Question

1. Cho sinx = \(\dfrac{2}{3}\)    , x ∈ (0,\(\dfrac{\Pi}{2}\))

Tính cosx, tanx , sin (x+\(\dfrac{\Pi}{4}\))

2. Cho cos = \(\dfrac{1}{4}\) . Tính sinx, cos2x

3. Cho tanx = 2 . Tính cosx, sinx

x ∈ (0,\(\dfrac{\Pi}{2}\))

4. Rút gọn a) A = cos2x - 2cos²x + sinx +1

                  b) B = \(\dfrac{cos3x+cos2x+cosx}{cos2x}\)

Answer 1

1.

\(0< x< \dfrac{\pi}{2}\Rightarrow cosx>0\)

\(\Rightarrow cosx=\sqrt{1-sin^2x}=\dfrac{\sqrt{5}}{3}\)

\(tanx=\dfrac{sinx}{cosx}=\dfrac{2}{\sqrt{5}}\)

\(sin\left(x+\dfrac{\pi}{4}\right)=\dfrac{\sqrt{2}}{2}\left(sinx+cosx\right)=\dfrac{\sqrt{10}+2\sqrt{2}}{6}\)

2.

Đề bài thiếu, cos?x

Và x thuộc khoảng nào?

3.

\(x\in\left(0;\dfrac{\pi}{2}\right)\Rightarrow sinx;cosx>0\)

\(\dfrac{1}{cos^2x}=1+tan^2x=5\Rightarrow cos^2x=\dfrac{1}{5}\Rightarrow cosx=\dfrac{\sqrt{5}}{5}\)

\(sinx=cosx.tanx=\dfrac{2\sqrt{5}}{5}\)

4.

\(A=\left(2cos^2x-1\right)-2cos^2x+sinx+1=sinx\)

\(B=\dfrac{cos3x+cosx+cos2x}{cos2x}=\dfrac{2cos2x.cosx+cos2x}{cos2x}=\dfrac{cos2x\left(2cosx+1\right)}{cos2x}=2cosx+1\)