题目

1.tan（π/8）-cot（π/8）=________
2.f（sinX）=3-cos2X,则f（cosX）=
A 3-cos2X
B 3+cos2X
C 3-sin2X
D 3+sin2X
3.化简（分子）1+sin4α+cos4α （分母）1+sin4α-cos4α
4.已知1/cosα - 1/sinα =1,求sin2α

提问时间：2020-11-09

答案

tanπ/8-cotπ/8
=(sinπ/8)/(cosπ/8)-(cosπ/8)/(sinπ/8)
=[(sinπ/8)^2-(cosπ/8)^2]/(sinπ/8*cosπ/8)
=(-cosπ/4)/[(1/2)*sinπ/4]
=-2
∵f(sinx)=3-cos2x=3-(1-2sinx*sinx)
∴f(x)=3-(1-2x*x)
=2+2x*x
或者：设sinx=t
cos2x=1-2(sinx)^2=1-2t^2
f(sinx)=3-cos2x
f(t)=3-(1-2t^2)=2t^2+2
把t改成x
f(x)=2x^2+2
f(cosx)=2cos^2 x+2=1+cos2x+2=3+cos2x
选B
(1+sin4α+cos4α)÷(1+sin4-cos4α)
=[2cos^2(2α)+2sin2αcos2α]/[2sin^2(2α)+2sin2αcos2α]
=2cos2α(sin2α+cos2α)/2sin2α(sin2α+cos2α)
=cot2α
=(1-tan^2 α)/2tanα
1/cosα - 1/sinα=1
(sinα-cosα)/sinαcosα=1
sinα-cosα=sinαcosα
(sinα)^2+(cosα)^2=1
(sinα-cosα)^2-2sinαcosα=1
即（sinαcosα)^2-2sinαcosα=1
得sinαcosα=1±√2
sin2α=2sinαcosα=2±2√2