TAN（3π/11）＋４SIN（2π/11）＝？★２
>15 下
cos(kθ)/(cosθ)^k = {(cosθ+i･sinθ)^k + (cosθ-i･sinθ)^k} / [2(cosθ)^k]
= {(1+it)^k + (1-it)^k} /2, （t≡tanθ とおいた.)
∑[k=0,n] cos(kθ)/(cosθ)^k = {(1+it)^(n+1) -1}/(2it) + {(1-it)^(n+1) -1}/(-2it)
= {(1+it)^(n+1) -(1-it)^(n+1)} / (2it)
= sin((n+1)θ) / [(cosθ)^(n+1) ･t]
= sin((n+1)θ) / [(cosθ)^n ･sinθ].
蛇足だが...
∑[k=1,n] sin(kθ)/(cosθ)^k = {2 -(1+it)^(n+1) -(1-it)^(n+1)} / (2t)
= 1/t -cos((n+1)θ)/[(cosθ)^(n+1) ･t]
= (1/tanθ) -cos((n+1)θ)/[(cosθ)^n ･sinθ].
cos(kθ)/(cosθ)^k = {(cosθ+i･sinθ)^k + (cosθ-i･sinθ)^k} / [2(cosθ)^k]
= {(1+it)^k + (1-it)^k} /2, （t≡tanθ とおいた.)
∑[k=0,n] cos(kθ)/(cosθ)^k = {(1+it)^(n+1) -1}/(2it) + {(1-it)^(n+1) -1}/(-2it)
= {(1+it)^(n+1) -(1-it)^(n+1)} / (2it)
= sin((n+1)θ) / [(cosθ)^(n+1) ･t]
= sin((n+1)θ) / [(cosθ)^n ･sinθ].
蛇足だが...
∑[k=1,n] sin(kθ)/(cosθ)^k = {2 -(1+it)^(n+1) -(1-it)^(n+1)} / (2t)
= 1/t -cos((n+1)θ)/[(cosθ)^(n+1) ･t]
= (1/tanθ) -cos((n+1)θ)/[(cosθ)^n ･sinθ].
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