H\cos(t)+2d}{\sqrt{H^2 + 4d^2 + 4dH\cos(t)}}$$ Let now $\sin^{-1}(a) = b \Rightarrow \sin(b) = a$. Also this means $\cos(b) = \sqrt{1-a^2}$. From all this we can conclude that: $$d\tan(b)=\frac{dsin(b)}{cos(b)}=\frac{da}{\sqrt{1-a^2}} = \frac{d\frac{\sin(t)*H}{\sqrt{H^2 + 4d^2 + 4dH\cos(t)}}}{\frac{H\cos(t)+2d}{\sqrt{H^2 + 4d^2 + 4dH\cos(t)}}} = \frac{d\sin(t)*H}{H\cos(t)+2d}$$ So yes it is true.

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Thank you very much! –  John Smith Apr 20 '13 at 14:13
    
Sorry, but there was a small typo in the statement, the fact is $d\tan(\arcsin ((\sin(180-t)*H)/ \sqrt(H² + 4d^2 - 4dH\cos(180-t))))$, not $d\tan(\arcsin ((\sin(180-t)*H)/ \sqrt(H² + 4d^2 - 4d\cos(180-t))))$. I already corrected the error in the question. Equality still true? –  John Smith Apr 20 '13 at 14:29
1  
Yes, in fact in my derivation the step on the 3rd line, where I factor out the square $H\cos(t)+2d$ it must be $4dH\cos(t)$, which is a bit of my mistake, but well it works out with the correct one. Sine I knew what the answer is I knew it has to factor out and overlooked it. I'll edit my solution. –  Belov Apr 20 '13 at 16:02
    
Thank you very much! –  John Smith Apr 20 '13 at 16:07

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