已知α是第二象限角,化简:√1+2sinα(5π-α)*(α-π)/sin(α-2分之3π)-√1-sin²(2分之3派+α):√1+2sinα(5π-α)*cos(α-π)—————————————————— = sin(α-2分之3π)-√1-sin²(2分之3派
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/05 23:36:36
![已知α是第二象限角,化简:√1+2sinα(5π-α)*(α-π)/sin(α-2分之3π)-√1-sin²(2分之3派+α):√1+2sinα(5π-α)*cos(α-π)—————————————————— = sin(α-2分之3π)-√1-sin²(2分之3派](/uploads/image/z/10142314-34-4.jpg?t=%E5%B7%B2%E7%9F%A5%CE%B1%E6%98%AF%E7%AC%AC%E4%BA%8C%E8%B1%A1%E9%99%90%E8%A7%92%2C%E5%8C%96%E7%AE%80%3A%E2%88%9A1%2B2sin%CE%B1%285%CF%80-%CE%B1%29%2A%28%CE%B1-%CF%80%29%2Fsin%28%CE%B1-2%E5%88%86%E4%B9%8B3%CF%80%29-%E2%88%9A1-sin%26sup2%3B%EF%BC%882%E5%88%86%E4%B9%8B3%E6%B4%BE%2B%CE%B1%EF%BC%89%3A%E2%88%9A1%2B2sin%CE%B1%285%CF%80-%CE%B1%29%2Acos%28%CE%B1-%CF%80%29%E2%80%94%E2%80%94%E2%80%94%E2%80%94%E2%80%94%E2%80%94%E2%80%94%E2%80%94%E2%80%94%E2%80%94%E2%80%94%E2%80%94%E2%80%94%E2%80%94%E2%80%94%E2%80%94%E2%80%94%E2%80%94+%3D+sin%28%CE%B1-2%E5%88%86%E4%B9%8B3%CF%80%29-%E2%88%9A1-sin%26%23178%3B%EF%BC%882%E5%88%86%E4%B9%8B3%E6%B4%BE)
已知α是第二象限角,化简:√1+2sinα(5π-α)*(α-π)/sin(α-2分之3π)-√1-sin²(2分之3派+α):√1+2sinα(5π-α)*cos(α-π)—————————————————— = sin(α-2分之3π)-√1-sin²(2分之3派
已知α是第二象限角,化简:√1+2sinα(5π-α)*(α-π)/sin(α-2分之3π)-√1-sin²(2分之3派+α)
:√1+2sinα(5π-α)*cos(α-π)
—————————————————— =
sin(α-2分之3π)-√1-sin²(2分之3派+α)
已知α是第二象限角,化简:√1+2sinα(5π-α)*(α-π)/sin(α-2分之3π)-√1-sin²(2分之3派+α):√1+2sinα(5π-α)*cos(α-π)—————————————————— = sin(α-2分之3π)-√1-sin²(2分之3派
分子√1+2sinα(5π-α)*cos(α-π)是不是有问题?
√1+2sin(5π-α)*cos(α-π)吧
显然sin(5π-α)=sin(π-α)=sinα,
而cos(α-π)= -cosα,
故√1+2sin(5π-α)*cos(α-π)=√(sin²α+cos²α-2sinα*cosα)=√(sinα-cosα)²
α是第二象限角,故sinα>cosα,
所以分子=√1+2sin(5π-α)*cos(α-π)=sinα-cosα
而sin(α-3π/2)=sin(α+π/2)=cosα,
1-sin²(3π/2+α)=cos²(3π/2+α)=sin²α
α是第二象限角即√[1-sin²(3π/2+α)] = sinα
所以
原式=[√1+2sin(5π-α)*cos(α-π)] / [sin(α-3π/2) - √[1-sin²(3π/2+α)] ]
=(sinα-cosα) / (cosα-sinα)
= -1