Tính:F=Cos(π/4+α) x cos(π/4-α)
G=Sin(π/3+α) x cos(π/3-α)
H=cos(π/2-α) x sin(π/2+α)
I=sin(π/4+α) - cos(π/4-α)
K=cos(π/6-x) - sin(π/3+x)
vì\(\dfrac{\pi}{2}< x< 0\Rightarrow sin< 0\)
Ta có: sin2x + cos2x=1\(\Rightarrow\)sin2x=1-cos2x=1-\(\left(\dfrac{3}{5}\right)^2=\dfrac{16}{25}\Rightarrow\sin x=\dfrac{-4}{5}\)
\(\sin\left(\dfrac{\pi}{2}-x\right)=\sin\dfrac{\pi}{2}\times\cos x-\cos\dfrac{\pi}{2}\times\sin x\)
\(=1\times\dfrac{3}{5}-0\times-\dfrac{4}{5}\)
=\(\dfrac{3}{5}\)
\(sin\left(\dfrac{\pi}{2}-x\right)=cosx=\dfrac{3}{5}\)