Формулы двойного и половинного аргумента. Задача 32math100admin44242025-03-23T21:03:16+03:00
Задача 32. Вычислите \(\sqrt {32} \left( {\cos \dfrac{\pi }{8}-\cos \dfrac{{3\pi }}{8}} \right)\left( {\cos \dfrac{\pi }{8} + \cos \dfrac{{3\pi }}{8}} \right)\)
Решение
\(\sqrt {32} \left( {\cos \dfrac{\pi }{8}-\cos \dfrac{{3\pi }}{8}} \right)\left( {\cos \dfrac{\pi }{8} + \cos \dfrac{{3\pi }}{8}} \right) = \)
\( = \sqrt {32} \left( {\cos \dfrac{\pi }{8}-\cos \left( {\dfrac{\pi }{2}-\dfrac{\pi }{8}} \right)} \right)\left( {\cos \dfrac{\pi }{8} + \cos \left( {\dfrac{\pi }{2}-\dfrac{\pi }{8}} \right)} \right) = \)
\( = \sqrt {32} \left( {\cos \dfrac{\pi }{8}-\sin \dfrac{\pi }{8}} \right)\left( {\cos \dfrac{\pi }{8} + \sin \dfrac{\pi }{8}} \right) = \sqrt {32} \left( {{{\cos }^2}\dfrac{\pi }{8}-{{\sin }^2}\dfrac{\pi }{8}} \right) = \)
\( = \sqrt {32} \cos \left( {2 \cdot \dfrac{\pi }{8}} \right) = \sqrt {32} \cos \dfrac{\pi }{4} = \sqrt {32} \cdot \dfrac{{\sqrt 2 }}{2} = \dfrac{8}{2} = 4.\)
Ответ: 4.