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已知sin=2√5/5,求tan(a+兀)+sin(5兀/2+a)/cos(5兀/2-a)的值
题目详情
已知sin=2√5/5,求tan(a+兀)+sin(5兀/2+a)/cos(5兀/2-a)的值
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答案和解析
tan(a+π)+sin(5π/2+a)/cos(5π/2-a)
=tana+sin(2π+π/2+a)/cos(2π+π/2-a)
=tana+sin(π/2+a)/cos(π/2-a)
=sina/cosa+cosa/sina
=[(sina)^2+(cosa)^2]/sinacosa
=1/sinacosa
=1/{sina{±√[1-(sina)^2]}}
=1/{(2√5/5){±√[1-(2√5/5)^2]}}
=1/{(2√5/5)[±√(1/5)]}
=1/[(2√5/5)(±√5/5)]
=1/(±2/5)
=±5/2
=tana+sin(2π+π/2+a)/cos(2π+π/2-a)
=tana+sin(π/2+a)/cos(π/2-a)
=sina/cosa+cosa/sina
=[(sina)^2+(cosa)^2]/sinacosa
=1/sinacosa
=1/{sina{±√[1-(sina)^2]}}
=1/{(2√5/5){±√[1-(2√5/5)^2]}}
=1/{(2√5/5)[±√(1/5)]}
=1/[(2√5/5)(±√5/5)]
=1/(±2/5)
=±5/2
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