求[3/(sin140)^2 -1/(cos140)^2]*[1/(
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求[3/(sin140)^2 -1/(cos140)^2]*[1/(
但老师说是-16
但老师说是-16
![求[3/(sin140)^2 -1/(cos140)^2]*[1/(](/uploads/image/z/20196040-40-0.jpg?t=%E6%B1%82%5B3%2F%28sin140%29%5E2+-1%2F%28cos140%29%5E2%5D%2A%5B1%2F%28)
[3/(sin140)^2 -1/(cos140)^2]*[1/(2sin10)]
原式=(3cos²40-sin²40)/[2sin²40cos²40sin10]
=2(√3cos-sin40)(√3cos40+sin40)/[sin²80sin10]
=8sin(60-40)sin(60+40)/(cos²10sin10)
=16sin20sin100/(cos10sin20)
=16cos10/cos10
=16
原式=(3cos²40-sin²40)/[2sin²40cos²40sin10]
=2(√3cos-sin40)(√3cos40+sin40)/[sin²80sin10]
=8sin(60-40)sin(60+40)/(cos²10sin10)
=16sin20sin100/(cos10sin20)
=16cos10/cos10
=16