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

GTLN của B = -x-2021+10√x là

\(A=sin^210^o+sin^220^o+sin^230^o+sin^240^o+sin^250^o+sin^260^o+sin^270^o+sin^280^o\)

\(A=\left(sin^210^o+sin^280^o\right)+\left(sin^220^o+sin^270^o\right)+\left(sin^230^o+sin^260^o\right)+\left(sin^240^o+sin^250^o\right)\)

\(A=\left(sin^210^o+cos^210^o\right)+\left(sin^220^o+cos^220^o\right)+\left(sin^230^o+cos^230^o\right)+\left(sin^240^o+cos^240^o\right)\)

\(A=1+1+1+1\)

\(A=4\)

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

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

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

\(A=0+0\)

\(A=0\)

A = ( sin² 10^o + sin² 80^o) + (sin² 20^o  + sin² 70^o) + ...+ (sin²40^o + sin² 50^o)

A =  ( sin² 10^o + cos² 10^o) + (sin² 20^o  + cos² 20^o) + ...+ (sin²40^o + cos² 40^o)

A = 1 + 1 + 1 + 1 = 4 ( Vì  ( sin² a + cos² a =  1 với mọi a)

Bài làm

A = ( sin² 10^o + sin² 80^o) + (sin² 20^o  + sin² 70^o) + ...+ (sin²40^o + sin² 50^o)

A =  ( sin² 10^o + cos² 10^o) + (sin² 20^o  + cos² 20^o) + ...+ (sin²40^o + cos² 40^o)

A = 1 + 1 + 1 + 1 = 4 

hok tốt

a) sin 40 - cos 50 =0

b) sin²30 + sin²40 + sin²50 + sin²60 = 2

c) cos²10 - cos²20 + cos²30 - cos²40 - cos²50 - cos²70 + cos²80 = - sin²30

\(a.sin40^o-cos50^o=sin40^o-sin40^o=0\)
\(b.sin^230^o+sin^240^o+sin^250^o+sin^260^o=\left(sin^230^0+sin^260^o\right)+\left(sin^240^0+sin^250^o\right)=\left(sin^230^0+cos^230^o\right)+\left(sin^240+cos^240^o\right)=1+1=2\)
\(c.\left(cos^210^o+cos^280^o\right)-\left(cos^220^o+cos^270^0\right)-\left(cos^240^o-cos^250^o\right)+cos^230^o=\left(cos^210^o+sin^210^o\right)-\left(cos^220^o+sin^220^o\right)-\left(cos^240^o+sin^240^0\right)+cos^230^0=1-1-1+\dfrac{3}{4}=-\dfrac{1}{4}\)

A=(sin​​​²20°+sin²70°)+(sin²30°+sin²60°)

+(sin²40°+sin²50°)-tan²45°

=(sin​​²20°+cos​²20°)+(sin²30°+cos²30°)+(sin²40°+cos²40°)-1

=1+1+1-1=2

kết quả là 2

\(\sin^210^o+\sin^220^o+\sin^230^o+\sin^240^o+\sin^250^o+\sin^260^o+\sin^270^o+\sin^280^o\)

\(=\cos^280^o+\cos^270^o+\cos^260^o+\cos^250^o+\sin^250^o+\sin^260^o+\sin^270^o+\sin^280^o\)

\(=\left(\sin^280^o+\cos^280^o\right)+\left(\sin^270^o+\cos^270^o\right)+\left(\sin^260^o+\cos^260^o\right)+\left(\sin^250^o+\cos^250^o\right)\)

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

\(=4\)

Vậy ....