a: \(cos70=sin20\)

20<25

=>\(sin20< sin25\)

=>\(cos70< sin25\)

b: \(\dfrac{sin50}{cos40}=\dfrac{cos\left(90-50\right)}{cos40}=\dfrac{cos40}{cos40}=1\)

a) Ta có: 

\(cos70^o=sin\left(90^o-70^o\right)=sin20^o\)

Ta so sánh \(sin25^o\) và \(sin20^o\)

\(25^o>20^o\Rightarrow sin25^o>sin20^o\)

\(\Rightarrow sin25^o>cos70^o\)

b) \(\dfrac{sin50^o}{cos40^o}\)

Ta có: 

\(cos40^o=sin\left(90^o-40^o\right)=sin50^o\)

\(\Rightarrow\dfrac{sin50^o}{cos40^o}=\dfrac{sin50^o}{sin50^o}=1\)

Vì với góc nhọn \(\widehat{A}\) và \(\widehat{B}\), ta có: \(\sin\widehat{A}< \sin\widehat{B}\Leftrightarrow\widehat{A}< \widehat{B}\)

Vậy: Vì 20⁰<70⁰ nên ta có: \(\sin20^0< \sin70^0\)

a: \(cos32=sin58;cos53=sin37;cos8=sin82\)

18<37<44<58<82

=>\(sin18< sin37< sin44< sin58< sin82\)

=>\(sin18< cos53< sin44< cos32< cos8\)

b: 20<45

=>\(sin20< tan20\)

\(cot8=tan82;cot37=tan53\)

20<40<53<82

=>\(tan20< tan40< tan53< tan82\)

=>\(tan20< tan40< cot37< cot8\)

=>\(sin20< tan20< tan40< cot37< cot8\)

\(\sin^212+\sin^220-\sin^235-\sin^255+\sin^270+\sin^278\)

\(=\cos^278+\cos^280-\cos^255-\sin^255+\sin^270+\sin^278\)

\(=1+1-1=1\)

a: \(A=sin^210^0+sin^280^0+cos^220^0+sin^270^0\)

\(=sin^210^0+cos^210^0+sin^270^0+sin^270^0\)

\(=2\cdot sin^270^0+1\)

b: \(=sin^215^0+sin^275^0+sin^235^0+sin^255^0\)

\(=sin^215^0+cos^215^0+sin^235^0+cos^235^0\)

=1+1

=2

\(A=sin^210^0+sin^280^0+cos^220^0+sin^270^0\)

\(=sin^210^0+cos^210^0+sin^270^0+sin^270^0\)

\(=2sin^270^0+1\)

\(B=sin^215^0+sin^275^0+sin^235^0+sin^255^0\)

\(=sin^215^0+cos^215^0+sin^235^0+cos^235^0\)

=1+1

=2

\(A=sin^210^o+cos^220^o+sin^280^o+cos^270^o\)

\(A=\left(sin^210^o+sin^280^o\right)+\left(cos^220^o+cos^270^o\right)\)

\(A=0+0\)

\(A=0\)

\(M=\left(\sin^210^0+\sin^280^0\right)+\left(\sin^220^0+\sin^270^0\right)-3\tan39^0\cdot\cot39^0\\ M=\left(\sin^210^0+\cos^210^0\right)+\left(\sin^220^0+\cos^220^0\right)-3\cdot1=1+1-3=-1\)

GTLN của B = -x-2021+10√x là