作业帮已知sinA+sin3A+sin5A=a,cosA+cos3A+cos5A=b.求(1+2cos2A)^2=a^2+b^2
来源：学生作业帮助网 编辑：作业帮 时间：2024/05/09 03:56:47
已知sinA+sin3A+sin5A=a,cosA+cos3A+cos5A=b.求(1+2cos2A)^2=a^2+b^2
已知sinA+sin3A+sin5A=a,cosA+cos3A+cos5A=b.求(1+2cos2A)^2=a^2+b^2
已知sinA+sin3A+sin5A=a,cosA+cos3A+cos5A=b.求(1+2cos2A)^2=a^2+b^2
.两角和与差的三角函数
sin(a+b)=sin(a)cos(b)+cos(α)sin(b)
cos(a+b)=cos(a)cos(b)-sin(a)sin(b)
sin(a-b)=sin(a)cos(b)-cos(a)sin(b)
cos(a-b)=cos(a)cos(b)+sin(a)sin(b)
tan(a+b)=tan(a)+tan(b)1-tan(a)tan(b)
tan(a-b)=tan(a)-tan(b)1+tan(a)tan(b)
3.和差化积公式 sin(a)+sin(b)=2sin(a+b2)cos(a-b2) sin(a)−sin(b)=2cos(a+b2)sin(a-b2) cos(a)+cos(b)=2cos(a+b2)cos(a-b2) cos(a)-cos(b)=-2sin(a+b2)sin(a-b2)
4.积化和差公式 (上面公式反过来就得到了) sin(a)sin(b)=-12⋅[cos(a+b)-cos(a-b)] cos(a)cos(b)=12⋅[cos(a+b)+cos(a-b)] sin(a)cos(b)=12⋅[sin(a+b)+sin(a-b)]
只是公式,比较麻烦…………………………………………
sinA+sin5A+sin3A = a => 2 sin3Acos2A + sin3A = a
cosA+cos5A+cos3A=b => 2 cos3Acos2A + cos3A = b
即 sin3A (2 cos2A + 1) = a, cos3A (2 cos2A + 1) = b,
∴ a² + b² = (2 cos2A + 1)²