We have cos‌α+cos‌β+cos‌γ=0 =sin‌α+sin‌β+sin‌γ Let a=cos‌α+isin‌α ‌‌‌‌‌b=cos‌β+isin‌β ‌c=cos‌γ+isin‌γ ‌∴a+b+c=(cos‌α+cos‌β+cos‌γ) ‌‌‌+i(sin‌α+sin‌β+sin‌γ) ‌a+b+c=0 ‌‌ and ‌a−1+b−1+c−1=0 ‌∴ab+bc+ca=0 ‌⇒cos(α+β)+cos(β+γ) ‌+cos(γ+α)=0 Now a2+b2+c2=(a+b+c)2 ‌−2(ab+bc+ca) ‌∴sin‌2α+sin‌2β+sin‌2γ=0
