Question

化简√(1-sin20°cos20°)/√(1-cos^2 20°-cos20°)

修改一下 1.化简（√1-sin20°cos20°）/[（√1-cos^2 20°）-cos20°]
2.若(2-5cos^2α)/（cos^2α-sin^2α）=-4，则tan=
3.若(4sinα-2cosα)/(5cosα+3sinα) =6/11，则sin^4 α-cos^4 α=

Answer 1

1.化简[√(1-2sin20°cos20°)]/[√(1-cos² 20°)-cos20°]【原题的分子上可能少写了个2倍】

解：原式=[√(cos20°-sin20°)²]/(sin20°-cos20°)=(cos20°-sin20°)/(cos20°-sin20°)=1
2.若(2-5cos²α)/(cos²α-sin²α)=-4，则tan2α=?

解：由(2-5cos²α)/(cos²α-sin²α)=-4，得：
[2-5(1+cos2α)/2]/cos2α=(-1-5cos2α)/(2cos2α)=-4
故得-1-5cos2α=-8cos2α，3cos2α=1，cos2α=1/3；sin2α=(2/3)√2；则tan2α=2√2；
tanα=(1-cos2α)/sin2α=(1-1/3)/[(2/3)√2]=1/√2=√2/2
3.若(4sinα-2cosα)/(5cosα+3sinα) =6/11，则sin⁴α-cos⁴ α=?

解·：由(4sinα-2cosα)/(5cosα+3sinα) =6/11，得
44sinα-22cosα=30cosα+18sinα；26sinα=52cosα，tanα=52/26=2
故sin⁴α-cos⁴ α=(sin²α+cos²α)(sin²α-cos²α)=-(cos²α-sin²α)=(sin²α-cos²α)/(sin²α+cos²α)
=(tan²α-1)/(tan²α+1)=(4-1)/(4+1)=3/5