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?
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