求sin²αsin²β+cos²αcos²β-1/2cos2αcos2β的值 化
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求sin²αsin²β+cos²αcos²β-1/2cos2αcos2β的值 化简:(sinθ+sin3θ+sin5θ)/
化简:(sinθ+sin3θ+sin5θ)/(cosθ+cos3θ+cos5θ)=
求sin²αsin²β+cos²αcos²β-1/2cos2αcos2β的值
化简:(sinθ+sin3θ+sin5θ)/(cosθ+cos3θ+cos5θ)=
求sin²αsin²β+cos²αcos²β-1/2cos2αcos2β的值
![求sin²αsin²β+cos²αcos²β-1/2cos2αcos2β的值 化](/uploads/image/z/3200480-8-0.jpg?t=%E6%B1%82sin%26sup2%3B%CE%B1sin%26sup2%3B%CE%B2%2Bcos%26sup2%3B%CE%B1cos%26sup2%3B%CE%B2-1%2F2cos2%CE%B1cos2%CE%B2%E7%9A%84%E5%80%BC+%E5%8C%96)
1.sin²αsin²β+cos²αcos²β-1/2cos2αcos2β
=(1-cos^2a)(1-cos^2b)+cos^2a*cos^2b-1/2cos2αcos2β
=1-cos^2a-cos^2b+2cos^2a*cos^2b-1/2cos2αcos2β
=1-(1+cos2a)/2-(1+cos2b)/2+2*(1+cos2a)/2*(1+cos2b)/2-1/2cos2αcos2β
=-cos2a/2-cos2b/2+1/2+cos2a/2+cos2b/2+1/2cos2αcos2β-1/2cos2αcos2β
1/2
2:(sinθ+sin3θ+sin5θ)/(cosθ+cos3θ+cos5θ)
=(sin(3θ-2θ)+sin3θ+sin(3θ+2θ))/(cos(3θ-2θ)+cos3θ+cos(3θ+2θ))
=(2sin3θcos2θ+sin3θ)/(2cos3θcos2θ+cos3θ)
=sin3θ(2cos2θ+1)/cos3θ(2cos2+1)
=tan3θ
=(1-cos^2a)(1-cos^2b)+cos^2a*cos^2b-1/2cos2αcos2β
=1-cos^2a-cos^2b+2cos^2a*cos^2b-1/2cos2αcos2β
=1-(1+cos2a)/2-(1+cos2b)/2+2*(1+cos2a)/2*(1+cos2b)/2-1/2cos2αcos2β
=-cos2a/2-cos2b/2+1/2+cos2a/2+cos2b/2+1/2cos2αcos2β-1/2cos2αcos2β
1/2
2:(sinθ+sin3θ+sin5θ)/(cosθ+cos3θ+cos5θ)
=(sin(3θ-2θ)+sin3θ+sin(3θ+2θ))/(cos(3θ-2θ)+cos3θ+cos(3θ+2θ))
=(2sin3θcos2θ+sin3θ)/(2cos3θcos2θ+cos3θ)
=sin3θ(2cos2θ+1)/cos3θ(2cos2+1)
=tan3θ
求sin²αsin²β+cos²αcos²β-1/2cos2αcos2β的值 化
化简:sin²αsin²β+cos²αcos²β-1/2cos2αcos2β
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