如何得出:2+sin2x+cos2x=根号2sinsin(2x+π/4)+2
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如何得出:2+sin2x+cos2x=根号2sinsin(2x+π/4)+2
还有如何得到:sin(x+π/6)+sin(x-π/6)+cosx=2sinxcosπ/6+cosx.....
还有如何得到:sin(x+π/6)+sin(x-π/6)+cosx=2sinxcosπ/6+cosx.....
1.你最后结果写错了.2+sin2x+cos2x=2+根号2(根号2/2*sin2x+根号2/2*cos2x)
=2+根号2(cosπ/4*sin2x+sinπ/4*cos2x)
=2+根号2*sin(2x+π/4)
2.用和差化积公式(在网上可以搜到)
sin(x+π/6)+sin(x-π/6)+cosx
=2sin【(x+π/6)/2+(x-π/6)/2】cos【(x+π/6)/2-(x-π/6)/2】+cosx
=2sinxcosπ/6+cosx
=2+根号2(cosπ/4*sin2x+sinπ/4*cos2x)
=2+根号2*sin(2x+π/4)
2.用和差化积公式(在网上可以搜到)
sin(x+π/6)+sin(x-π/6)+cosx
=2sin【(x+π/6)/2+(x-π/6)/2】cos【(x+π/6)/2-(x-π/6)/2】+cosx
=2sinxcosπ/6+cosx
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