Question

Докажите тождества
Даю 60 баллов

Answer 1

Ответ:

a)

$\sin(2x) = 2 \sin(x) \cos(x) = \\ = 2 \sin(x) \cos(x) + 1 - 1 = \\ = 2 \sin(x) \cos(x) + \cos {}^{2} (x) + \sin {}^{2} ( {}^{} ) - 1 = \\ = ( \sin(x) + \cos(x)) {}^{2} - 1$

б)

$\cos {}^{4} (x) - \sin {}^{4} (x) = ( \cos {}^{2} (x) - \sin {}^{2} (x) )(\cos {}^{2} (x) + \sin {}^{2} (x)) = \\ = \cos(2x) \times 1 = \cos(2x)$

в)

$\frac{2 \sin(2x) - \sin(4x) }{2 \sin(2x) + \sin(4x) } = \frac{2 \sin(2x) - 2 \sin(2x) \cos(2x) }{ 2\sin(2x) + 2\sin(2x) \cos(2x) } = \\ = \frac{2 \sin(2x)(1 - \cos(2x) ) }{2 \sin(2x)(1 + \cos(2 {x}^{} ) ) } = \frac{1 - \cos {}^{2} (x) + \sin {}^{2} (x) }{ 1 + \cos {}^{2} (x) - \sin {}^{2} (x) } = \\ = \frac{2 \sin {}^{2} (x) }{2 \cos {}^{2} (x) } = {tg}^{2} x$

г)

$\cos(x - \frac{2\pi}{3} ) = \cos(x - (\pi - \frac{\pi}{3} ) ) = \\ = \cos( - \pi + x + \frac{\pi}{3} ) = - \cos(x + \frac{\pi}{3} )$

д)

$3 \cos(2x) - \sin {}^{2} (x) - \cos {}^{2} (x) = \\ = 3 \cos {}^{2} (x) - 3\sin {}^{2} (x) - \cos {}^{2} (x) - \sin {}^{2} (x) = \\ = 2 \cos {}^{2} (x) - 2 \sin {}^{2} (x) = 2 \cos(2x)$

е)

$\frac{ \sin(5x) - \sin(3x) }{ 2\cos(4x) } = \frac{ 2\sin( \frac{5x - 3x}{2} ) \cos( \frac{5x + 3x}{2} ) }{2 \cos(4x) } = \\ = \frac{2 \sin(x) \cos(4x) }{2 \cos(4x) } = \sin(x)$