bạn phải ns rõ là bài này có được dùng máy tính hay ko .

mình làm theo cách ko bấm máy nhé

 Ta có : khi góc \(\alpha\)tăng từ 0 -> 90 độ thì : \(\hept{\begin{cases}\sin\alpha\\\tan\alpha\end{cases}}\)tăng ; \(\hept{\begin{cases}\cos\alpha\\\cot\alpha\end{cases}}\)tăng

a) \(\sin15^o=\cos75^o>\cos80^o\)   ;\(\tan25^o=\cot65^o>\cot75^o\)

\(\cot75^o=\tan15^o=\frac{\sin15^o}{\cos15^o}>\sin15^o\)( vì \(0< \cos15^o< 1\) )

tóm lại : \(\cos80^o< \sin15^o< \cot75^o< \tan25^o\)

b) tương tự 

\(A=cos10+cos170+cos40+cos140+cos70+cos110\)

\(A=cos10+cos\left(180-10\right)+cos40+cos\left(180-40\right)+cos70+cos\left(180-70\right)\)

\(A=cos10-cos10+cos40-cos40+cos70-cos70\)

\(A=0\)

\(B=sin5+sin355+sin10+sin350+...+sin175+sin185+sin360\)

\(B=sin5+sin\left(360-5\right)+sin10+sin\left(360-10\right)+...+sin175+sin\left(360-175\right)+sin360\)

\(B=sin5-sin5+sin10-sin10+...+sin175-sin175+sin360\)

\(B=sin360=0\)

\(C=cos^22+cos^288+cos^24+cos^284+...+cos^244+cos^246\)

\(C=cos^22+cos^2\left(90-2\right)+cos^24+cos^2\left(90-4\right)+...+cos^244+cos^2\left(90-44\right)\)

\(C=cos^22+sin^22+cos^24+sin^24+...+cos^244+sin^244\)

\(C=1+1+...+1\) (có \(\frac{44-2}{2}+1=22\) số 1)

\(\Rightarrow C=22\)

A=sin²40+cos²10+2sin40cos10-cos²40-sin²10-2sin10cos40+cos(90+50)

A=(sin²40-cos²40)+(cos²10-sin²10)+2(sin40cos10-cos40sin10)-sin50

A=(sin40-cos40)(sin40+cos40)-(sin10-cos10)(sin10+cos10)+1-sin50

A=\(\sqrt{2}\) sin(40-\(\frac{\pi}{4}\))\(\sqrt{2}\) cos(40-\(\frac{\pi}{4}\))-\(\sqrt{2}\)sin(10-\(\frac{\pi}{4}\))\(\sqrt{2}\) cos(10-\(\frac{\pi}{4}\))+1-sin50

A=-2sin5cos5+2sin35cos35+1-sin50

A= - sin10+sin70+1-sin50

A= 2cos40sin30-sin(90-40)+1

A=cos40-cos40+1 =1