1B

2A

3A

4C

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.

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

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

xét vế phải

( cosa+1-sina)^2

= cos^2 +1+ sin^2+2cosa-2sina-2sinacosa

= 2( 1+ cosa-sina-sinacosa)

= 2( 1-sina) ( 1+cosa)

hộ vs ae ơi

đáp án không giống lắm 

2: \(\left(sinx+cosx\right)^2=1+2\cdot sinx\cdot cosx=1+2\cdot\dfrac{\sqrt{3}}{4}=1+\dfrac{\sqrt{3}}{2}=\dfrac{2+\sqrt{3}}{2}\)

=>\(sinx+cosx=\dfrac{\sqrt{3}+1}{2}\)

mà sin x*cosx=căn 3/4

nên sinx,cosx là các nghiệm của phương trình là:

\(a^2-\dfrac{\sqrt{3}+1}{2}\cdot a+\dfrac{\sqrt{3}}{4}=0\)

=>\(\left[{}\begin{matrix}a=\dfrac{\sqrt{3}}{2}\\a=\dfrac{1}{2}\end{matrix}\right.\)

Ta sẽ có hai trường hợp:

TH1: sin x=căn 3/2; cosx=1/2

tan x=sinx/cosx=căn 3

cot x=1/căn 3

TH2: sin x=1/2; cosx=căn 3/2

tan x=sin x/cosx=1/căn 3

cot x=1:1/căn 3=căn 3

    1 + sinx + cosx + sin2x + cos2x = 0

<=> sin^2x+ cos^2 x + ( sinx+cosx) + 2.sinx.cosx + ( cos^2 x - sin^2 x)=0

<=> 2 cos^2 x + 2sinx.cosx + sinx + cosx =0

<=> 2cosx ( cos x + sinx) + sinx + cosx = 0

<=> ( cosx + sinx ) (2 cos x + 1 ) = 0

<=> cosx + sinx = 0 hoặc 2cosx + 1 =0

\(cosx=cos2.\left(\dfrac{x}{2}\right)=cos^2\dfrac{x}{2}-sin^2\dfrac{x}{2}\)

\(sinx=sin2\left(\dfrac{x}{2}\right)=2sin\dfrac{x}{2}cos\dfrac{x}{2}\)

\(\Rightarrow\dfrac{sinx+cosx}{sinx}=\dfrac{sinx+cos^2\dfrac{x}{2}-sin^2\dfrac{x}{2}}{2sin\dfrac{x}{2}cos\dfrac{x}{2}}\)

a/

\(\left(\frac{sin2x}{cos2x}-\frac{sinx}{cosx}\right)cos2x=\left(\frac{sin2x.cosx-cos2x.sinx}{cos2x.cosx}\right).cos2x\)

\(=\frac{sin\left(2x-x\right)}{cosx}=\frac{sinx}{cosx}=tanx\)

b/

\(2\left(1-sinx\right)\left(1+cosx\right)=2+2cosx-2sinx-2sinxcosx\)

\(=1+sin^2x+cos^2x-2sinx+2cosx-2sinx.cosx\)

\(=\left(1-sinx+cosx\right)^2\)

c/

\(1+cotx+cot^2x+cot^3x=1+cotx+cot^2x\left(1+cotx\right)\)

\(=\left(1+cotx\right)\left(1+cot^2x\right)=\left(1+\frac{cosx}{sinx}\right)\left(1+\frac{cos^2x}{sin^2x}\right)=\frac{sinx+cosx}{sin^3x}\)

d/

\(\frac{cos3x}{sinx}+\frac{sin3x}{cosx}=\frac{cos3x.cosx+sin3x.sinx}{sinx.cosx}=\frac{cos\left(3x-x\right)}{\frac{1}{2}2sinx.cosx}=\frac{2cos2x}{sin2x}=2cot2x\)