Prove as seguintes identidades trigonométricas:
\sin \theta \pm \sin \varphi=2 \sin \left(\frac{\theta \pm \varphi}{2}\right) \cos \left(\frac{\theta \mp \varphi}{2}\right)

Passo 1

O enunciado pede para provar as fórmulas de soma em produto para senos:
\sin\theta\pm\sin\varphi =2\sin\Bigl(\frac{\theta\pm\varphi}{2}\Bigr)\cos\Bigl(\frac{\theta\mp\varphi}{2}\Bigr).
O que vamos usar aqui são as fórmulas de adição e subtração de ângulos para o seno:
\sin(A\pm B)=\sin A\cos B\pm\cos A\sin B,
além da ideia de escrever \theta e \varphi em termos de média e metade da diferença:
\theta=\frac{\theta+\varphi}{2}+\frac{\theta-\varphi}{2},\qquad \varphi=\frac{\theta+\varphi}{2}-\frac{\theta-\varphi}{2}.

Passo 2

Vamos primeiro expressar \sin\theta e \sin\varphi em função dessas somas e diferenças de meios:
\sin\theta =\sin\Bigl(\tfrac{\theta+\varphi}{2}+\tfrac{\theta-\varphi}{2}\Bigr) =\sin\!\tfrac{\theta+\varphi}{2}\,\cos\!\tfrac{\theta-\varphi}{2} +\cos\!\tfrac{\theta+\varphi}{2}\,\sin\!\tfrac{\theta-\varphi}{2},
\sin\varphi =\sin\Bigl(\tfrac{\theta+\varphi}{2}-\tfrac{\theta-\varphi}{2}\Bigr) =\sin\!\tfrac{\theta+\varphi}{2}\,\cos\!\tfrac{\theta-\varphi}{2} -\cos\!\tfrac{\theta+\varphi}{2}\,\sin\!\tfrac{\theta-\varphi}{2}.

Passo 3

Agora somamos as duas expressões:
\sin\theta+\sin\varphi =\bigl[\sin\!\tfrac{\theta+\varphi}{2}\cos\!\tfrac{\theta-\varphi}{2} +\cos\!\tfrac{\theta+\varphi}{2}\sin\!\tfrac{\theta-\varphi}{2}\bigr]
\quad+\bigl[\sin\!\tfrac{\theta+\varphi}{2}\cos\!\tfrac{\theta-\varphi}{2} -\cos\!\tfrac{\theta+\varphi}{2}\sin\!\tfrac{\theta-\varphi}{2}\bigr] =2\,\sin\!\tfrac{\theta+\varphi}{2}\,\cos\!\tfrac{\theta-\varphi}{2}.

Passo 4

Para a diferença, subtraímos em vez de somar:
\sin\theta-\sin\varphi =\bigl[\sin\!\tfrac{\theta+\varphi}{2}\cos\!\tfrac{\theta-\varphi}{2} +\cos\!\tfrac{\theta+\varphi}{2}\sin\!\tfrac{\theta-\varphi}{2}\bigr]
\quad-\bigl[\sin\!\tfrac{\theta+\varphi}{2}\cos\!\tfrac{\theta-\varphi}{2} -\cos\!\tfrac{\theta+\varphi}{2}\sin\!\tfrac{\theta-\varphi}{2}\bigr] =2\,\cos\!\tfrac{\theta+\varphi}{2}\,\sin\!\tfrac{\theta-\varphi}{2}.
Note que 2\cos\frac{\theta+\varphi}{2}\sin\frac{\theta-\varphi}{2} =2\sin\frac{\theta-\varphi}{2}\cos\frac{\theta+\varphi}{2}.

Resposta

\displaystyle \sin\theta+\sin\varphi =2\sin\Bigl(\tfrac{\theta+\varphi}{2}\Bigr)\cos\Bigl(\tfrac{\theta-\varphi}{2}\Bigr) \displaystyle \sin\theta-\sin\varphi =2\sin\Bigl(\tfrac{\theta-\varphi}{2}\Bigr)\cos\Bigl(\tfrac{\theta+\varphi}{2}\Bigr)

Exercícios relacionados