Usando as relações trigonométricas
\left\{\begin{aligned} \cos (a+b) & =\cos a \cos b-\sin a \sin b \\\ \cos (a-b) & =\cos a \cos b+\sin a \sin b \\\ \sin (a+b) & =\sin a \cos b+\sin b \cos a \\\ \sin (a-b) & =\sin a \cos b-\sin b \cos a \end{aligned}\right.
mostre as seguintes relações:
  • 2 \sin \frac{x}{2} \sin k x=\cos \left(\frac{x}{2}-k x\right)-\cos \left(\frac{x}{2}+k x\right)=\cos \left(\frac{2 k-1}{2}\right) x-\cos \left(\frac{2 k+1}{2}\right) x.
  • 2 \sin \frac{x}{2} \sum_{k=1}^{n} \sin k x=\cos \left(\frac{x}{2}\right)-\cos \left(n+\frac{1}{2}\right) x.
  • \cos \left(\frac{x}{2}\right)-\cos \left(n+\frac{1}{2}\right) x=\cos \left(\frac{x}{2}\right)(1-\cos n x)+\sin \left(\frac{x}{2}\right) \sin n x.
  • (1-\cos n x)=2 \sin ^{2}\left(\frac{n x}{2}\right).
  • 2 \sin \frac{x}{2} \sum_{k=1}^{n} \sin k x=2 \sin \left(\frac{n x}{2}\right) \sin \left(\frac{n+1}{2}\right) x.

Passo 1

O enunciado pede para mostrar várias identidades trigonométricas envolvendo produtos e somas de senos, usando as fórmulas de soma e diferença
\begin{aligned} \cos(a\pm b)&=\cos a\cos b\mp\sin a\sin b,\\\ \sin(a\pm b)&=\sin a\cos b\pm\sin b\cos a, \end{aligned}
além de telescopar somas e usar identidades pitagóricas.

Passo 2

Começamos com a identidade produto-para-soma
2\sin A\sin B=\cos(A-B)-\cos(A+B).
Tomando A=kx e B=\tfrac{x}{2}, temos
2\sin\!\bigl(\tfrac{x}{2}\bigr)\,\sin(kx) =\cos\bigl(kx-\tfrac{x}{2}\bigr)-\cos\bigl(kx+\tfrac{x}{2}\bigr) =\cos\!\bigl(\tfrac{2k-1}{2}x\bigr)-\cos\!\bigl(\tfrac{2k+1}{2}x\bigr).

Passo 3

Agora somamos para k=1,2,\dots,n:
2\sin\!\bigl(\tfrac{x}{2}\bigr)\sum_{k=1}^n\sin(kx) =\sum_{k=1}^n\Bigl[\cos\!\bigl(\tfrac{2k-1}{2}x\bigr)-\cos\!\bigl(\tfrac{2k+1}{2}x\bigr)\Bigr].
Essa soma telescopa, pois para cada k o -\cos(\tfrac{2k+1}{2}x) cancela o +\cos(\tfrac{2(k+1)-1}{2}x). Sobra
\cos\!\bigl(\tfrac{x}{2}\bigr)-\cos\!\bigl(\tfrac{2n+1}{2}x\bigr) =\cos\!\bigl(\tfrac{x}{2}\bigr)-\cos\!\bigl(n+\tfrac12\bigr)x.

Passo 4

Expandimos \cos\bigl((n+\tfrac12)x\bigr)=\cos(nx+x/2) pela fórmula de soma:
\cos(nx+x/2) =\cos(nx)\cos\!\bigl(\tfrac{x}{2}\bigr)-\sin(nx)\sin\!\bigl(\tfrac{x}{2}\bigr).
Logo
\cos\!\bigl(\tfrac{x}{2}\bigr)-\cos\!\bigl(n+\tfrac12\bigr)x =\cos\!\bigl(\tfrac{x}{2}\bigr)\bigl[1-\cos(nx)\bigr] +\sin\!\bigl(\tfrac{x}{2}\bigr)\,\sin(nx).

Passo 5

Usamos a identidade pitagórica
1-\cos(nx)=2\sin^2\!\bigl(\tfrac{nx}{2}\bigr).

Passo 6

Substituindo em \cos(\tfrac{x}{2})(1-\cos(nx))+\sin(\tfrac{x}{2})\sin(nx), obtemos
2\cos\!\bigl(\tfrac{x}{2}\bigr)\sin^2\!\bigl(\tfrac{nx}{2}\bigr) +2\sin\!\bigl(\tfrac{x}{2}\bigr)\sin\!\bigl(\tfrac{nx}{2}\bigr)\cos\!\bigl(\tfrac{nx}{2}\bigr).
Fatorando 2\sin(\tfrac{nx}{2}) e reconhecendo \sin(A+B)=\sin A\cos B+\cos A\sin B com A=\tfrac{nx}{2}, B=\tfrac{x}{2}, vem
2\sin\!\bigl(\tfrac{nx}{2}\bigr)\,\sin\!\bigl(\tfrac{n+1}{2}x\bigr).
Assim concluímos
2\sin\!\bigl(\tfrac{x}{2}\bigr)\sum_{k=1}^n\sin(kx) =2\sin\!\bigl(\tfrac{nx}{2}\bigr)\sin\!\bigl(\tfrac{n+1}{2}x\bigr).

Resposta

  • 2\sin\!\bigl(\tfrac{x}{2}\bigr)\sin(kx) =\cos\!\bigl(\tfrac{2k-1}{2}x\bigr)-\cos\!\bigl(\tfrac{2k+1}{2}x\bigr).
  • 2\sin\!\bigl(\tfrac{x}{2}\bigr)\sum_{k=1}^n\sin(kx) =\cos\!\bigl(\tfrac{x}{2}\bigr)-\cos\!\bigl(n+\tfrac12\bigr)x.
  • \cos\!\bigl(\tfrac{x}{2}\bigr)-\cos\!\bigl(n+\tfrac12\bigr)x =\cos\!\bigl(\tfrac{x}{2}\bigr)\bigl[1-\cos(nx)\bigr] +\sin\!\bigl(\tfrac{x}{2}\bigr)\sin(nx).
  • 1-\cos(nx)=2\sin^2\!\bigl(\tfrac{nx}{2}\bigr).
  • 2\sin\!\bigl(\tfrac{x}{2}\bigr)\sum_{k=1}^n\sin(kx) =2\sin\!\bigl(\tfrac{nx}{2}\bigr)\sin\!\bigl(\tfrac{n+1}{2}x\bigr).

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