cos(A+B)cosB+sin(A+B)sinB=1/3,且A∈(3π/2,2π),求cos(2A+(π/4))
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cos(A+B)cosB+sin(A+B)sinB=1/3,且A∈(3π/2,2π),求cos(2A+(π/4))
cos(A+B)cosB+sin(A+B)sinB=1/3,且A∈(3π/2,2π),求cos(2A+(π/4))
cos(A+B)cosB+sin(A+B)sinB=1/3,且A∈(3π/2,2π),求cos(2A+(π/4))
cos(A+B)cosB+sin(A+B)sinB
=cos[(A+B)-B]
=cosA=1/3
A∈(3π/2,2π),
第四象限
所以sinA
cos(a-b)cosb-sin(a-b)sinb等于什么
化简cos(a+b)cosb+sin(a+b)sinb
cos(a-b)cosb-sin(a-b)sinb
化简:cos(A-B)cosB-sin(A-B)sinB
cosb=cos[(a+b)-a]=cos(a+b)cosa+sin(a+b)sina
化简cos(a-b)cosb-sin(a-b)cosb的 结果是
sin²a+sin²b-sina²sin²b+cos²cosb=1
cos a+cosb=-1/3 sin a+sin b=1/2 求cos(a+b)
为什么sinb=sin(a+b-b)=sin(a+b)cosb+cos(a+b)sinb,这是公式吗?
[sin(2a+b)cosb+cos(2a+b)sinb]/[cos(2a+b)cosb]=[sin(2a+2b)]/[cos(2a+b)cosb]是怎么来的,看不懂
sin(a-b)cosb+cos(b-a)sinb=cos(b-a)怎么变形,为什么?
化简sin(a-b)sina+cos(a-b)cosa答案为COSB,
sin(a-b)cosb+cos(a-b)sinb=?
sin(a+b)cos(a-b)=sina*cosa+sinb*cosb
求证:sin(a+b)cos(a-b)=sina*cosa+sinb*cosb麻烦个位 帮个忙~
sinA(cos(2A+B)+cosB)=cosA(sin(2A+B)-sinB)证明
证明下列恒等式:sin(a+b)*cos(a-b)=sina*cosa+sinb*cosb
cos(A+B)+sin(A-B)=(cosA+sinA)(cosB-sinB)