Prove sin 2 ⁡ ( 10 ∘ ) − sin 2 ⁡ ( 20 ∘ ) − sin 2 ⁡ ( 40 ∘ ) = − 1 2 identity

Question

Prove       sin    2    ⁡  (      10    ∘    )  −      sin    2    ⁡  (      20    ∘    )  −      sin    2    ⁡  (      40    ∘    )  =  −      1    2   identity

vikafa4g · Accepted Answer

Let me show the first identity. By the double-angle identity:

\begin{aligned} \sin^{2} (10^{\circ}) - \sin^{2} (20^{\circ}) - \sin^{2} (40^{\circ}) \\ = & \frac{1 - \cos (20^{\circ})}{2} - \frac{1 - \cos (40^{\circ})}{2} - \frac{1 - \cos (80^{\circ})}{2} \\ = & - \frac{1}{2} - \frac{1}{2} (\cos (20^{\circ}) - \cos (40^{\circ}) - \cos (80^{\circ})) . \end{aligned}

So to show the result, it suffices to show that

\cos (20^{\circ}) - \cos (40^{\circ}) - \cos (80^{\circ}) = 0

. Indeed,

\begin{aligned} \cos (80^{\circ}) = \cos (60^{\circ} + 20^{\circ}) = \frac{1}{2} \cos (20^{\circ}) - \frac{\sqrt{3}}{2} \sin (20^{\circ}), \\ \cos (40^{\circ}) = \cos (60^{\circ} - 20^{\circ}) = \frac{1}{2} \cos (20^{\circ}) + \frac{\sqrt{3}}{2} \sin (20^{\circ}) . \end{aligned}

Adding these two equations gives that

\cos (20^{\circ}) = \cos (80^{\circ}) + \cos (40^{\circ})

, thus the result follows.

Prove sin 2 </msup> ⁡ ( 10 ∘

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