设P(-3t,-4t)是角α终边上不同与原点O的一点,求sin α,cos α,tan α三个三角函数值
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![设P(-3t,-4t)是角α终边上不同与原点O的一点,求sin α,cos α,tan α三个三角函数值](/uploads/image/z/5194839-39-9.jpg?t=%E8%AE%BEP%EF%BC%88-3t%2C-4t%EF%BC%89%E6%98%AF%E8%A7%92%CE%B1%E7%BB%88%E8%BE%B9%E4%B8%8A%E4%B8%8D%E5%90%8C%E4%B8%8E%E5%8E%9F%E7%82%B9O%E7%9A%84%E4%B8%80%E7%82%B9%2C%E6%B1%82sin+%CE%B1%2Ccos+%CE%B1%2Ctan+%CE%B1%E4%B8%89%E4%B8%AA%E4%B8%89%E8%A7%92%E5%87%BD%E6%95%B0%E5%80%BC)
设P(-3t,-4t)是角α终边上不同与原点O的一点,求sin α,cos α,tan α三个三角函数值
设P(-3t,-4t)是角α终边上不同与原点O的一点,求sin α,cos α,tan α三个三角函数值
设P(-3t,-4t)是角α终边上不同与原点O的一点,求sin α,cos α,tan α三个三角函数值
P(-3t,-4t)是角α终边上不同与原点O的一点,令t>0
OP=√((-3t)^2+(-4t)^2) = 5t
sinα=-4t/(5t)=-4/5
cosα=-3t/(5t)=-3/5
tanα=-4t/(-3t)=4/3
以下sqrt是开方,abs是取绝对值:
sin(alpha) = -4t / sqrt (9t^2 + 16t^2) = -4t / (5 abs (t)) ;
cos(alpha) = -3t / sqrt (9t^2 + 16t^2) = -3t / (5 abs (t)) ;
tan(alpha) = -4t / (-3t) = 4/3;
所以,当t > 0时...
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以下sqrt是开方,abs是取绝对值:
sin(alpha) = -4t / sqrt (9t^2 + 16t^2) = -4t / (5 abs (t)) ;
cos(alpha) = -3t / sqrt (9t^2 + 16t^2) = -3t / (5 abs (t)) ;
tan(alpha) = -4t / (-3t) = 4/3;
所以,当t > 0时,t = abs(t),
sin(alpha) = -4/5,cos(alpha) = -3/4,tan(alpha) = 4/3;
当t < 0时,t = -abs(t),
sin(alpha) = 4/5,cos(alpha) = 3/4,tan(alpha) = 4/3。
一个点P(x,y)的正弦就是 y / sqrt (x^2 + y^2),这是通过作过P点的x轴的垂线(垂足为Q),得到直角三角形OPQ得到的。
楼上的答案不尽完整,得讨论两段才行。如果要表达为一个统一的式子,答案应该是:
sin(alpha) = -4t / sqrt (9t^2 + 16t^2) = -4t / (5 abs (t)) ;
cos(alpha) = -3t / sqrt (9t^2 + 16t^2) = -3t / (5 abs (t)) ;
tan(alpha) = -4t / (-3t) = 4/3.
楼主居然把错误答案看成是满意的,严重怀疑其数学水平。
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sin α=4/5或-4/5。cos α=3/5或-3/5。tan α=4/3