• Предмет: Алгебра
  • Автор: MrGlyouk
  • Вопрос задан 1 год назад

(sin(7pi/24)cos(pi/24)-cos(7pi/24)sin(pi/24))/(cos(pi/7)cos(4pi/21)-sin(pi/7)sin(4pi/21))

Ответы

Ответ дал: 12FunGirl12
0

\frac{\sin \left(\frac{7\pi }{24}\right)\cos \left(\frac{\pi }{24}\right)-\cos \left(\frac{7\pi }{24}\right)\sin \left(\frac{\pi }{24}\right)}{\cos \left(\frac{\pi }{7}\right)\cos \left(\frac{4\pi }{21}\right)-\sin \left(\frac{\pi }{7}\right)\sin \left(\frac{4\pi }{21}\right)}

=\frac{\sin \left(\frac{7\pi }{24}\right)\cos \left(\frac{\pi }{24}\right)-\cos \left(\frac{7\pi }{24}\right)\sin \left(\frac{\pi }{24}\right)}{\cos \left(\frac{\pi }{7}+\frac{4\pi }{21}\right)}

=\frac{\cos \left(\frac{\pi }{24}\right)\sin \left(\frac{7\pi }{24}\right)-\sin \left(\frac{\pi }{24}\right)\cos \left(\frac{7\pi }{24}\right)}{\cos \left(\frac{\pi }{3}\right)}

=\frac{\frac{\sqrt{2+\sqrt{2+\sqrt{3}}}}{2}\cdot \frac{\sqrt{2}\sqrt{4-\sqrt{2}+\sqrt{6}}}{4}-\frac{\sqrt{2-\sqrt{2+\sqrt{3}}}}{2}\cdot \frac{\sqrt{2}\sqrt{4+\sqrt{2}-\sqrt{6}}}{4}}{\frac{1}{2}}

=\frac{\sqrt{2}\left(\sqrt{\sqrt{2+\sqrt{3}}+2}\sqrt{4+\sqrt{6}-\sqrt{2}}-\sqrt{-\sqrt{2+\sqrt{3}}+2}\sqrt{4+\sqrt{2}-\sqrt{6}}\right)}{4}

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