sin(a-b)cosa-cos(b-a)sina=12/13,求cos(5π/4-b)
和cos(π/3+2b)
人气:141 ℃ 时间:2020-04-13 02:38:58
解答
sin(a-b)cosa-cos(a-b)sina=12/13
sin(a-b-a)=-sinb=12/13
sinb=-12/13
cos2b=1-2sin^2b=(169-288)/169=-119/169
cosb=±5/13
sin2b=2sinbcosb=120/169
cos(5π/4-b)
=cos5π/4cosb+sin5π/4sinb
=-√2/2cosb-√2/2sinb
.
cos(π/3+2b)=cosπ/3cos2b-sin(π/3)sin2b
,.
推荐
- 90°<b<a<270°,cos(a-b)=12/13,sin(a+b=-3/5),sina+cosa=?
- 若cos(a+B)*cosa+sin(a+B)*sina=-4/5,则sin(π/2-B)
- sin(π/2+a)=cosa,cos(π/2+a)=-sina,sin(π/2-a)=cosa,cos(π/2-a)=sina,这里的
- sin(a+b)cosa-cos(a+b)sina=5/13 则sin(π-B)=?
- 比较大小:cosa,cos(sina),sin(cosa)
- the above paper chair is the 连词成句
- 英语翻译
- 标准状况下,2mol CO2占有体积是多少,质量是多少,质子数是多少个?
猜你喜欢