Ta có \(\sin x=\cos\left(90^0-x\right)\)

\(\Rightarrow M=\left(\sin^242^0+\sin^248^0\right)+\left(\sin^243^0+\sin^247^0\right)+\left(\sin^244^0+\sin^246^0\right)+\sin^245^0\)

\(=\left(\sin^242^0+\cos^242^0\right)+\left(\sin^243^0+\cos^243^0\right)+\left(\sin^244^0+\cos^244^0\right)+\sin^245^0\)

\(=1+1+1+\left(\frac{\sqrt{2}}{2}\right)^2=3+\frac{1}{2}=\frac{7}{2}\)

\(ADCT:\sin^2\alpha+\cos^2\alpha=1\)

\(A=\left(\sin^242^0+\sin^248^0\right)+\left(\sin^243^0+\sin^247^0\right)+\left(\sin^244^0+\sin^246^0\right)+\sin45^0\)

\(A=\left(\sin^242^0+\cos^242^0\right)+\left(\sin^243^0+\cos^243^0\right)+\left(\sin^244^0+\cos^244^0\right)+\frac{\sqrt{2}}{2}\)

\(A=1+1+1+\frac{\sqrt{2}}{2}=\frac{6+\sqrt{2}}{2}\)

Câu b lm tương tự

Ta có: \(\sin^226^0+2\cdot cos^215^0+2\cdot cos^275^0+\sin^264^0\)

\(=\left(\sin^226^0+cos^226^0\right)+2\left(\sin^215^0+cos^215^0\right)\)

=1+2

=3

Ta có: \(\sin^255^0+\sin^235^0+\sin^242^0+\sin^248^0\)

\(=\left(\sin^255^0+cos^255^0\right)+\left(\sin^242^0+cos^242^0\right)\)

=1+1

=2

Ta có: \(A=\frac{\sin^226^0+2\cdot cos^215^0+2\cdot cos^275^0+\sin^264^0}{\sin^255^0+\sin^235^0+\sin^242^0+\sin^248^0}-\frac{\tan81^0}{2\cdot\cot9^0}\)

\(=\frac32-\frac12=\frac22=1\)

\(H=cot15^o.cot35^o.cot55^o.cot75^o\)

\(=\left(cot15^o.cot75^o\right).\left(cot35^o.cot55^o\right)\)

\(=\left(cot15^o.tan15^o\right).\left(cot35^o.tan35^o\right)\)

\(=1\)

:D hết đc khong 

\(A=\left(sin^247^0+cos^247^0\right)-2+1=1+1-2=0\)

a) \(sin40^o-cos50^o=cos50^o-cos50^o=0\)

b) \(sin^230^o+sin^240^o+sin^250^o+sin^260^o\)

= \(sin^230^o+sin^260^o+sin^240^o+sin^250^o\)

= \(sin^230^o+cos^230^o+sin^240^o+cos^240^o\)

= \(1+1=2\)

a) Gợi ý: Hai góc phụ nhau thì có sin góc này bằng cos góc kia.

vd: \(sin30^o=cos70^o\)

b) Gợi ý: \(sin^2+cos^2=1\)