Usando as fórmulas:
\cos (\alpha \pm \beta)=\cos \alpha \cos \beta \mp \operatorname{sen} \alpha \operatorname{sen} \beta
\operatorname{sen}(\alpha \pm \beta)=\operatorname{sen} \alpha \cos \beta \pm \cos \alpha \operatorname{sen} \beta
encontrar:
\operatorname{sen} \pi, \cos \pi, \tan \pi

Passo 1

O enunciado pede para calcular \operatorname{sen}\pi, \cos\pi e \tan\pi usando as fórmulas de ângulo soma:
\cos(\alpha+\beta)=\cos\alpha\cos\beta-\operatorname{sen}\alpha\operatorname{sen}\beta, \qquad \operatorname{sen}(\alpha+\beta)=\operatorname{sen}\alpha\cos\beta+\cos\alpha\operatorname{sen}\beta.
Uma escolha natural é escrever \pi=\tfrac{\pi}{2}+\tfrac{\pi}{2}, pois conhecemos \operatorname{sen}\tfrac{\pi}{2}=1 e \cos\tfrac{\pi}{2}=0.

Passo 2

Calcular \operatorname{sen}\pi via soma de ângulos:
\operatorname{sen}\pi = \operatorname{sen}\Bigl(\tfrac{\pi}{2}+\tfrac{\pi}{2}\Bigr) = \operatorname{sen}\tfrac{\pi}{2}\cos\tfrac{\pi}{2} + \cos\tfrac{\pi}{2}\operatorname{sen}\tfrac{\pi}{2} = 1\cdot0 + 0\cdot1 = 0.

Passo 3

Calcular \cos\pi via soma de ângulos:
\cos\pi = \cos\Bigl(\tfrac{\pi}{2}+\tfrac{\pi}{2}\Bigr) = \cos\tfrac{\pi}{2}\cos\tfrac{\pi}{2} - \operatorname{sen}\tfrac{\pi}{2}\operatorname{sen}\tfrac{\pi}{2} = 0\cdot0 - 1\cdot1 = -1.

Passo 4

Calcular \tan\pi como quociente:
\tan\pi = \frac{\operatorname{sen}\pi}{\cos\pi} = \frac{0}{-1} = 0.

Resposta

\operatorname{sen}\pi = 0 \cos\pi = -1 \tan\pi = 0

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