Question

高一数学………第五第七题求解求过程…………

Answer 1

5. sina+sinβ= √2cos40°
sinα *sinβ= (cos40°)^2-1/2= (1+cos80°-1)/2= cos80°/2
联立两式得 sinα^2+sinβ^2=1- sinα^2=1-sinβ^2=cosβ^2
0°< α< β<90° 所以sinα=cosβ α+ β=90° sinα*sinβ=sinα*cos α=cos^2 40°-1/2 sin2α＝2cos^2 40°-1＝cos80°＝sin10° 2α＝10°
α＝5°
β＝85°
2α- β= -75°
cos (-75°)= cos(45+30)= cos45cos30-sin45sin30=√2/2*√3/2-√2/2*1/2= (√6-√2) /4
选A

7.
1/cos2α+ tan2α= (1+sin2α)/cos2α= ((sinα)^2+ (cosα)^2+2sinαcosα))/((cos^a)^2-(sinα)^2)
= (cosα+sinα)^2/((cos^a)^2-(sinα)^2)= (cosα+sinα)/(cosα-sinα)
= (1+tana)/(1-tana)=2013

很巧妙的两题啊

追问

是呵…………

好难的说