已知函数f(x)=sinx/2cosx/2+cos^2(x/2)-1/2f(x)=sin(x/2)cos(x/2)+cos^2(x/2)-(1/2)=(1/2)×sinx+(1/2)×(cosx+1)-(1/2)=(1/2)×(sinx+cosx)=(√2/2)×sin(x+(π/4))(1)若f(a)=√2/4=(√2/2)×sin(a+(π/4))则sin(a+(π/4))=1/2,a+(π/4)=2kπ+(π/2)±((2π)/
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/24 05:16:21
![已知函数f(x)=sinx/2cosx/2+cos^2(x/2)-1/2f(x)=sin(x/2)cos(x/2)+cos^2(x/2)-(1/2)=(1/2)×sinx+(1/2)×(cosx+1)-(1/2)=(1/2)×(sinx+cosx)=(√2/2)×sin(x+(π/4))(1)若f(a)=√2/4=(√2/2)×sin(a+(π/4))则sin(a+(π/4))=1/2,a+(π/4)=2kπ+(π/2)±((2π)/](/uploads/image/z/2524479-15-9.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3Dsinx%2F2cosx%2F2%2Bcos%5E2%28x%2F2%29-1%2F2f%28x%29%3Dsin%28x%2F2%29cos%28x%2F2%29%2Bcos%5E2%28x%2F2%29-%281%2F2%29%3D%281%2F2%29%C3%97sinx%2B%281%2F2%29%C3%97%28cosx%2B1%29-%281%2F2%29%3D%281%2F2%29%C3%97%28sinx%2Bcosx%29%3D%28%E2%88%9A2%2F2%29%C3%97sin%28x%2B%28%CF%80%2F4%29%29%281%29%E8%8B%A5f%28a%29%3D%E2%88%9A2%2F4%3D%28%E2%88%9A2%2F2%29%C3%97sin%28a%2B%28%CF%80%2F4%29%29%E5%88%99sin%28a%2B%28%CF%80%2F4%29%29%3D1%2F2%2Ca%2B%28%CF%80%2F4%29%3D2k%CF%80%2B%28%CF%80%2F2%29%C2%B1%28%282%CF%80%29%2F)
已知函数f(x)=sinx/2cosx/2+cos^2(x/2)-1/2f(x)=sin(x/2)cos(x/2)+cos^2(x/2)-(1/2)=(1/2)×sinx+(1/2)×(cosx+1)-(1/2)=(1/2)×(sinx+cosx)=(√2/2)×sin(x+(π/4))(1)若f(a)=√2/4=(√2/2)×sin(a+(π/4))则sin(a+(π/4))=1/2,a+(π/4)=2kπ+(π/2)±((2π)/
已知函数f(x)=sinx/2cosx/2+cos^2(x/2)-1/2
f(x)=sin(x/2)cos(x/2)+cos^2(x/2)-(1/2)
=(1/2)×sinx+(1/2)×(cosx+1)-(1/2)
=(1/2)×(sinx+cosx)
=(√2/2)×sin(x+(π/4))
(1)若f(a)=√2/4=(√2/2)×sin(a+(π/4))
则sin(a+(π/4))=1/2,a+(π/4)=2kπ+(π/2)±((2π)/3);
又a∈(0,π),则a=π/12或(5π)/12;
为什么(1/2)×(sinx+cosx)
=(√2/2)×sin(x+(π/4))
为什么还要乘以(1/2)*√2*(1/2)啊
已知函数f(x)=sinx/2cosx/2+cos^2(x/2)-1/2f(x)=sin(x/2)cos(x/2)+cos^2(x/2)-(1/2)=(1/2)×sinx+(1/2)×(cosx+1)-(1/2)=(1/2)×(sinx+cosx)=(√2/2)×sin(x+(π/4))(1)若f(a)=√2/4=(√2/2)×sin(a+(π/4))则sin(a+(π/4))=1/2,a+(π/4)=2kπ+(π/2)±((2π)/
这个简单
(1/2)×(sinx+cosx)
=(1/2)*√2*(1/√2)(sinx+cosx)
=(√2/2)(√2/2*sinx+√2/2*cosx)
而sin(π/4)=cos(π/4)=√2/2
所以就变成了
上式=(√2/2)(cos(π/4)*sinx+sin(π/4)*cosx)
根据和角公式
Sin(A+B)=SinA*CosB+SinB*CosA
这里A=x,B=π/4
于是有结果
上式=(√2/2)×sin(x+(π/4))
因为我提了个根号2出来,但是笔误少加一个根号,应该没那么难看出来吧Orz.已经修正了
sinx+cosx=√2*[sinx(√2/2)+cosx(√2/2)]=√2*[sinxcos(π/4)+cosxsin(π/4)]=√2*sin(x+(π/4)