Question

[Maga]

tính giá trị biểu thức P=(1+3 sin2 x) (1+4 cos2 x) biết cosx= \(-\frac{2}{3}\)                 

P = \(\frac{35}{6}\)

Answer 1

[Nguyễn Quang Trung]

Ta có: \(sin^2x+cos^2x=1\Rightarrow sinx=\sqrt{1-\left(-\frac{2}{3}\right)^2}=\frac{\sqrt{5}}{3}\)

  \(P=\left(1+3sin2x\right)\left(1+4cos2x\right)=\left(1+3.2sinx.cosx\right)\left[1+4.\left(cos^2x-sin^2x\right)\right]\)

     \(=\left(1+6.\frac{\sqrt{5}}{3}.\left(-\frac{2}{3}\right)\right).\left[1+4\left(\left(-\frac{2}{3}\right)^2-\left(\frac{\sqrt{5}}{3}\right)^2\right)\right]\) \(=\frac{15-20\sqrt{5}}{27}\)