y=sin(x-π/3)cosx=[sin(2x-π/3)+sin(-π/3)]/2为什么y=sin(x-π/3)cosx,就说明y=[sin(2x-π/3)+sin(-π/3)]/2,望会此题者给小妹说一下,
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y=sin(x-π/3)cosx=[sin(2x-π/3)+sin(-π/3)]/2为什么y=sin(x-π/3)cosx,就说明y=[sin(2x-π/3)+sin(-π/3)]/2,望会此题者给小妹说一下,
y=sin(x-π/3)cosx=[sin(2x-π/3)+sin(-π/3)]/2
为什么y=sin(x-π/3)cosx,就说明y=[sin(2x-π/3)+sin(-π/3)]/2,望会此题者给小妹说一下,
y=sin(x-π/3)cosx=[sin(2x-π/3)+sin(-π/3)]/2为什么y=sin(x-π/3)cosx,就说明y=[sin(2x-π/3)+sin(-π/3)]/2,望会此题者给小妹说一下,
这是积化和差公式:
sinαcosβ=[sin(α+β)+sin(α-β)]/2
还有
sinαsinβ=-[cos(α+β)-cos(α-β)]/2
cosαcosβ=[cos(α+β)+cos(α-β)]/2
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