a: 

b: \(B=3-sin^290^0+2\cdot cos^260^0-3\cdot tan^245^0\)

\(=3-1+2\cdot\left(\dfrac{1}{2}\right)^2-3\cdot1^2\)

\(=2-3+2\cdot\dfrac{1}{4}=-1+\dfrac{1}{2}=-\dfrac{1}{2}\)

c: \(C=sin^245^0-2\cdot sin^250^0+3\cdot cos^245^0-2\cdot sin^240^0+4\cdot tan55\cdot tan35\)

\(=\left(\dfrac{\sqrt{2}}{2}\right)^2+3\cdot\left(\dfrac{\sqrt{2}}{2}\right)^2-2\cdot\left(sin^250^0+sin^240^0\right)+4\)

\(=\dfrac{1}{2}+3\cdot\dfrac{1}{2}-2+4\)

\(=2-2+4=4\)

\(B=tan^267^0-cot^223^0+2\cdot\left(sin^216^0+cos^216^0\right)-2\)

\(=0+2\cdot1-2=0\)

\(A=cot67\cdot tan67-2\left(\dfrac{\sqrt{2}}{2}\cdot sin64\right)^2-2\cdot\dfrac{sin23}{3\cdot sin23}-sin^226^0\)

\(=1-2\cdot\dfrac{1}{2}\cdot sin^264^0-\dfrac{2}{3}-sin^226^0\)

\(=1-1-\dfrac{2}{3}=-\dfrac{2}{3}\)

Ta có : \(\sin^244+\sin^246+tan18+tan72\)

\(=\sin^244+\cos^244+\tan18+\cot18\)

\(=1+\dfrac{\sin^218+\cos^218}{\sin18.\cos18}=1+\dfrac{1}{\sin18.\cos18}\)

Vậy ...

- Ấn vô đây để soạn công thức toán nha .

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

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

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

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

a:\(a\cdot sin0+b\cdot cos0+c\cdot sin90\)

\(=a\cdot0+b\cdot1+c\cdot1\)

=b+c

b: \(a\cdot cos90+b\cdot sin90+c\cdot sin180\)

\(=a\cdot0+b\cdot1+c\cdot0\)

=b

c: \(a^2\cdot sin90+b^2\cdot cos90+c^2\cdot cos180\)

\(=a^2\cdot1+b^2\cdot0+c^2\left(-1\right)\)

\(=a^2-c^2\)

Thứ tự tăng dần :

1) cos 62 độ 25 phút; sin 35 độ; cos 47 độ; sin 53 độ 30 phút; sin 74 độ.

2) tan 11 độ; cot 63 độ = tan 27 độ; cot 57 độ 30 phút; tan 55 độ