sin(a-B)cosa-1/2[sin(2a+B)-sinB]=?已知cos(a-π/6)+sina=4√3/5则sin( a+π/6)的值?
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![sin(a-B)cosa-1/2[sin(2a+B)-sinB]=?已知cos(a-π/6)+sina=4√3/5则sin( a+π/6)的值?](/uploads/image/z/1116724-4-4.jpg?t=sin%28a-B%29cosa-1%2F2%5Bsin%282a%2BB%29-sinB%5D%3D%3F%E5%B7%B2%E7%9F%A5cos%28a-%CF%80%2F6%29%2Bsina%3D4%E2%88%9A3%2F5%E5%88%99sin%EF%BC%88+a%2B%CF%80%2F6%29%E7%9A%84%E5%80%BC%3F)
sin(a-B)cosa-1/2[sin(2a+B)-sinB]=?已知cos(a-π/6)+sina=4√3/5则sin( a+π/6)的值?
sin(a-B)cosa-1/2[sin(2a+B)-sinB]=?已知cos(a-π/6)+sina=4√3/5则sin( a+π/6)的值?
sin(a-B)cosa-1/2[sin(2a+B)-sinB]=?已知cos(a-π/6)+sina=4√3/5则sin( a+π/6)的值?
sin(a-B)cosa-1/2[sin(2a+B)-sinB]=sin(a-B)cosa-1/2[2cos(a+b)sina]=sin(a-b)cosa-cos(a+b)sina=sinacosbcosa-cosasinbcosa-cosacosbsina+sinasinbsina=sinb(sinasina-sinasina)=-sinbcos2a
已知cos(a-π/6)+sina=4√3/5展开得
则cosacos(π/6)+sinasin(π/6)+sina=4√3/5
即(√3/2)cosa+(3/2)sina=4√3/5
两边同除以√3得
(√3/2)sina+(1/2)cosa=4/5,由于cos(π/6)=
(√3/2),sin(π/6)=1/2.最后再合并得sin(a+π/6)=4/5
sin(a-B)cosa-1/2[sin(2a+B)-sinB]
=(1/2)[sin(2a-B)+sin(-B)]-1/2[sin(2a+B)-sinB]
=(1/2)[sin(2a-B)-sin(2a+B)]
=(1/2)*2cos(2a)sin(-B)
=-sinBcos(2a)
已知cos(a-π/6)+sina=4√3/5
则cosaco...
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sin(a-B)cosa-1/2[sin(2a+B)-sinB]
=(1/2)[sin(2a-B)+sin(-B)]-1/2[sin(2a+B)-sinB]
=(1/2)[sin(2a-B)-sin(2a+B)]
=(1/2)*2cos(2a)sin(-B)
=-sinBcos(2a)
已知cos(a-π/6)+sina=4√3/5
则cosacos(π/6)+sinasin(π/6)+sina=4√3/5
(√3/2)cosa+(3/2)sina=4√3/5
两边同乘以√3
(√3/2)sina+(1/2)cosa=4/5
所以sin(a+π/6)=4/5
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