1、计算;27^1/3-(-1/2)^log₂⁴+sinπ/3 tanπ/3+(2010)^lgl2、已知tan(3π+α)=2,求(1)(sinα+cosα)²(2)(sinα-cosα)/(2sinα+cosα)
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![1、计算;27^1/3-(-1/2)^log₂⁴+sinπ/3 tanπ/3+(2010)^lgl2、已知tan(3π+α)=2,求(1)(sinα+cosα)²(2)(sinα-cosα)/(2sinα+cosα)](/uploads/image/z/9310602-66-2.jpg?t=1%E3%80%81%E8%AE%A1%E7%AE%97%3B27%5E1%2F3-%28-1%2F2%29%5Elog%26%238322%3B%26%238308%3B%2Bsin%CF%80%2F3+tan%CF%80%2F3%2B%282010%29%5Elgl2%E3%80%81%E5%B7%B2%E7%9F%A5tan%283%CF%80%2B%CE%B1%29%3D2%2C%E6%B1%82%281%29%28sin%CE%B1%2Bcos%CE%B1%29%26%23178%3B%282%29%28sin%CE%B1-cos%CE%B1%29%2F%282sin%CE%B1%2Bcos%CE%B1%29)
1、计算;27^1/3-(-1/2)^log₂⁴+sinπ/3 tanπ/3+(2010)^lgl2、已知tan(3π+α)=2,求(1)(sinα+cosα)²(2)(sinα-cosα)/(2sinα+cosα)
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)
1、计算;27^1/3-(-1/2)^log₂⁴+sinπ/3 tanπ/3+(2010)^lgl2、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
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
=21/4
1、
即tanα=2
所以sinαcosα
=sin αcosα/(sin²α+cos²α)
上下除以co...
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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
=21/4
1、
即tanα=2
所以sinαcosα
=sin αcosα/(sin²α+cos²α)
上下除以cos²α
因为sinα/cosα=tanα
所以sinαcosα=tan²α/(tan²α+1)=4/5
所以(sinα+cosα)²
=sin²α+cos²α+2sinαcosα
=1+2sinαcosα
=13/5
2、
上下除以cosα
因为sinα/cosα=tanα
所以(sinα-cosα)/(2sinα+cosα)
= (tanα-1)/(2tanα+1)
=1/5
收起
(1)原式
=3-(-1/2)²+√3/2x√3+(2010)º
=3-1/4+3/2+1
=4-1/4+3/2
=21/4
二题:
tan(3π+α)=2
tanα=2
(1)
(sinα+cosα)²
=sin²α+cos²α+2sinαcosα
=(sin...
全部展开
(1)原式
=3-(-1/2)²+√3/2x√3+(2010)º
=3-1/4+3/2+1
=4-1/4+3/2
=21/4
二题:
tan(3π+α)=2
tanα=2
(1)
(sinα+cosα)²
=sin²α+cos²α+2sinαcosα
=(sin²α+cos²α+2sinαcosα)/(sin²α+cos²α)
=(tan²α+1+2tanα)/(tan²α+1) (分子分母同时除以cos²α而得)
=(4+1+2x2)/(4+1)
=9/5
(2)(sinα-cosα)/(2sinα+cosα)
=(tanα-1)/(2tanα+1) (分子分母同时除以cosα而得)
=(2-1)/(2x2+1)
=1/5
收起