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1/sin10° -√3/cos10°=
人气:332 ℃ 时间:2020-03-27 12:21:52
解答
1/sin(10°) - √3/cos(10°)
= [cos(10°) - √3sin(10°)]/[sin(10°)cos(10°)]
= 2[(1/2)cos(10°) - (√3/2)sin(10°)]/[(1/2)sin(2 * 10°)]
= 2[cos(60°)cos(10°) - sin(60°)sin(10°)]/[(1/2)sin(20°)]
= 4cos(60° + 10°)/sin(20°)
= 4cos(70°)/sin(20°) = 4sin(20°)/sin(20°)
= 4
公式:
sin(2A) = 2sinAcosA
cosAcosB - sinAsinB = cos(A + B)
sin(90° - A) = cosA cos(90° - A) = sinA
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