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Задача Задача №255353
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Аноним
29 января 2024 15:06 · публичный вопрос
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Аноним

Вычислите cos 2alpha 2) sin alpha = - 3/5

Нейросеть29 января 2024 15:06

cos 2alpha = cos^2 alpha - sin^2 alpha cos 2alpha = (cos^2 alpha) - (1 - cos^2 alpha) cos 2alpha = 2cos^2 alpha - 1

Given sin alpha = - 3/5, we can use the Pythagorean identity to find cos alpha: cos^2 alpha = 1 - sin^2 alpha cos^2 alpha = 1 - (-3/5)^2 cos^2 alpha = 1 - 9/25 cos^2 alpha = 16/25

Substituting this into the formula for cos 2alpha: cos 2alpha = 2(16/25) - 1 cos 2alpha = 32/25 - 1 cos 2alpha = 32/25 - 25/25 cos 2alpha = 7/25

Ответ: cos 2alpha = 7/25

Вычислите cos 2alpha 2) sin alpha = - 3/5
cos 2alpha = cos^2 alpha - sin^2 alpha cos 2alpha = (cos^2 alpha) - (1 - cos^2 alpha) cos 2alpha = 2cos^2 alpha - 1 Given sin alpha = - 3/5, we can use the Pythagorean identity to find cos alpha: cos^2 alpha = 1 - sin^2 alpha cos^2 alpha = 1 - (-3/5)^2 cos^2 alpha = 1 - 9/25 cos^2 alpha = 16/25 Substituting this into the formula for cos 2alpha: cos 2alpha = 2(16/25) - 1 cos 2alpha = 32/25 - 1 cos 2alpha = 32/25 - 25/25 cos 2alpha = 7/25 Ответ: cos 2alpha = 7/25