Question

√3csc20度-sec20度

√3csc20度-sec20度=
(1-tan7度-tan8度-tan7度tan8度)/(1+tan7度+tan8度-tan7度tan8度）=
sina+sanb=√2/2，则cosa+cosb的取值范围是
三题最好都有过程

Answer 1

√3csc20°-sec20°
=√3/sin20°-1/cos20°
=(√3cos20°-sin20°)/sin20°cos20°
=2(√3cos20°/2-sin20°/2)/(sin40°)/2
=4(sin60°cos20°-sin20°cos60°)/sin40°
=4sin40°/sin40°
=4

(1-tan7°-tan8°-tan7°tan8°)/(1+tan7°+tan8°-tan7°tan8°)
=[1-(tan7°+tan8°)/(1-tan7°tan8°)]/[1+(tan7°+tan8°)/(1-tan7°tan8°)]
=(1-tan15°)/(1+tan15°)
=(1-sin15°/cos15°)/(1+sin15°/cos15°)
=(cos15°-sin15°)/(cos15°+sin15°)
=(cos15°-sin15°)²/(cos15°+sin15°)(cos15°-sin15°)
=(sin²15°-2sin15°cos15°+sin²15°)/(cos²15°-sin²15°)
=(1-sin30°)/cos30°
=(1-1/2)/(√3/2)
=√3/3

sinα+sinβ=√2/2
(sinα+sinβ)²=1/2
sin²α+2sinαsinβ+sin²β=1/2

设cosα+cosβ=m
(cosα+cosβ)²=m²
cos²α+2cosαcosβ+cos²β=m²

两式相加得
sin²α+2sinαsinβ+sin²β+cos²α+2cosαcosβ+cos²β=1/2+m²
2+2cos(α-β)=1/2+m²
m²=3/2+2cos(α-β)
m²≤7/2
-√14/2≤m≤√14/2

cosα+cosβ的取值范围是[-√14/2，√14/2]