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A=(sin210+sin280)+(sin220+sin70)+(sin230+sin260)+(sin240+sin250)
Lại có: sin80=cos10; sin70=cos20; sin60=cos30; sin50=cos40
=> sin280=cos210; sin270=cos220; sin260=cos230; sin250=cos240
=>A=(sin210+cos210)+(sin220+cos220)+(sin230+cos230)+(sin240+cos240)
=>A=1+1+1+1=4
a, cos220o + cos240o + cos250o + cos270o
= (cos220o + cos270o) + (cos240o + cos250o)
= (cos220o + sin220o) + (cos240o + sin240o)
= 1 + 1 = 2
Mình nghĩ chắc sin285o là sin255o
b, sin225o + sin245o + sin265o + sin255o
= (sin225o + sin265o) + (sin245o + sin255o)
= (sin225o + cos225o) + (sin245o + cos245o)
= 1 + 1 = 2
Chúc bn học tốt!
a: \(=\left(\cos^215^0+\cos^275^0\right)+\left(\cos^225^0+\cos^265^0\right)+\left(\cos^235^0+\cos^255^0\right)+\cos^245^0\)
=1+1+1+1/2
=3,5
b: \(=\left(\sin^210^0+\sin^280^0\right)-\left(\sin^220^0+\sin^270^0\right)+\left(\sin^230^0\right)-\left(\sin^240^0+\sin^250^0\right)\)
=1-1-1+1/4
=-1+1/4=-3/4
c: \(=\left(\sin15^0-\cos75^0\right)+\left(\sin75^0-\cos15^0\right)+\sin30^0\)
=1/2
\(\sin^210^o+\sin^220^o+\sin^230^o+\sin^240^o+\sin^250^o+\sin^260^o+\sin^270^o+\sin^280^o\)
\(=\cos^280^o+\cos^270^o+\cos^260^o+\cos^250^o+\sin^250^o+\sin^260^o+\sin^270^o+\sin^280^o\)
\(=\left(\sin^280^o+\cos^280^o\right)+\left(\sin^270^o+\cos^270^o\right)+\left(\sin^260^o+\cos^260^o\right)+\left(\sin^250^o+\cos^250^o\right)\)
\(=1+1+1+1\)
\(=4\)
Vậy ....
Ta có:
\(C=sin^22^0+sin^24^0+...+sin^288^0\)
\(C=\left(sin^22^0+sin^288^0\right)+\left(sin^24^0+sin^286^0\right)+...+\left(sin^244^0+sin^246^0\right)\)
\(C=\left(sin^22^0+cos^22^0\right)+\left(sin^24^0+cos^24^0\right)+...+\left(sin^244^0+cos^244^0\right)\)
\(C=1+1+...+1\) \(C=22\)
áp dụng công thức sin2a+cos2a=1
A= sin2a +cos2a-2sina.cosa-sin2a-cos2a+2sina.cosa = 0
B=(sỉn2a+cos2a)2 =12 =1
C= cos2a(cos2a+sin2a)+ sin2a=cos2a+sin2a=1
D=sin2a(sin2p+cos2p)+cos2a=sin2a+cos2a=1
E= (sin2a+cos2a)(sin4a-sin2a.cos2a+cos4a)+3sin2a.cos2a
=sin4a+2sin2a.cos2a+ cos4a=(sin2a+cos2a)2=1