Question

cho cos x=3/5; -pi/2<x<0. Tính giá trị của sin(pi/2-x)

Answer 1

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)

Answer 2

vì\(\dfrac{\pi}{2}< x< 0\Rightarrow sin< 0\)

Ta có: sin²x + cos²x=1\(\Rightarrow\)sin²x=1-cos²x=1-\(\left(\dfrac{3}{5}\right)^2=\dfrac{16}{25}\Rightarrow\sin x=\dfrac{-4}{5}\)

\(\sin\left(\dfrac{\pi}{2}-x\right)=\sin\dfrac{\pi}{2}\times\cos x-\cos\dfrac{\pi}{2}\times\sin x\)

\(=1\times\dfrac{3}{5}-0\times-\dfrac{4}{5}\)

=\(\dfrac{3}{5}\)

Answer 3

\(sin\left(\dfrac{\pi}{2}-x\right)=cosx=\dfrac{3}{5}\)