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a) sin 40 - cos 50 =0
b) sin230 + sin240 + sin250 + sin260 = 2
c) cos210 - cos220 + cos230 - cos240 - cos250 - cos270 + cos280 = - sin230
\(a.sin40^o-cos50^o=sin40^o-sin40^o=0\)
\(b.sin^230^o+sin^240^o+sin^250^o+sin^260^o=\left(sin^230^0+sin^260^o\right)+\left(sin^240^0+sin^250^o\right)=\left(sin^230^0+cos^230^o\right)+\left(sin^240+cos^240^o\right)=1+1=2\)
\(c.\left(cos^210^o+cos^280^o\right)-\left(cos^220^o+cos^270^0\right)-\left(cos^240^o-cos^250^o\right)+cos^230^o=\left(cos^210^o+sin^210^o\right)-\left(cos^220^o+sin^220^o\right)-\left(cos^240^o+sin^240^0\right)+cos^230^0=1-1-1+\dfrac{3}{4}=-\dfrac{1}{4}\)
\(A=\cos^215^o-\cos^225^o+\cos^235^o-\cos^245^o+\cos^255^o-\cos^265^o+\cos^275^o\)
\(A=\sin^275^o-\sin^265^o+\sin^255^o-\sin^245^o+\cos^255^o-\cos^265^o+\cos^275^o\)
\(A=\left(\sin^275^o+\cos^275^o\right)-\left(\sin^265^o+\cos^265^o\right)+\left(\sin^255^o+\cos^255^o\right)-\sin^245^o\)
\(A=1-1+1-\frac{1}{2}\)
\(A=\frac{1}{2}\)
A = cos2 100 + cos2 200 + cos2 300 + .... + cos2 800
= ( cos2 100 + cos2 800 ) + ( cos2 200 + cos2 700 ) + ( cos2 300 + cos2 600 ) + ( cos2 400 + cos2 500 )
= ( cos2 100 + sin2 100 ) + ( cos2 200 + sin2 700 ) + ( cos2 300 + sin2 300 ) + ( cos2 400 + sin2 400 )
= 1 + 1 + 1 +1 = 4
Tính:
a) A= cos2 20 độ + cos2 40 độ + cos2 50 độ + cos2 70 độ
b) B= sin4 a + cos4 a + 2sin2 a . cos2 a
\(A=sin^210^o+cos^220^o+sin^280^o+cos^270^o\)
\(A=\left(sin^210^o+sin^280^o\right)+\left(cos^220^o+cos^270^o\right)\)
\(A=0+0\)
\(A=0\)
cos220 + cos240 + cos250 + cos270
= sin270 + sin250 + cos250 + cos270
= (sin270 + cos270) +( sin250 + cos250)
=1+1=2
Lời giải:
Ta có: \(\cos a=\sin (90-a)\Rightarrow \cos ^2a=\sin ^2(90-a)\)
Do đó:
\(\cos ^210=\sin ^280\)
\(\cos ^220=\sin ^270\)
\(\cos ^230=\sin ^260\)
\(\cos ^240=\sin ^250\)
\(\Rightarrow \cos ^210+\cos ^220+..+\cos ^280=(\sin ^280+\cos ^280)+(\sin ^270+\cos ^270)+(\sin ^260+\cos ^260)+(\sin ^250+\cos ^250)\)
\(=1+1+1+1=4\)