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\(B=\sin^230^0+\sin^240^0+\sin^250^0+\sin^260^0\)
\(B=\sin^230^0+\sin^240^0+\cos^2\left(90^0-50^0\right)+\cos^2\left(90^0-60^0\right)\)
\(B=\sin^230^0+\sin^240^0+\cos^240^0+\cos^230^0\)
\(B=\left(\sin^230^0+\cos^230^0\right)\left(\sin^240^0+\cos^240^0\right)\)
\(B=1+1\)
\(B=2\)
Chúc bạn hok tốt!!! vvvvvvvv
Ta có :\(\sin\left(60\right)=\cos\left(30\right)\)(phụ nhau)
\(\Leftrightarrow sin^2\left(60\right)=\cos^2\left(30\right)\)
và :\(sin^2\left(50\right)=\cos^2\left(40\right)\)(tương tự như trên nha bạn)
Thay vào biểu thức B ta có :
\(B=\sin^2\left(30\right)+sin^2\left(40\right)+\cos^2\left(30\right)+\cos^2\left(40\right)\)
\(B=1+1\)
\(B=2\)
chúc bạn học tốt :)
a) sin 40 - cos 50 =0
b) sin230 + sin240 + sin250 + sin260 = 2
c) cos210 - cos220 + cos230 - cos240 - cos250 - cos270 + cos280 = - sin230
\(a.sin40^o-cos50^o=sin40^o-sin40^o=0\)
\(b.sin^230^o+sin^240^o+sin^250^o+sin^260^o=\left(sin^230^0+sin^260^o\right)+\left(sin^240^0+sin^250^o\right)=\left(sin^230^0+cos^230^o\right)+\left(sin^240+cos^240^o\right)=1+1=2\)
\(c.\left(cos^210^o+cos^280^o\right)-\left(cos^220^o+cos^270^0\right)-\left(cos^240^o-cos^250^o\right)+cos^230^o=\left(cos^210^o+sin^210^o\right)-\left(cos^220^o+sin^220^o\right)-\left(cos^240^o+sin^240^0\right)+cos^230^0=1-1-1+\dfrac{3}{4}=-\dfrac{1}{4}\)
P=sin2200+sin2400+sin2450+sin2500+sin2700
đổi sin2500 thành cos2400,sin2700 thành cos2200 rồi thay vào ta được:
sin2200+cos2200+sin2400+cos2400+\(\left(\dfrac{\sqrt{2}}{2}\right)^2\)
=\(2+\dfrac{1}{2}=\dfrac{5}{2}=2,5\)
a: \(=\left(sin^210^0+sin^280^0\right)+\left(sin^220^0+sin^270^0\right)+sin^245^0\)
\(=1+1+\dfrac{1}{2}=\dfrac{5}{2}\)
b: \(=\left(sin^242^0+sin^248^0\right)+\left(sin^243^0+sin^247^0\right)+...+sin^245^0\)
=1+1+1+1/2
=3,5
c: \(=tan35^0\cdot tan55^0\cdot tan40^0\cdot tan50^0\cdot tan45^0=1\)
d: \(=\left(cos^215^0+cos^275^0\right)-\left(cos^225^0+cos^265^0\right)+\left(cos^235^0+cos^255^0\right)-\dfrac{1}{2}\)
=1-1+1-1/2
=1/2
Bài 1 :
\(D=cos^220^0+cos^230^0+cos^240^0+cos^250^0+cos^260^0+cos^270^0\)
\(=\left(cos^220^0+cos^270^0\right)+\left(cos^230^0+cos^260^0\right)+\left(cos^240^0+cos^250^0\right)\)
\(=1+1+1=3\)
Bài 2 :
\(E=sin^25^0+sin^225^0+sin^245^0+sin^265^0+sin^285^0\)
\(=\left(sin^25^0+sin^285^0\right)+\left(sin^225^0+sin^265^0\right)+sin^245^0\)
\(=1+1+\dfrac{1}{2}=\dfrac{5}{2}\)
Bài 3 :
\(F=sin^6\alpha+cos^6\alpha+3sin^2\alpha.cos^2\alpha\)
\(=1-3sin^2\alpha.cos^2\alpha+3sin^2a.cos^2\alpha\)
\(=1\)
\(B=\dfrac{1-4\sin^2x\cdot\cos^2x}{\sin^2x+2\sin x\cdot\cos x+\cos^2}+2\sin x\cdot\cos x\\ B=\dfrac{1-4\sin^2x\cdot\cos^2x}{2\sin x\cdot\cos x}+2\sin x\cdot\cos x\\ B=\dfrac{1-4\sin^2x\cdot\cos^2x+4\sin^2x\cdot\cos^2x}{2\sin x\cdot\cos x}=\dfrac{1}{2\sin x\cdot\cos x}\)
d)
sin 2 20 0 + cos 2 30 0 - sin 2 40 0 - sin 2 50 0 + cos 2 60 0 + sin 2 70 0
= sin 2 20 0 + cos 2 30 0 - sin 2 40 0 - c o s 2 40 0 + sin 2 30 0 + cos 2 20 0
= sin 2 20 0 + cos 2 20 0 + cos 2 30 0 + sin 2 30 0 - sin 2 40 0 + cos 2 40 0= 1 + 1 - 1
= 1