1/2[cos(2x)cos(π/6)+sin(2x)sin(π/6)]-1/2[cos(2x)cos(π/6)-sin
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1/2[cos(2x)cos(π/6)+sin(2x)sin(π/6)]-1/2[cos(2x)cos(π/6)-sin(2x)sin(π/6)]=1/2sin(2x)
请问怎么化得1/2sin(2x)?
请问怎么化得1/2sin(2x)?
前後两式提出1/2相减後结果为2sin(2x)sin(π/3)
sin(π/3)=1/2
所以结果为1/2sin(2x)
再问: cos(2x)cos(π/6)消掉,
sin(2x)sin(π/6)+sin(2x)sin(π/6)=2sin(2x)sin(π/6)吗?那sin(π/3)哪来的?
再答: 那是不小心敲错....是sin(π/6)
sin(π/3)=1/2
所以结果为1/2sin(2x)
再问: cos(2x)cos(π/6)消掉,
sin(2x)sin(π/6)+sin(2x)sin(π/6)=2sin(2x)sin(π/6)吗?那sin(π/3)哪来的?
再答: 那是不小心敲错....是sin(π/6)
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