જો cos^3 80^{circ} + cos^3 40^{circ} - cos^3 | vedclass

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Question

(B) આપણે નિત્યસમ $\cos 3	heta = 4\cos^3 	heta - 3\cos 	heta$ નો ઉપયોગ કરીએ છીએ,જેનો અર્થ છે $\cos^3 	heta = \frac{1}{4}(\cos 3	heta + 3\cos 	het

Vedclass · Accepted Answer

$\cos \left(\frac{2\pi}{3}\right)$

Vedclass · Answer

(B) આપણે નિત્યસમ $\cos 3	heta = 4\cos^3 	heta - 3\cos 	heta$ નો ઉપયોગ કરીએ છીએ,જેનો અર્થ છે $\cos^3 	heta = \frac{1}{4}(\cos 3	heta + 3\cos 	heta)$.આ કિંમત $k = \cos^3 80^{\circ} + \cos^3 40^{\circ} - \cos^3 20^{\circ}$ માં મૂકતા:$k = \frac{1}{4}(\cos 240^{\circ} + 3\cos 80^{\circ}) + \frac{1}{4}(\cos 120^{\circ} + 3\cos 40^{\circ}) - \frac{1}{4}(\cos 60^{\circ} + 3\cos 20^{\circ})$$k = \frac{1}{4} [(\cos 240^{\circ} + \cos 120^{\circ} - \cos 60^{\circ}) + 3(\cos 80^{\circ} + \cos 40^{\circ} - \cos 20^{\circ})]$$\cos 240^{\circ} = -\frac{1}{2}$,$\cos 120^{\circ} = -\frac{1}{2}$,અને $\cos 60^{\circ} = \frac{1}{2}$ હોવાથી,પ્રથમ ભાગ $(-\frac{1}{2} - \frac{1}{2} - \frac{1}{2}) = -\frac{3}{2}$ થાય.બીજા ભાગ માટે,$\cos 80^{\circ} + \cos 40^{\circ} = 2\cos 60^{\circ} \cos 20^{\circ} = 2(\frac{1}{2})\cos 20^{\circ} = \cos 20^{\circ}$.આમ,$3(\cos 80^{\circ} + \cos 40^{\circ} - \cos 20^{\circ}) = 3(\cos 20^{\circ} - \cos 20^{\circ}) = 0$.તેથી,$k = \frac{1}{4}(-\frac{3}{2} + 0) = -\frac{3}{8}$.હવે $\frac{4k}{3} = \frac{4}{3} 	imes (-\frac{3}{8}) = -\frac{1}{2}$.વિકલ્પો સાથે સરખાવતા: $\cos(\frac{2\pi}{3}) = \cos(120^{\circ}) = -\frac{1}{2}$.તેથી,સાચો વિકલ્પ $B$ છે.

જો $\cos^3 80^{\circ} + \cos^3 40^{\circ} - \cos^3 20^{\circ} = k$ હોય,તો $\frac{4k}{3} =$

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