设b=2a/7
则原式=cosb+cos2b+cos3b
∵cos2b=cos²b-sin²b
∵cos3b=cos(b+2b)=cosbcos2b-sinbsin2b=cos³b-cosbsin²b-2sin²bcosb
代入原式得:cosb(1+cos²b-sin²b-2sin²b)=2cosb(cos²b-sin²b)=2cosbcos2b
将b=2a/7代入得:2cos(2a/7)cos(4a/7)
设b=2a/7
则原式=cosb+cos2b+cos3b
∵cos2b=cos²b-sin²b
∵cos3b=cos(b+2b)=cosbcos2b-sinbsin2b=cos³b-cosbsin²b-2sin²bcosb
代入原式得:cosb(1+cos²b-sin²b-2sin²b)=2cosb(cos²b-sin²b)=2cosbcos2b
将b=2a/7代入得:2cos(2a/7)cos(4a/7)