A geometry problem by Bala vidyadharan

Geometry Level 3

{ a = cos ⁡ α + cos ⁡ β b = sin ⁡ α + sin ⁡ β 2 θ = α − β \begin{cases} a= \cos\alpha +\cos\beta \\ b = \sin\alpha +\sin\beta \\ 2\theta = \alpha -\beta \\ \end{cases}

If α , β , a , b \alpha , \beta, a,b and θ \theta satisfy the system of equations above, state the value of cos ⁡ ( 3 θ ) cos ⁡ ( θ ) \dfrac{\cos(3\theta)}{\cos(\theta)} in terms of a a and b b .

a 2 + b 2 4 a^2 + \frac{b^2}4 a 2 + b 2 − 2 a^2+b^2-2 a 2 + b 2 − 3 a^2+b^2-3 3 − a 2 − b 2 3-a^2-b^2

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Bala Vidyadharan by Bala vidyadharan , 20, India

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1 solution

Kenny Lau
Aug 1, 2015

Seriously. I can't see how this can be solved by a geometrical approach at all.

  • Note that cos ⁡ 3 θ cos ⁡ θ = 4 cos ⁡ 2 θ − 3 \dfrac{\cos3\theta}{\cos\theta}=4\cos^2\theta-3 .
  • Also:
  • a 2 + b 2 a^2+b^2
  • = cos ⁡ 2 α + 2 cos ⁡ α cos ⁡ β + cos ⁡ 2 β + sin ⁡ 2 α + 2 sin ⁡ α sin ⁡ β + sin ⁡ 2 β =\cos^2\alpha + 2\cos\alpha\cos\beta + \cos^2\beta + \sin^2\alpha + 2\sin\alpha\sin\beta + \sin^2\beta
  • = 2 + 2 cos ⁡ α cos ⁡ β + 2 sin ⁡ α sin ⁡ β =2 + 2\cos\alpha\cos\beta + 2\sin\alpha\sin\beta
  • = 2 + 2 cos ⁡ ( α − β ) =2 + 2\cos(\alpha-\beta)
  • = 2 + 2 cos ⁡ 2 θ =2 + 2\cos2\theta
  • Then:
  • cos ⁡ 3 θ cos ⁡ θ \dfrac{\cos3\theta}{\cos\theta}
  • = 4 cos ⁡ 2 θ − 3 =4\cos^2\theta-3
  • = 2 + 2 cos ⁡ 2 θ − 3 =2 + 2\cos2\theta - 3
  • = a 2 + b 2 − 3 =a^2 + b^2 - 3
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