化簡a²sin3/π+b²cosπ/6-absinπ/4cosπ/4tanπ/3+abcosπ/6= 過 程

化簡a²sin3/π+b²cosπ/6-absinπ/4cosπ/4tanπ/3+abcosπ/6= 過 程

a²sinπ/3+b²cosπ/6-absinπ/4cosπ/4tanπ/3+abcosπ/6
=a²(√3/2)+b²(√3/2)-ab(√2/2)(√2/2)(√3)+ab(√3/2)
=a²(√3/2)+b²(√3/2)-ab(√3/2)+ab(√3/2)
=(√3/2)(a²+b²)