A simple problem

Geometry Level 3

Let a = sin ⁡ 1 0 ∘ a= \sin 10^\circ , b = sin ⁡ 5 0 ∘ b= \sin 50^\circ , c = sin ⁡ 7 0 ∘ c= \sin 70^\circ , then relation between a, b and c is:

2 a + 4 b = c 2a + 4b = c 2 a + b = c 2a + b = c None of these. a − b = c a - b = c 2 a − 4 b = c 2a - 4b = c a − 2 b = c a - 2b = c a + b = c a + b = c

Sachin Vishwakarma by Sachin Vishwakarma , 22, India

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2 solutions

We have that b = sin ⁡ ( 5 0 ∘ ) = sin ⁡ ( 6 0 ∘ − 1 0 ∘ ) = sin ⁡ ( 6 0 ∘ ) cos ⁡ ( 1 0 ∘ ) − cos ⁡ ( 6 0 ∘ ) sin ⁡ ( 1 0 ∘ ) b = \sin(50^{\circ}) = \sin(60^{\circ} - 10^{\circ}) = \sin(60^{\circ})\cos(10^{\circ}) - \cos(60^{\circ})\sin(10^{\circ})

and that c = sin ⁡ ( 7 0 ∘ ) = sin ⁡ ( 6 0 ∘ + 1 0 ∘ ) = sin ⁡ ( 6 0 ∘ ) cos ⁡ ( 1 0 ∘ ) + cos ⁡ ( 6 0 ∘ ) sin ⁡ ( 1 0 ∘ ) . c = \sin(70^{\circ}) = \sin(60^{\circ} + 10^{\circ}) = \sin(60^{\circ})\cos(10^{\circ}) + \cos(60^{\circ})\sin(10^{\circ}).

Thus c − b = 2 cos ⁡ ( 6 0 ∘ ) sin ⁡ ( 1 0 ∘ ) = 2 ∗ 1 2 ∗ a = a ⟹ a + b = c . c - b = 2\cos(60^{\circ})\sin(10^{\circ}) = 2*\dfrac{1}{2}*a = a \Longrightarrow \boxed{a + b = c}.

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Using the identity below, we have:

cos ⁡ 4 0 ∘ + cos ⁡ 8 0 ∘ + cos ⁡ 12 0 ∘ + cos ⁡ 16 0 ∘ = − 1 2 cos ⁡ θ ∘ = sin ⁡ ( 90 − θ ) ∘ cos ⁡ θ ∘ = − cos ⁡ ( 180 − θ ) ∘ sin ⁡ 5 0 ∘ + sin ⁡ 1 0 ∘ − 1 2 − cos ⁡ 2 0 ∘ = − 1 2 sin ⁡ 5 0 ∘ + sin ⁡ 1 0 ∘ − sin ⁡ 7 0 ∘ = 0 sin ⁡ 1 0 ∘ + sin ⁡ 5 0 ∘ = sin ⁡ 7 0 ∘ \begin{aligned} \color{#3D99F6}{\cos 40^\circ} + \color{#3D99F6}{\cos 80^\circ} + \cos 120^\circ + \color{#D61F06}{\cos 160^\circ} & = -\frac{1}{2} \quad \quad \small \color{#3D99F6}{\cos \theta^\circ = \sin (90-\theta)^\circ} \quad \color{#D61F06}{\cos \theta^\circ = - \cos (180-\theta)^\circ} \\ \color{#3D99F6}{\sin 50^\circ} + \color{#3D99F6}{\sin 10^\circ} - \frac{1}{2} - \color{#D61F06}{\cos 20^\circ} & = -\frac{1}{2} \\ \color{#3D99F6}{\sin 50^\circ} + \color{#3D99F6}{\sin 10^\circ} - \color{#D61F06}{\sin 70^\circ} & = 0 \\ \color{#3D99F6}{\sin 10^\circ} + \color{#3D99F6}{\sin 50^\circ} & = \color{#D61F06}{\sin 70^\circ} \end{aligned}

a + b = c \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \space \boxed{\color{#3D99F6}{a} + \color{#3D99F6}{b} = \color{#D61F06}{c}}

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