求证sin^2α+cosαcos(π/3+α)-sin^(π/6-α)的值与α无关求证:sin^2α+cosαcos(π/3+a)-sin^2(π/6-α)的值是与α无关的定值
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![求证sin^2α+cosαcos(π/3+α)-sin^(π/6-α)的值与α无关求证:sin^2α+cosαcos(π/3+a)-sin^2(π/6-α)的值是与α无关的定值](/uploads/image/z/2758325-5-5.jpg?t=%E6%B1%82%E8%AF%81sin%5E2%CE%B1%2Bcos%CE%B1cos%28%CF%80%2F3%2B%CE%B1%29-sin%5E%EF%BC%88%CF%80%2F6-%CE%B1%EF%BC%89%E7%9A%84%E5%80%BC%E4%B8%8E%CE%B1%E6%97%A0%E5%85%B3%E6%B1%82%E8%AF%81%3Asin%5E2%CE%B1%2Bcos%CE%B1cos%28%CF%80%2F3%2Ba%29-sin%5E2%28%CF%80%2F6-%CE%B1%29%E7%9A%84%E5%80%BC%E6%98%AF%E4%B8%8E%CE%B1%E6%97%A0%E5%85%B3%E7%9A%84%E5%AE%9A%E5%80%BC)
求证sin^2α+cosαcos(π/3+α)-sin^(π/6-α)的值与α无关求证:sin^2α+cosαcos(π/3+a)-sin^2(π/6-α)的值是与α无关的定值
求证sin^2α+cosαcos(π/3+α)-sin^(π/6-α)的值与α无关
求证:sin^2α+cosαcos(π/3+a)-sin^2(π/6-α)的值是与α无关的定值
求证sin^2α+cosαcos(π/3+α)-sin^(π/6-α)的值与α无关求证:sin^2α+cosαcos(π/3+a)-sin^2(π/6-α)的值是与α无关的定值
利用积化和差公式和二倍角公式:
sin^2α+cosαcos(π/3+a)-sin^2(π/6-α)
=(1-cos2a)/2+[cos(2a+π/3)+cos(π/3)]/2 -[1-cos(π/3-2a)]/2
=1/2-(cos2a)/2+cos(2a+π/3)/2+1/4-1/2+cos(π/3-2a)]/2
=1/4-(cos2a)/2+cos2a*1/4-sin2a*√3/2*1/2+cos2a*1/4+sin2a*√3/2 *1/2
=1/4-(cos2a)/2+cos2a*1/2+sin2a*√3/4-sin2a*√3/4
=1/4
所以sin^2α+cosαcos(π/3+α)-sin^(π/6-α)的值与α无关
sin^2α是什么意思?是(sinα)^2吗?
如果是的话:
证:
(sinα)^2+cosαcos(π/3+α)-[sin(π/6-α)]^2
=(sinα)^2+cosα[cos(π/3)cosα-sin(π/3)sinα]-[sin(π/6)cosα-cos(π/6)sinα)]^2
=(sinα)^2+cosα(cosα-√3sinα)/2-[(co...
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sin^2α是什么意思?是(sinα)^2吗?
如果是的话:
证:
(sinα)^2+cosαcos(π/3+α)-[sin(π/6-α)]^2
=(sinα)^2+cosα[cos(π/3)cosα-sin(π/3)sinα]-[sin(π/6)cosα-cos(π/6)sinα)]^2
=(sinα)^2+cosα(cosα-√3sinα)/2-[(cosα-√3sinα)/2]^2
=(sinα)^2+(1/2)(cosα)^2-(√3/2)sinαcosα-(1/4)(cosα)^2+(√3/2)sinαcosα-(3/4)(sinα)^2
=(1/4)(sinα)^2+(1/4)(cosα)^2
=(1/4)[(sinα)^2+(cosα)^2]
=1/4
可见:原式的值与α无关。
证毕。
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