Question

26.7

Вычислите sin2a, ctg2a, если tga = 2,4 и π<а<3π/2
Помогитее срочно, пожалуйста
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Answer 1

Ответ:

$tga=2,4\\\\\pi <a<\dfrac{3\pi}{2}\ \ \ \to \ \ \ 2\pi <2a<3\pi \\\\sin2a=\dfrac{2tga}{1+tg^2a}=\dfrac{2\cdot 2,4}{1+2,4^2}=\dfrac{4,8}{1+5,76}=\dfrac{4,8}{6,76}=\dfrac{480}{676}=\dfrac{120}{169}\\\\\\cos2a=\dfrac{1-tg^2a}{1+tg^2a}=\dfrac{1-5,76}{1+5,76}=\dfrac{-4,76}{6,76}=-\dfrac{476}{676}=-\dfrac{119}{169}\\\\\\ctg2a=\dfrac{cos2a}{sin2a}=-\dfrac{119}{120}$

Answer 2

Объяснение:

$tg\alpha =2,4\ \ \ \ \pi <\alpha <\frac{3\pi }{2} \ \ \ \ sin2\alpha =?\ \ \ \ ctg2\alpha =?\\tg\alpha =\frac{sin\alpha }{cos\alpha } =2,4\\(\frac{sin\alpha }{cos\alpha })^2 =(2\frac{2}{5}) ^2\\(\frac{sin\alpha }{cos\alpha })^2 =(\frac{12}{5})^2\\ (\frac{sin\alpha }{cos\alpha })^2 =\frac{12^2}{5^2}\\\frac{sin^2\alpha }{cos^2\alpha } =\frac{144}{25} \\25*sin^2\alpha =144*cos^2\alpha \\ 25*sin^2\alpha+25*cos^2\alpha =144*cos^2\alpha +25*cos^2\alpha\\25*(sin^2\alpha +cos^2\alpha )=169*cos^2\alpha$

$169*cos^2\alpha =25\ |:169\\cos^2\alpha =\frac{25}{169} \\cos\alpha =б\sqrt{\frac{25}{169} } =б\frac{5}{13} \\\pi <\alpha <\frac{3\pi }{2}\ \ \ \ \Rightarrow\\ cos\alpha =-\frac{5}{13} \\sin^2\alpha =1-cos^2\alpha =1-(-\frac{5}{13} )^2=1-\frac{25}{169}=\frac{169-25}{169}=\frac{144}{169} .\\sin\alpha =б\sqrt{\frac{144}{169} }=б\frac{12}{13} \\ \pi <\alpha <\frac{3\pi }{2}\ \ \ \ \Rightarrow\\sin\alpha =-\frac{12}{13}.$

$sin2\alpha =2*sin\alpha *cos\alpha =2*(-\frac{12}{13})*(-\frac{5}{13} ) =\frac{120}{169} .\\cos2\alpha =cos^2\frac{x}{y} -sin^2\alpha =(-\frac{5}{13})^2+(-\frac{12}{13} )^2=\frac{25}{169}-\frac{144}{169}=-\frac{119}{169}\ \ \ \ \Rightarrow\\ ctg2\alpha =\frac{cos2\alpha }{sin2\alpha }=\frac{-\frac{119}{169} }{\frac{120}{169} } =-\frac{119}{120}.$

Ответ: sin2α=120/169,   ctg2α=-119/120.