已知α为锐角,且sin^2α+sinαcosα-2cos^2=0.(1)求tanα的值.(2)求sin[α-(π/3)]
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![已知α为锐角,且sin^2α+sinαcosα-2cos^2=0.(1)求tanα的值.(2)求sin[α-(π/3)]](/uploads/image/z/3152149-61-9.jpg?t=%E5%B7%B2%E7%9F%A5%CE%B1%E4%B8%BA%E9%94%90%E8%A7%92%2C%E4%B8%94sin%5E2%CE%B1%EF%BC%8Bsin%CE%B1cos%CE%B1-2cos%5E2%3D0.%281%29%E6%B1%82tan%CE%B1%E7%9A%84%E5%80%BC.%282%29%E6%B1%82sin%5B%CE%B1-%28%CF%80%2F3%29%5D)
已知α为锐角,且sin^2α+sinαcosα-2cos^2=0.(1)求tanα的值.(2)求sin[α-(π/3)]
已知α为锐角,且sin^2α+sinαcosα-2cos^2=0.(1)求tanα的值.(2)求sin[α-(π/3)]
已知α为锐角,且sin^2α+sinαcosα-2cos^2=0.(1)求tanα的值.(2)求sin[α-(π/3)]
(1)
sin^2α+sinαcosα-2cos^2=0
(sinα+2cosα)(sinα-cosα)=0
因为α为锐角
所以
sinα=cosα
tanα=1
所以
α=π/4
(2)
sin(α-π/3)
=sinαcosπ/3-cosαsinπ/3
=√2/2*1/2-√2/2*√3/2
=(√2-√6)/4
sin^2α=2sin^αcosα,cos^2α=......,代入上式与cos平方+sin平方组成方程 求出cos sin 后面的就迎刃而解了!
1)两边同除以cosα ,得:tan^2α+tanα-2=0
解得:tanα=1 或 tanα=-2(舍去)
2)因为:α为锐角 且 tanα=1
所以,可知:α=π/4
所以,sin[α-(π/3)]=sin(π/4)cos(π/3)-cos(π/4)sin(π/3)
...
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1)两边同除以cosα ,得:tan^2α+tanα-2=0
解得:tanα=1 或 tanα=-2(舍去)
2)因为:α为锐角 且 tanα=1
所以,可知:α=π/4
所以,sin[α-(π/3)]=sin(π/4)cos(π/3)-cos(π/4)sin(π/3)
=(√2/2)*(1/2)-(√2/2)*(√3/2)
=(√2-√6)/4
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