三角函数问题打包150分

1.
tan100度=t, tan80度=-t, tan10度=-1/t

cos20度=[1-(tan10度)^2]/[1+(tan10度)^2]=(1-1/t^2)/(1+1/t^2)=(t^2-1)/(t^2+1)

2.
a/2∈(3π/4,π） cos（a/2）<0 sin（a/2）<0

根据cos2α=2cosα平方-1得
√(1+cosα)=√(1+2cos(α/2)平方-1)=-√2 * cos(α/2)
根据cos2α=1-2sinα平方得
√(1-cosα)=√(1-1+2sin(α/2)平方)=-√2 * sin(α/2)
故√(1+cosα)+√(1-cosα)=-√2 * cos(α/2)+(-√2 * sin(α/2))
=-√2 ( sin(α/2)+cos(α/2))

3.
sin2α=2sinαcosα;
cos2α=1-2(sinα/2)^2=2(cosα)^2-1
sinα=2sinα/2cosα/2;
cosα=1-2(sinα/2)^2=2(cosα/2)^2-1
根据以上公式得
[(sin2α/1+cos2α)]*[(cosα/(1-cosα)]
=[(2sinαcosα/(1+2(cosα)^2-1)]*[(cosα/(1-cosα)]
=[(2sinαcosα/(2(cosα)^2]*[(cosα/(1-cosα)]
=[sinα/cosα]*[(cosα/(1-cosα)]
=[sinα/(1-cosα)]
=[2sin(α/2)cos(α/2)/(1-(1-2(sinα)^2)]
=[2sin(α/2)cos(α/2)/(2(sinα/2)^2)]
=[cos(α/2)/sinα/2]
=ctan(α/2)

4.
sin(∏/4-α)=5/13
==>cos(∏/2-(∏/4-α))=5/13
==>cos(∏/4+α)=5/13