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cum se calculeaza sin^8 pi/12 - cos^8 pi/12

Răspuns :

sin^8 pi/12 - cos^8 pi/12 =
= (sin^4 pi/12 + cos^4 pi/12)(sin^4 pi/12 - cos^4 pi/12) =
= (sin^4 pi/12 + cos^4 pi/12)(sin^2 pi/12 + cos^2 pi/12)( sin^2 pi/12 - cos^2 pi/12) =
= (sin^2^2 pi/12 + cos^2^2 pi/12)×1×(-cos(2pi/12)) =
= ((sin^2 pi/12+cos^2 pi/12)^2-2sin^2 pi/12 cos^2 pi/12))×(-√3/2) =
= (1^2 - 1/2× sin^2 pi/6)×(-√3/2) =
= (1-1/8)×(-√3/2) =
= (-7√3)/16
[tex]\sin ^8\left(\frac{\pi }{12}\right)-\cos ^8\left(\frac{\pi }{12}\right) \\ \sin \left(\frac{\pi }{12}\right)=\sin \left(\frac{\frac{\pi }{6}}{2}\right)=\sqrt{\frac{1-\cos \left(\frac{\pi }{6}\right)}{2}}=\frac{\sqrt{2-\sqrt{3}}}{2} \\ \\ \cos \left(\frac{\pi }{12}\right)=\cos \left(\frac{\frac{\pi }{6}}{2}\right) =\sqrt{\frac{1+\cos \left(\frac{\pi }{6}\right)}{2}}=\frac{\sqrt{2+\sqrt{3}}}{2} \\ \\ =\left(\frac{\sqrt{2-\sqrt{3}}}{2}\right)^8-\left(\frac{\sqrt{2+\sqrt{3}}}{2}\right)^8[/tex][tex]=\frac{\left(2-\sqrt{3}\right)^4}{2^8}-\frac{\left(2+\sqrt{3}\right)^4}{2^8}=\frac{\left(2-\sqrt{3}\right)^4-\left(2+\sqrt{3}\right)^4}{256}=\frac{-112\sqrt{3}}{256}=-\frac{7\sqrt{3}}{16}[/tex]