1、计算;27^1/3-(-1/2)^log₂⁴+sinπ/3 tanπ/3+(2010)^lg
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1、计算;27^1/3-(-1/2)^log₂⁴+sinπ/3 tanπ/3+(2010)^lgl
2、已知tan(3π+α)=2,求(1)(sinα+cosα)²(2)(sinα-cosα)/(2sinα+cosα)
2、已知tan(3π+α)=2,求(1)(sinα+cosα)²(2)(sinα-cosα)/(2sinα+cosα)
答:
1)
27^1/3-(-1/2)^log₂⁴+sinπ/3 tanπ/3+(2010)^lgl
=3-(-1/2)^2+(√3/2)*(√3)+2010^0
=3-1/4+3/2+1
=4+5/4
=21/4
2)
tan(3π+α)=2
tana=2
sina=2cosa,代入sin²a+cos²a=1有:
4cos²a+cos²a=1
cos²a=1/5
2.1)
(sinα+cosα)²
=1+2sinacosa
=1+2*2cos²a
=1+4*(1/5)
=9/5
2.2)
(sinα-cosα)/(2sinα+cosα) 分子分母同除以cosa:
=(tana-1)/(2tana+1)
=(2-1)/(2*2+1)
=1/5
1)
27^1/3-(-1/2)^log₂⁴+sinπ/3 tanπ/3+(2010)^lgl
=3-(-1/2)^2+(√3/2)*(√3)+2010^0
=3-1/4+3/2+1
=4+5/4
=21/4
2)
tan(3π+α)=2
tana=2
sina=2cosa,代入sin²a+cos²a=1有:
4cos²a+cos²a=1
cos²a=1/5
2.1)
(sinα+cosα)²
=1+2sinacosa
=1+2*2cos²a
=1+4*(1/5)
=9/5
2.2)
(sinα-cosα)/(2sinα+cosα) 分子分母同除以cosa:
=(tana-1)/(2tana+1)
=(2-1)/(2*2+1)
=1/5
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