高一数学三角函数
化简根号
解:
(1 + 2sin290cos430)/(sin250 + cos790)
= [1 + 2sin(360 - 70)cos(360 + 70)]/[sin(180 + 70) + cos(720 + 70)]
= √[1 + 2(-sin70)cos70]/[-sin70 + cos70]
= √[1 - 2sin70cos70]/[cos70 - sin70]
= √[(sin70)^2 + (cos70)^2 - 2sin70cos70]/[cos70 - sin70]
= |sin70 - cos70|/[cos70 - sin70]
= (sin70 - cos7...全部
化简根号
解:
(1 + 2sin290cos430)/(sin250 + cos790)
= [1 + 2sin(360 - 70)cos(360 + 70)]/[sin(180 + 70) + cos(720 + 70)]
= √[1 + 2(-sin70)cos70]/[-sin70 + cos70]
= √[1 - 2sin70cos70]/[cos70 - sin70]
= √[(sin70)^2 + (cos70)^2 - 2sin70cos70]/[cos70 - sin70]
= |sin70 - cos70|/[cos70 - sin70]
= (sin70 - cos70)/[cos70 - sin70]
= -1
* 因为sin70＞cos70，所以|sin70 - cps70| = sin70 - cos70
。
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