Can you Divide?

Geometry Level 3

Simplify :

2 cos ⁡ ( 1 0 ∘ ) − sin ⁡ ( 4 0 ∘ ) cos ⁡ ( 4 0 ∘ ) \large \dfrac{2\cos(10^{\circ})-\sin(40^{\circ})}{\cos(40^{\circ})}

cos ⁡ ( 10 ) \cos(10) π 2 \dfrac{\pi}{2} None of the above 3 \sqrt{3} π 3 \dfrac{\pi}{3}

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Anish Harsha by Anish Harsha , 20, India

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2 solutions

We have that

cos ⁡ ( 1 0 ∘ ) = sin ⁡ ( 10 0 ∘ ) = sin ⁡ ( 6 0 ∘ ) cos ⁡ ( 4 0 ∘ ) + cos ⁡ ( 6 0 ∘ ) sin ⁡ ( 4 0 ∘ ) = 3 cos ⁡ ( 4 0 ∘ ) 2 + sin ⁡ ( 4 0 ∘ ) 2 \cos(10^{\circ}) = \sin(100^{\circ}) = \sin(60^{\circ})\cos(40^{\circ}) + \cos(60^{\circ})\sin(40^{\circ}) = \dfrac{\sqrt{3}\cos(40^{\circ})}{2} + \dfrac{\sin(40^{\circ})}{2}

⟹ 2 cos ⁡ ( 1 0 ∘ ) − sin ⁡ ( 4 0 ∘ ) = 3 cos ⁡ ( 4 0 ∘ ) ⟹ 2 cos ⁡ ( 1 0 ∘ ) − sin ⁡ ( 4 0 ∘ ) cos ⁡ ( 4 0 ∘ ) = 3 . \Longrightarrow 2\cos(10^{\circ}) - \sin(40^{\circ}) = \sqrt{3}\cos(40^{\circ}) \Longrightarrow \dfrac{2\cos(10^{\circ}) - \sin(40^{\circ})}{\cos(40^{\circ})} = \boxed{\sqrt{3}}.

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Pulkit Gupta
Dec 5, 2015

We write sin ⁡ 40 \sin 40 as 2 sin ⁡ 30 2\sin 30 sin ⁡ 40 \sin 40 and then apply 2 sin ⁡ A 2\sin A sin ⁡ B \sin B = cos ⁡ ( A − B ) − cos ⁡ ( A + B ) \cos (A-B) - \cos (A+B) .

This simplifies the expression to an easily tack able form.

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