Đề yêu cầu gì bạn nhỉ?

ta có : \(sin^3a+cos^3a=\left(sina+cosa\right)^3-3sina.cosa\left(sina+cosa\right)\)

\(=2^3-3sina.cosa\left(2\right)=8-6sina.cosa\)

\(=11-3sin^2a-6sina.cosa-3cos^2a=11-3\left(sin+cos\right)^2=11-3.2^2=11-12=-1\)

ĂN CHO CÒN NÓNG:NGON.

\(sina+cosa=\frac{5}{4}\Rightarrow\left(sina+cosa\right)^2=\frac{25}{16}\)

\(\Rightarrow sin^2a+cos^2a+2sina.cosa=\frac{25}{16}\)

\(sina.cosa=\frac{\frac{25}{16}-1}{2}=\frac{9}{32}\)

b/ \(\left(sina-cosa\right)^2=sin^2a+cos^2a-2sinacosa\)

\(\left(sina-cosa\right)^2=1-2.\frac{9}{32}=\frac{7}{16}\)

\(\Rightarrow sina-cosa=\pm\frac{\sqrt{7}}{4}\)

c/ \(sin^3a-cos^3a=\left(sina-cosa\right)\left(sin^2a+cos^2a+sina.cosa\right)\)

\(=\left(sina-cosa\right)\left(1+\frac{9}{32}\right)=\pm\frac{41\sqrt{7}}{128}\)

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

\(\frac{sin^3a-cos^3a}{sina-cosa}-sina.cosa=\frac{\left(sina-cosa\right)\left(sin^2a+cos^2a+sina.cosa\right)}{sina-cosa}-sina.cosa\)

\(=sin^2a+cos^2a+sina.cosa-sina.cosa=1\)

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

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

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

\(\frac{1-cosx+cos2x}{sin2x-sinx}=\frac{1-cosx+2cos^2x-1}{2sinx.cosx-sinx}=\frac{cosx\left(2cosx-1\right)}{sinx\left(2cosx-1\right)}=\frac{cosx}{sinx}=cotx\)

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

\(=1+sina.cosa+sina+cosa\)

\(=\left(sina+1\right)\left(cosa+1\right)\)