设a=cos50cos127+cos40cos37,b=根号2/2(sin56-cos56),设a=cos50cos127+cos40cos37,b=根号2/2(sin56-cos56),c=1-tan^2*39/1+tan^2*39,d=1/2(cos80-2cos^50+1),则abcd的大小为?
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![设a=cos50cos127+cos40cos37,b=根号2/2(sin56-cos56),设a=cos50cos127+cos40cos37,b=根号2/2(sin56-cos56),c=1-tan^2*39/1+tan^2*39,d=1/2(cos80-2cos^50+1),则abcd的大小为?](/uploads/image/z/14898700-28-0.jpg?t=%E8%AE%BEa%3Dcos50cos127%2Bcos40cos37%2Cb%3D%E6%A0%B9%E5%8F%B72%2F2%EF%BC%88sin56-cos56%EF%BC%89%2C%E8%AE%BEa%3Dcos50cos127%2Bcos40cos37%2Cb%3D%E6%A0%B9%E5%8F%B72%2F2%EF%BC%88sin56-cos56%EF%BC%89%2Cc%3D1-tan%5E2%2A39%2F1%2Btan%5E2%2A39%2Cd%3D1%2F2%28cos80-2cos%5E50%2B1%29%2C%E5%88%99abcd%E7%9A%84%E5%A4%A7%E5%B0%8F%E4%B8%BA%3F)
设a=cos50cos127+cos40cos37,b=根号2/2(sin56-cos56),设a=cos50cos127+cos40cos37,b=根号2/2(sin56-cos56),c=1-tan^2*39/1+tan^2*39,d=1/2(cos80-2cos^50+1),则abcd的大小为?
设a=cos50cos127+cos40cos37,b=根号2/2(sin56-cos56),
设a=cos50cos127+cos40cos37,b=根号2/2(sin56-cos56),c=1-tan^2*39/1+tan^2*39,d=1/2(cos80-2cos^50+1),则abcd的大小为?
设a=cos50cos127+cos40cos37,b=根号2/2(sin56-cos56),设a=cos50cos127+cos40cos37,b=根号2/2(sin56-cos56),c=1-tan^2*39/1+tan^2*39,d=1/2(cos80-2cos^50+1),则abcd的大小为?
a=cos50cos127+cos40cos37=cos40cos37-sin40sin37°=cos77;
b=√2/2(sin56-cos56)=sin45sin56-cos45cos56=-cos101=cos(180-101)=cos79;
c=1-tan^2*39/1+tan^2*39=(cos²39-sin²39)/(cos²39+sin²39)=cos(39*2)=cos78;
d=1/2(cos80-2cos^2 50+1)=1/2(cos80-cos100)=1/2(cos80+cos80)=cos80;
所以 a>c>b>d.
啧啧
看着就麻烦
用90度神马的倒吧
靠,3角函数关系式,10 年前还记得,现在记不得了
c的完整过程