Question

100 раз прошу хоть кто то
Вычислить:
cos4a×ctg2a если tga=2

Answer 1

$\displaystyle tg\alpha =2; \\\\tg^2a+1=\frac{1}{cos^2a}; 4+1=\frac{1}{cos^2a}; cos^2a=\frac{1}{5}; sin^2a=\frac{4}{5}$

$\displaystyle cos4a*ctg2a=(1-2sin^22a)*\frac{cos2a}{sin2a}=(1-2(sin2a)^2)*\frac{cos^2a-sin^2a}{2sina*cosa}=\\\\=(1-2(2sina*cosa)^2)*\frac{\frac{1}{5}-\frac{4}{5}}{2*\frac{1}{\sqrt{5}}*\frac{2}{\sqrt{5}}}=\\\\=(1-8sin^2a*cos^2a)*(\frac{-\frac{3}{5}}{\frac{4}{5}})=\\\\=(1-8*\frac{4}{5}*\frac{1}{5})*(-\frac{3}{4})=(1-\frac{32}{25})*\frac{-3}{4}=\frac{-7}{25}*\frac{-3}{4}=0.21$