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设f(a)=sin^2a+cosacos(π/3+a)-sin^2(π/6-a)π.请问不用和差化积能做吗,这个老师说不

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设f(a)=sin^2a+cosacos(π/3+a)-sin^2(π/6-a)π.请问不用和差化积能做吗,这个老师说不要求,
f(a)=sin^2a+cosacos(π/3+a)-sin^2(π/6-a)
=(1-cos2a)/2+ cosacos(π/3+a)-(1-cos(π/3-2a))/2
=(1-cos2a)/2-(1-cos(π/3-2a))/2+ cosacos(π/3+a)
= cos(π/3-2a)/2-cos2a/2+ cosacos(π/3+a)
=[cosπ/3cos2a +sinπ/3sin2a]/2-cos2a/2+ cosacos(π/3+a)
=[1/2*cos2a +√3/2*sin2a]/2-cos2a/2+ cosacos(π/3+a)
=[-1/2*cos2a +√3/2*sin2a]/2+ cosacos(π/3+a)
= -cos(2a +π/3)/2+cosacos(π/3+a)
=-cos[a+(a +π/3)/2+cosacos(π/3+a)
=-[cosacos(π/3+a)-sinasin(π/3+a)] /2+cosacos(π/3+a)
=[cosacos(π/3+a)+sinasin(π/3+a)] /2
= cos[(a +π/3)-a]/2
= cos(π/3)/2
=1/4.