№9 Вычислиme sin 2d+CDS21 25ти 22-00122 зная что 80032+1-38inad=0

Вычислить \(\frac{sin2\alpha + cos2\alpha}{2sin^2 \alpha - cos2\alpha}\), зная что cos3\(\alpha\) + 1 - 3sin2\(\alpha\) = 0

cos3\(\alpha\) = 4cos³\(\alpha\) - 3cos\(\alpha\) sin3\(\alpha\) = 3sin\(\alpha\) - 4sin³\(\alpha\) cos3\(\alpha\) + 1 = 3sin2\(\alpha\) 4cos³\(\alpha\) - 3cos\(\alpha\) + 1 = 6sin\(\alpha\)cos\(\alpha\)

sin2\(\alpha\) + cos2\(\alpha\) = 2sin\(\alpha\)cos\(\alpha\) + cos²\(\alpha\) - sin²\(\alpha\) 2sin²\(\alpha\) - cos2\(\alpha\) = 2sin²\(\alpha\) - cos²\(\alpha\) + sin²\(\alpha\) = 3sin²\(\alpha\) - cos²\(\alpha\)

Пусть tg\(\alpha\) = x, sin\(\alpha\) = \(\frac{x}{\sqrt{1+x^2}}\) и cos\(\alpha\) = \(\frac{1}{\sqrt{1+x^2}}\) , тогда cos3\(\alpha\) = \(\frac{1-3x^2}{(1+x^2)^{3/2}}\) и sin3\(\alpha\) = \(\frac{3x-x^3}{(1+x^2)^{3/2}}\)

cos3\(\alpha\) + 1 - 3sin2\(\alpha\) = \(\frac{1-3x^2}{(1+x^2)^{3/2}}\) + 1 - \(\frac{6x}{1+x^2}\) = 0

Ответ: -1