Do \(0< a< \frac{\pi}{2}\Rightarrow sina>0\)

\(sin^2a+cos^2a=1\Rightarrow sina=\sqrt{1-cos^2a}=\sqrt{1-\left(\frac{15}{17}\right)^2}=\frac{8}{17}\)

\(cos2a=2cos^2a-1=2.\left(\frac{15}{17}\right)^2-1=\frac{161}{289}\)

\(\pi< a< \frac{3\pi}{2}\Rightarrow\left\{{}\begin{matrix}sina< 0\\cosa< 0\end{matrix}\right.\) \(\Rightarrow sin2a=2sina.cosa>0\)

\(\Rightarrow sin2a=\sqrt{1-cos^22a}=\frac{3\sqrt{7}}{8}\)

\(cos2a=1-2sin^2a=\frac{1}{8}\)

\(\Leftrightarrow sin^2a=\frac{7}{16}\Rightarrow sina=-\frac{\sqrt{7}}{4}\)

\(\Rightarrow M=\frac{-\frac{\sqrt{7}}{4}-\frac{3\sqrt{7}}{8}}{-\frac{\sqrt{7}}{4}+\frac{3\sqrt{7}}{8}}=...\)

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

\(=sinx\left(\frac{1+cosx-\left(1-cosx\right)}{1+cosx}\right)=\frac{2sinx.cosx}{1+cosx}\)

\(1-sin2x.sin3x-cos2x.cos3x=1-\left(cos3x.cos2x+sin3x.sin2x\right)=1-cos\left(3x-2x\right)=1-cosx\)

\(\Rightarrow\frac{1-sin2x.sin3x-cos2x.cos3x}{sinx\left(1-tan^2\frac{x}{2}\right)}=\frac{1-cosx}{\frac{2sinx.cosx}{1+cosx}}=\frac{\left(1-cosx\right)\left(1+cosx\right)}{2sinx.cosx}\)

\(=\frac{1-cos^2x}{2sinx.cosx}=\frac{sin^2x}{2sinx.cosx}=\frac{sinx}{2cosx}=\frac{1}{2}tanx\)

\(\frac{sina+sin3a+sin2a}{cosa+cos3a+cos2a}=\frac{2sin2a.cosa+sin2a}{2cos2a.cosa+cos2a}=\frac{sin2a\left(2cosa+1\right)}{cos2a\left(2cosa+1\right)}=\frac{sin2a}{cos2a}=tan2a\)

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

\(=cos\left(\frac{\pi}{2}-2a\right)=sin2a\)

theo giả thiết: \(\sin x=\frac{1}{3}\Rightarrow\left(1-\cos^2x\right)=\frac{1}{9}\Rightarrow cosx=\frac{\pm2\sqrt{2}}{3}\)

mà \(0< x< \frac{\pi}{2}\) nên \(cosx=\frac{2\sqrt{2}}{3}\)

ta có: \(\sin\left(a+\frac{\pi}{3}\right)=\sin a\cos\frac{\pi}{3}+\cos a\sin\frac{\pi}{3}=\frac{1}{6}+\frac{\sqrt{6}}{3}\)

Bạn @Nhók Lì Lợm giải đúng rồi nhưng bị nhầm phần biến x (lẽ ra theo đề là a)

có A=\(\dfrac{1-cosa+2cos^2a-1}{2sina.cosa-sina}=\dfrac{cosa\left(2cosa-1\right)}{sina\left(2cosa-1\right)}=\dfrac{cosa}{sina}=cota\)

\(A=cos^2a+cos^2b+2cosa.cosb+sin^2a+sin^2b+2sina.sinb\)

\(=cos^2a+sin^2a+cos^2b+sin^2b+2\left(cosa.cosb+sina.sinb\right)\)

\(=2+2cos\left(a-b\right)=2+2cos\frac{\pi}{3}=3\)

\(\left(cosa+sina\right)^2=\frac{36}{25}\Leftrightarrow1+2sina.cosa=\frac{36}{25}\)

\(\Rightarrow sin2a=\frac{36}{25}-1=\frac{11}{25}\)

\(cos2a=cos^2a-sin^2a=\left(cosa-sina\right)\left(cosa+sina\right)>0\)

\(\Rightarrow cos2a=\sqrt{1-sin^22a}=\frac{6\sqrt{14}}{25}\)

\(A=\frac{cos^2a-sin^2a}{2sin^2a+3sina.cosa}=\frac{\frac{cos^2a}{cos^2a}-\frac{sin^2a}{sin^2a}}{\frac{2sin^2a}{cos^2a}+\frac{3sina.cosa}{cos^2a}}=\frac{1-tan^2a}{2tan^2a+3tana}=\frac{1-2^2}{2.2^2+3.2}=...\)

\(\frac{1-2sin2a+cos2a}{1+2sin2a+cos2a}=\frac{1-4sina.cosa+2cos^2a-1}{1+4sina.cosa+2cos^2a-1}=\frac{2cosa\left(cosa-2sina\right)}{2cosa\left(cosa+2sina\right)}\)

\(=\frac{cosa-2sina}{cosa+2sina}\)

Bạn coi lại đề, muốn ra được biểu thức vế phải thì trước sin2a không được có số 2