Assume $A = \cos 20^{\circ} + \cos 100^{\circ} + \cos 140^{\circ}$ . Prove that value of $A$ is zero.
My try : $A = 2\cos 60^{\circ} \cos 40^{\circ} + \cos 140^{\circ}$ and I'm stuck here
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See also: http://math.stackexchange.com/questions/2135044/prove-that-cos2-20-cos2-40-cos2-80-sin2-20-sin2-40-si/2135388#2135388 – lab bhattacharjee Mar 02 '17 at 16:56
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$\cos 20^{\circ} + \cos 100^{\circ} = \cos (60^{\circ}-40^{\circ}) + \cos (60^{\circ}+40^{\circ}) = \cos 40^{\circ}$ by using the addition formulae for $\cos$.
Then $\cos 140^{\circ} = \cos (180^{\circ} - 40^{\circ}) = -\cos (40^{\circ})$. So $$\cos 20^{\circ} + \cos 100^{\circ} + \cos 140^{\circ} = \cos 40^{\circ} - \cos 40^{\circ} = 0$$
Zain Patel
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Very good answer . Is there any general method for computing sum of cosines or sines ? – S.H.W Mar 02 '17 at 15:18
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$\cos 20^{\circ}+\cos 100^{\circ}+\cos 140^{\circ}=\\ =\cos 20^{\circ}+\cos (120^{\circ}-20^{\circ})+\cos (120^{\circ}+20^{\circ})=\\ =\cos 20^{\circ}+\cos 120^{\circ}\cos 20^{\circ}+\sin 120^{\circ}\sin 20^{\circ} +\cos 120^{\circ}\cos 20^{\circ}-\sin 120^{\circ}\sin 20^{\circ} =\\ =\cos 20^{\circ}+2\cos 120^{\circ}\cos 20^{\circ} =\\ =\cos 20^{\circ}-2\cos 60^{\circ}\cos 20^{\circ} =\\ =\cos 20^{\circ}-2\frac{1}{2}\cos 20^{\circ} =\\ =\cos 20^{\circ}-\cos 20^{\circ} = 0$
Jaroslaw Matlak
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