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Geometry Level 4

sin ⁡ 2 2 0 ∘ sin ⁡ 4 0 ∘ sin ⁡ 1 0 ∘ \large \frac{\sin^2 20^{\circ} \sin 40^{\circ}}{\sin 10^{\circ}}

If the above expression can be written as a b \dfrac{\sqrt{a}}{b} , where a a and b b are positive integers, and a a is square free, find a + b a+b .


The answer is 7.

Vilakshan Gupta by Vilakshan Gupta , 19, India

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1 solution

Steven Yuan
Mar 3, 2018

We use the product-to-sum identities sin ⁡ x sin ⁡ y = 1 2 ( cos ⁡ ( x − y ) − cos ⁡ ( x + y ) ) \sin x \sin y = \dfrac{1}{2}(\cos(x - y) - \cos(x + y)) and cos ⁡ x cos ⁡ y = 1 2 ( cos ⁡ ( x + y ) + cos ⁡ ( x − y ) ) \cos x \cos y = \dfrac{1}{2}(\cos(x + y) + \cos(x - y)) :

sin ⁡ 2 2 0 ∘ sin ⁡ 4 0 ∘ sin ⁡ 1 0 ∘ = sin ⁡ 2 0 ∘ ( sin ⁡ 2 0 ∘ sin ⁡ 4 0 ∘ ) sin ⁡ 1 0 ∘ = sin ⁡ 2 0 ∘ ( cos ⁡ 2 0 ∘ − cos ⁡ 6 0 ∘ ) 2 sin ⁡ 1 0 ∘ = ( 2 sin ⁡ 1 0 ∘ cos ⁡ 1 0 ∘ ) ( cos ⁡ 2 0 ∘ − 1 2 ) 2 sin ⁡ 1 0 ∘ = cos ⁡ 1 0 ∘ ( cos ⁡ 2 0 ∘ − 1 2 ) = cos ⁡ 1 0 ∘ cos ⁡ 2 0 ∘ − 1 2 cos ⁡ 1 0 ∘ = 1 2 ( cos ⁡ 3 0 ∘ + cos ⁡ 1 0 ∘ ) − 1 2 cos ⁡ 1 0 ∘ = 3 4 + 1 2 cos ⁡ 1 0 ∘ − 1 2 cos ⁡ 1 0 ∘ = 3 4 . \begin{aligned} \dfrac{\sin^2 20^{\circ} \sin 40^{\circ}}{\sin 10^{\circ}} &= \dfrac{\sin 20^{\circ} (\sin 20^{\circ} \sin 40^{\circ})}{\sin 10^{\circ}} \\ &= \dfrac{\sin 20^{\circ} (\cos 20^{\circ} - \cos 60^{\circ})}{2 \sin 10^{\circ}} \\ &= \dfrac{(2 \sin 10^{\circ} \cos 10^{\circ})(\cos 20^{\circ} - \frac{1}{2})}{2 \sin 10^{\circ}} \\ &= \cos 10^{\circ} \left (\cos 20^{\circ} - \dfrac{1}{2} \right) \\ &= \cos 10^{\circ} \cos 20^{\circ} - \dfrac{1}{2} \cos 10^{\circ} \\ &= \dfrac{1}{2} (\cos 30^{\circ} + \cos 10^{\circ}) - \dfrac{1}{2} \cos 10^{\circ} \\ &= \dfrac{\sqrt{3}}{4} + \dfrac{1}{2} \cos 10^{\circ} - \dfrac{1}{2} \cos 10^{\circ} \\ &= \dfrac{\sqrt{3}}{4}. \end{aligned}

Thus, a + b = 3 + 4 = 7 . a + b = 3 + 4 = \boxed{7}.

3 Helpful 0 Interesting 0 Brilliant 0 Confused

Nice, Actually, i have a bad habit,Whenever is see sin square or cos square , i generally use 2 cos ⁡ 2 θ = 1 + cos ⁡ 2 θ 2\cos^2 \theta = 1+\cos 2\theta or the one for the sin :D.

So That way also, answer is coming, and a bit easily i guess,

Md Zuhair - 3 years, 3 months ago

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Can you post your solution?

Vilakshan Gupta - 3 years, 3 months ago

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Ya sure. Till tomorrow I will try posting

Md Zuhair - 3 years, 3 months ago

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