题目内容
求sin210°+cos240°+sin10°cos40°的值.
解法1:因为40°=30°+10°,于是
原式=sin210°+cos2(30°+10°)+sin10°cos(30°+10°)=sin210°+
2+sin10°·
cos10°-
sin10°=
(sin210°+cos210°)=
.
解法2:令sin10°=a+b,cos40°=a-b,则
a=
(sin10°+cos40°)=
(sin10°+sin50°)
=sin30°cos20°=
cos20°,
b=
(sin10°-cos40°)=
(sin10°-sin50°)
=cos30°sin(-20°)=-
sin20°.
原式=(a+b)2+(a-b)2+(a+b)(a-b)
=3a2+b2
=
cos220°+
sin220°=
.
解法3:设x=sin210°+cos240°+sin10°cos40°,
y=cos210°+sin240°+cos10°sin40°.则
x+y=1+1+sin10°cos40°+cos10°sin40°=2+sin50°=2+cos40°
x-y=cos80°-cos20°-
=-sin50°-![]()
=-cos40°-
,因此,2x=
,x=
.
练习册系列答案
相关题目