高一数学三角恒等变形
sin(x+π/3)+2sin(x-π/3)-根号3cos(2π/3-x)用三角恒等变形怎么解？...
sin(x+π/3)+2sin(x-π/3)-根号3cos(2π/3-x) 用三角恒等变形怎么解？
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原式
=sin(x+π/3)+根号3cos(π/3+x)+2sin(x-π/3)
=2sin(x+2π/3)+2sin(x-π/3)
=2sin(π-(x+2π/3))+2sin(x-π/3)
=2sin(π/3-x)+2sin(x-π/3)
=0
=sin(x+π/3)+根号3cos(π/3+x)+2sin(x-π/3)
=2sin(x+2π/3)+2sin(x-π/3)
=2sin(π-(x+2π/3))+2sin(x-π/3)
=2sin(π/3-x)+2sin(x-π/3)
=0
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