Question

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Answer 1

Объяснение:

1.

$sin^2\frac{\pi}{8} + cos^2\frac{3\pi}{8} + sin^2\frac{5\pi}{8} + cos^2\frac{7\pi}{8} = sin^2\frac{\pi}{8} + cos^2\frac{3\pi}{8} + sin^2(\pi - \frac{3\pi}{8}) + cos^2(\pi - \frac{\pi}{8}) = sin^2\frac{\pi}{8} + cos^2\frac{3\pi}{8} + sin^2\frac{3\pi}{8} + cos^2\frac{\pi}{8} = 1 + 1 = 2.$

2.

$cos(2x + \frac{7\pi}{4}) = cos(2x)cos(\frac{7\pi}{4}) - sin(2x)sin(\frac{7\pi}{4}) = cos(2x)(cos(2\pi - \frac{\pi}{4}) - sin(2x)sin(2\pi - \frac{\pi}{4}) = cos(2x)cos\frac{\pi}{4} - sin(2x)sin(-\frac{\pi}{4}) = cos(2x)cos\frac{\pi}{4} + sin(2x)sin\frac{\pi}{4} = \frac{1}{\sqrt{2}}(sin(2x)+cos(2x))\\ctg(x) = \frac{2}{3}\\ctg^2(x) = \frac{1}{sin^2(x)} - 1 \\sin^2(x) = \frac{1}{ctg^2(x) + 1} = \frac{1}{\frac{4}{9} + 1} = \frac{9}{13}\\sin(x) = \frac{3}{\sqrt{13}}\\$

$cos(x) = \sqrt{1-sin^2(x)} = \sqrt{1 - \frac{9}{13}} = \sqrt{\frac{4}{13}} = \frac{2}{\sqrt{13}}\\sin(2x) = 2sin(x)cos(x) = 2 * \frac{3}{\sqrt{13}} * \frac{2}{\sqrt{13}} = \frac{12}{13}\\cos(2x) = 2cos^2(x) - 1 = 2 * \frac{4}{13} - 1 = -\frac{5}{13}\\cos(2x + \frac{7\pi}{4}) = \frac{1}{\sqrt{2}}(sin(2x) + cos(2x)) = \frac{1}{\sqrt{2}}(\frac{12}{13} -\frac{5}{13}) = \frac{7}{13\sqrt{2}}\\Answer:\frac{7}{13\sqrt{2}}$