神马品牌网韩国隆胸豆瓣:一道简单的高一数学题，谢谢
来源：百度文库 编辑：神马品牌网 时间：2024/04/29 03:07:17
A、B为钝角sinA=5^(1/2)/5,sinB=10^(1/2)/10,求A+B
sinA=√5/5
(sinA)^2=1/5
(cosA)^2=4/5
cosA=-2√5/5
sinB=√10/10
(sinB)^2=1/10
(cosB)^2=9/10
cosB=-3√10/10
cos(A+B)
=cosA*cosB-sinA*sinB
=(-2√5/5)*(-3√10/10)-(√10/10)*(√5/5)
=√2/2
A+B=315度
sin(a+b)=sinA*cosB+sinB*cosA
=5^(1/2)/5*3*10^(1/2)/10+10^(1/2)/10*2*5^(1/2)/5
=5*2^(1/2)/10
A+B=arcsin5*2^(1/2)/10
有点难度！我不太清楚！不好意思！
∵sinA=5^(1/2)/5 sinB=10^(1/2)/10
∴cosA=-2/5^(1/2) cosB=-3/5^(1/2)
cos(A+B)=cosA*cosB-sinA*sinB
=[-2/5^(1/2)]*[-3/5^(1/2)]-[10^(1/2)/10]*[5^(1/2)/5]
=1/2^(1/2)
A+B=315度