另, 請問以下答案是如何變出來的? 請詳細解釋, THX!
1. sin(-120°) = -sin60°
2. tan(-480°) = tan60°
3. cos(570°) = -cos30°
4. sin(-300°) = sin60°
恆等式 "sin(-θ) = -sinθ" 中的θ是不是只限銳角?
2016-10-21 6:27 am
回答 (2)
2016-10-21 6:52 am
Sol
Sin(－θ)=－Sinθ θ所有角階正確
1
Sin(－120°)=Sin(360°－120°)=Sin240°<0
Sin(－120°)=－Sin(240°－180°)=－Sin60
2.
Tan(－480°)=Tan(720°－480°)=Tan240°>0
Tan(－480°)=Tan(240°－180°)=Tan60°
3. Cos(570°)=Cos(570°－360°)=Cos210°<0
Cos(570°)=－Cos(210°－180°)=－Cos30°
4. Sin(－300°)=Sin60°
Sin(－300°)=Sin(360°－300°)=Sin60°
Sin(－θ)=－Sinθ θ所有角階正確
1
Sin(－120°)=Sin(360°－120°)=Sin240°<0
Sin(－120°)=－Sin(240°－180°)=－Sin60
2.
Tan(－480°)=Tan(720°－480°)=Tan240°>0
Tan(－480°)=Tan(240°－180°)=Tan60°
3. Cos(570°)=Cos(570°－360°)=Cos210°<0
Cos(570°)=－Cos(210°－180°)=－Cos30°
4. Sin(－300°)=Sin60°
Sin(－300°)=Sin(360°－300°)=Sin60°
2016-10-22 4:11 am
TO 螞蟻雄兵:
又比唔到意見, 希望你會見到依個回覆然後覆我.
我想問, 既然"Sin(－θ)=－Sinθ θ所有角皆正確 ", 咁點解第1和第4題你唔用依條恆等式去SOLVE? 你冇解釋, 淨係得STEPS好難明
又比唔到意見, 希望你會見到依個回覆然後覆我.
我想問, 既然"Sin(－θ)=－Sinθ θ所有角皆正確 ", 咁點解第1和第4題你唔用依條恆等式去SOLVE? 你冇解釋, 淨係得STEPS好難明
收錄日期: 2021-04-20 16:33:31
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20161020222756AAuIrVX