f(x)=4cosx/2cos(x/2+π/6) =2[cos(x+π/6)+cosπ/6] =2cos(x+π/3)+√3是怎么化简的?

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f(x)=4cosx/2cos(x/2+π/6) =2[cos(x+π/6)+cosπ/6] =2cos(x+π/3)+√3是怎么化简的?
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f(x)=4cosx/2cos(x/2+π/6) =2[cos(x+π/6)+cosπ/6] =2cos(x+π/3)+√3是怎么化简的?
f(x)=4cosx/2cos(x/2+π/6) =2[cos(x+π/6)+cosπ/6] =2cos(x+π/3)+√3是怎么化简的?

f(x)=4cosx/2cos(x/2+π/6) =2[cos(x+π/6)+cosπ/6] =2cos(x+π/3)+√3是怎么化简的?
是第一步不懂吧?
积化和差公式:cosαcosβ=[cos(α+β)+cos(α-β)]/2
所以,cosx/2cos(x/2+π/6)=[cos(x+π/6)+cos(π/6)]/2
所以,4cosx/2cos(x/2+π/6) =2[cos(x+π/6)+cosπ/6]
=2cos(x+π/6)+√3
ps:貌似最后一步是错的,应该是2cos(x+π/6)+√3