Lời giải:

$A=\frac{2\cos \frac{2x+y}{2}\sin \frac{x}{2}}{2\sin \frac{2x+y}{2}.\cos \frac{x}{2}}-\frac{2\cos \frac{2x+y}{2}\cos \frac{x}{2}}{-2\sin \frac{2x+y}{2}\sin \frac{x}{2}}$

$=\tan \frac{x}{2}.\cot \frac{2x+y}{2}+\cot \frac{x}{2}.\cot \frac{2x+y}{2}=\cot \frac{2x+y}{2}(\tan \frac{x}{2}+\cot \frac{x}{2})$

\(cos^2t=cos^2\left(2\pi-x-y-z\right)=cos^2\left(x+y+z\right)\)

\(VT=cos^2x+cos^2y-cos^2z-cos^2\left(x+y+z\right)\)

\(=\frac{1}{2}+\frac{1}{2}cos2x+\frac{1}{2}+\frac{1}{2}cos2y-\frac{1}{2}-\frac{1}{2}cos2z-\frac{1}{2}-\frac{1}{2}cos\left(2x+2y+2z\right)\)

\(=\frac{1}{2}\left(cos2x-cos2z\right)+\frac{1}{2}\left(cos2y-cos\left(2x+2y+2z\right)\right)\)

\(=-sin\left(x+z\right)sin\left(x-z\right)-sin\left(x+2y+z\right)sin\left(-x-z\right)\)

\(=sin\left(x+z\right)\left[sin\left(x+2y+z\right)-sin\left(x-z\right)\right]\)

\(=2sin\left(x+z\right).sin\left(y+z\right).cos\left(x+y\right)\)

Không giống :D

a) Biểu thức \(\frac{{1 - \cos x}}{{\sin x}}\) có nghĩa khi \(\sin x \ne 0\), tức là \(x \ne k\pi \;\left( {k\; \in \;\mathbb{Z}} \right)\).

Vậy tập xác định của hàm số đã cho là \(\mathbb{R}/{\rm{\{ }}k\pi {\rm{|}}\;k\; \in \;\mathbb{Z}\} \;\)

b) Biểu thức \(\sqrt {\frac{{1 + \cos x}}{{2 - \cos x}}} \) có nghĩa khi \(\left\{ {\begin{array}{*{20}{c}}{\frac{{1 + \cos x}}{{2 - \cos x}} \ge 0}\\{2 - \cos x \ne 0}\end{array}} \right.\) 

Vì \( - 1 \le \cos x \le 1 ,\forall x \in \mathbb{R}\)

 Vậy tập xác định của hàm số là \(D = \mathbb{R}\)

\(A=\frac{2sinx.cosx+sinx}{1+2cos^2x-1+cosx}=\frac{sinx\left(2cosx+1\right)}{cosx\left(2cosx+1\right)}=\frac{sinx}{cosx}=tanx\)

\(B=\frac{cosa}{sina}\left(\frac{1+sin^2a}{cosa}-cosa\right)=\frac{cosa}{sina}\left(\frac{1+sin^2a-cos^2a}{cosa}\right)=\frac{cosa}{sina}.\frac{2sin^2a}{cosa}=2sina\)

\(C=\frac{1+cos2x+cosx+cos3x}{2cos^2x-1+cosx}=\frac{1+2cos^2x-1+2cos2x.cosx}{cos2x+cosx}=\frac{2cosx\left(cosx+cos2x\right)}{cos2x+cosx}=2cosx\)

\(D=\frac{2sinx.cosx.\left(-tanx\right)}{-tanx.sinx}-2cosx=2cosx-2cosx=0\)

\(E=cos^2x.cot^2x-cot^2x+cos^2x+2cos^2x+2sin^2x\)

\(E=cot^2x\left(cos^2x-1\right)+cos^2x+2=\frac{cos^2x}{sin^2x}\left(-sin^2x\right)+cos^2x+2=2\)

\(F=\frac{sin^2x\left(1+tan^2x\right)}{cos^2x\left(1+tan^2x\right)}=\frac{sin^2x}{cos^2x}=tan^2x\)

Câu G mẫu số có gì đó sai sai, sao lại là \(2sina-sina?\)

\(H=sin^4\left(\frac{\pi}{2}+a\right)-cos^4\left(\frac{3\pi}{2}-a\right)+1=cos^4a-sin^4a+1\)

\(=\left(cos^2a-sin^2a\right)\left(cos^2a+sin^2a\right)+1=cos^2a-\left(1-cos^2a\right)+1=2cos^2a\)

Bài 2:

\(sin\frac{A+B}{2}=sin\left(\frac{180^0-C}{2}\right)=sin\left(90^0-\frac{C}{2}\right)=cos\frac{C}{2}\)

b/

\(A=cosx+cos\left(x+\frac{2\pi}{3}\right)+cos\left(x+\frac{4\pi}{3}\right)=cosx+2cos\left(x+\pi\right).cos\frac{\pi}{3}\)

\(=cosx-2cosx.\frac{1}{2}=0\)

c/

\(sinA+sinB+sinC=2sin\frac{A+B}{2}cos\frac{A-B}{2}+2sin\frac{C}{2}cos\frac{C}{2}=2cos\frac{C}{2}.cos\frac{A-B}{2}+2sin\frac{C}{2}cos\frac{C}{2}\)

\(=2cos\frac{C}{2}\left(cos\frac{A-B}{2}+sin\frac{C}{2}\right)=2cos\frac{C}{2}\left(cos\frac{A-B}{2}+cos\frac{A+B}{2}\right)=4cos\frac{A}{2}cos\frac{B}{2}cos\frac{C}{2}\)

d/ \(\frac{cos2a}{1+sin2a}=\frac{cos^2a-sin^2a}{cos^2a+sin^2a+2sina.cosa}=\frac{\left(cosa-sina\right)\left(cosa+sina\right)}{\left(cosa+sina\right)^2}=\frac{cosa-sina}{cosa+sina}\)

e/

\(E=\frac{sina+cosa}{cos^3a}=\frac{1}{cos^2a}\left(tana+1\right)=\left(1+tan^2a\right)\left(tana+1\right)\)

\(E=tan^3a+tan^2a+tana+1\)

Câu 1:

\(y=S\left(\frac{3-S^2}{2}\right)=\frac{3}{2}S-\frac{1}{2}S^3\)

Khi \(S\rightarrow+\infty\) thì \(y\rightarrow-\infty\)

Khi \(S\rightarrow-\infty\) thì \(y\rightarrow+\infty\)

Hàm số không có GTLN và GTNN

Câu 2:

\(y=sin^4x+cos^4x+2sin^2x.cos^2x-2sin^2x.cos^2x\)

\(y=\left(sin^2x+cos^2x\right)^2-\frac{1}{2}\left(2sinx.cosx\right)^2\)

\(y=1-\frac{1}{2}sin^22x\)

Do \(0\le sin^22x\le1\)

\(\Rightarrow y_{max}=1\) khi \(sin2x=0\)

\(y_{min}=\frac{1}{2}\) khi \(sin2x=\pm1\)

Câu 3:

\(y=sin^6x+cos^6x+3sin^2x.cos^2x\left(sin^2x+cos^2x\right)-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)\)

\(y=\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\)

\(y=1-\frac{3}{4}sin^22x\)

Do \(0\le sin^22x\le1\)

\(\Rightarrow y_{max}=1\) khi \(sin2x=0\)

\(y_{min}=\frac{1}{4}\) khi \(sin2x=\pm1\)

Câu 4:

\(y=\frac{cosx+2sinx+3}{2cosx-sinx+4}\)

\(\Leftrightarrow2y.cosx-y.sinx+4y=cosx+2sinx+3\)

\(\Leftrightarrow\left(y+2\right)sinx+\left(1-2y\right)cosx=4y-3\)

Theo điều kiện có nghiệm của pt lượng giác bậc nhất:

\(\left(y+2\right)^2+\left(1-2y\right)^2\ge\left(4y-3\right)^2\)

\(\Leftrightarrow11y^2-24y+4\le0\)

\(\Leftrightarrow\frac{2}{11}\le y\le2\)

Đặt \(\frac{x}{2}=a\)

\(\frac{sin2a+sina}{1+cos2a+cosa}=\frac{2sina.cosa+sina}{1+2cos^2a-1+cosa}=\frac{sina\left(2cosa+1\right)}{cosa\left(2cosa+1\right)}=\frac{sina}{cosa}=tana=tan\frac{x}{2}\)

a) \(cos^4x-sin^4x=\left(cos^2x+sin^2x\right)\left(cos^2x-sin^2x\right)=cos^2x-sin^2x\)

b) \(\frac{1}{1+tanx}+\frac{1}{1+cotx}=\frac{1}{1+tanx}+\frac{tanxcotx}{tanxcotx+cotx}=\frac{1}{1+tanx}+\frac{tanx}{tanx+1}\)

\(=\frac{1+tanx}{1+tanx}=1\)

c) Ta có: \(1+tan^2x=1+\frac{sin^2x}{cos^2x}=\frac{cos^2x+sin^2x}{cos^2x}=\frac{1}{cos^2x}\)

\(\Rightarrow\frac{1}{1+tan^2x}=cos^2x\)

Tương tự \(\frac{1}{1+tan^2y}=cos^2y\)

\(\Rightarrow cos^2x-cos^2y=\frac{1}{1+tan^2x}-\frac{1}{1+tan^2y}\)

\(cos^2x-cos^2y=\left(1-sin^2x\right)-\left(1-sin^2y\right)=sin^2y-sin^2x\)

d) \(\frac{1+sin^2x}{1-sin^2x}=\frac{cos^2x+sin^2x+sin^2x}{cos^2x+sin^2x-sin^2x}=\frac{cos^2x+2sin^2x}{cos^2x}=1+2\left(\frac{sinx}{cosx}\right)^2=1+2tan^2x\)