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Geometry Level 2

{ x = a sec ⁡ 3 β tan ⁡ β y = b tan ⁡ 3 β sec ⁡ β \large \begin{cases} x = a \sec^3 \beta \ \tan \beta \\ y = b \tan^3 \beta \ \sec \beta \end{cases}

Given the above, find sin ⁡ 2 β \sin^2 \beta .

Details and assumptions:

  • a a , b b , x x , y y , sin ⁡ β \sin\beta , cos ⁡ β ≠ 0 \cos\beta ≠ 0 .
x y a b \frac {xy}{ab} a y b x \frac {ay}{bx} x a − y b \frac xa - \frac yb x a + y b \frac xa + \frac yb

Hritesh Mourya by Hritesh Mourya , 21, India

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2 solutions

Hung Woei Neoh
Jun 30, 2016

x = a sec ⁡ 3 β tan ⁡ β = a ( 1 cos ⁡ 3 β ) ( sin ⁡ β cos ⁡ β ) = a sin ⁡ β cos ⁡ 4 β x = a \sec^3 \beta \tan \beta\\ =a\left(\dfrac{1}{\cos^3 \beta}\right)\left(\dfrac{\sin \beta}{\cos \beta}\right)\\ =\dfrac{a \sin \beta}{\cos^4 \beta}

⟹ cos ⁡ 4 β = a sin ⁡ β x ⟹ \implies \cos^4 \beta = \dfrac{a \sin \beta}{x}\implies Eq.(1)

y = b tan ⁡ 3 β sec ⁡ β = b ( sin ⁡ 3 β cos ⁡ 3 β ) ( 1 cos ⁡ β ) = b sin ⁡ 3 β cos ⁡ 4 β y=b\tan^3 \beta \sec \beta\\ =b\left(\dfrac{\sin^3 \beta}{\cos^3 \beta}\right)\left(\dfrac{1}{\cos \beta}\right)\\ =\dfrac{b\sin^3 \beta}{\cos^4 \beta}

⟹ cos ⁡ 4 β = b sin ⁡ 3 β y ⟹ \implies \cos^4 \beta = \dfrac{b\sin^3 \beta}{y}\implies Eq.(2)

Substitute Eq.(1) into Eq.(2):

a sin ⁡ β x = b sin ⁡ 3 β y sin ⁡ 2 β = a y b x \dfrac{a \sin \beta}{x} = \dfrac{b\sin^3 \beta}{y}\\ \sin^2 \beta =\boxed{\dfrac{ay}{bx}}

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Seems like we need a , b , x , y ≠ 0 a, b, x, y \neq 0 , in order for these to work out. Likely sin ⁡ β , cos ⁡ β ≠ 0 \sin \beta, \cos \beta \neq 0 too.

Calvin Lin Staff - 4 years, 11 months ago

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Let's see......

For the original equations to hold true, cos ⁡ β ≠ 0 \cos \beta \neq0

For the final answer to hold true, b , x ≠ 0 b,x \neq 0

If x ≠ 0 x \neq 0 , then a , sin ⁡ β ≠ 0 a, \sin \beta \neq 0 as well

b , sin ⁡ β ≠ 0 b,\sin \beta \neq 0 implies that y ≠ 0 y\neq 0 as well...

So yeah, I guess all the variables here cannot be 0 0

Hung Woei Neoh - 4 years, 11 months ago
Chew-Seong Cheong
Jun 30, 2016

{ x = a sec ⁡ 3 β tan ⁡ β y = b tan ⁡ 3 β sec ⁡ β \begin{cases} x = a \sec^3 \beta \ \tan \beta \\ y = b \tan^3 \beta \ \sec \beta \end{cases}

⟹ x y = a sec ⁡ 2 β b tan ⁡ 2 β b x a y = 1 + tan ⁡ 2 β tan ⁡ 2 β = 1 tan ⁡ 2 β + 1 b x a y − 1 = cos ⁡ 2 β sin ⁡ 2 β = 1 − sin ⁡ 2 β sin ⁡ 2 β = 1 sin ⁡ 2 β − 1 ⟹ b x a y = 1 sin ⁡ 2 β sin ⁡ 2 β = a y b x \begin{aligned} \implies \frac xy & = \frac {a\sec^2 \beta}{b\tan^2 \beta} \\ \frac {bx}{ay} & = \frac {1+\tan^2 \beta}{\tan^2 \beta} \\ & = \frac 1{\tan^2 \beta} + 1 \\ \frac {bx}{ay} - 1 & = \frac {\cos^2 \beta}{\sin^2 \beta} \\ & = \frac {1 - \sin^2 \beta}{\sin^2 \beta} \\ & = \frac 1{\sin^2 \beta} - 1 \\ \implies \frac {bx}{ay} & = \frac 1{\sin^2 \beta} \\ \sin^2 \beta & = \boxed{\dfrac {ay}{bx}} \end{aligned}

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