Sapendo che $\sin \alpha=\frac{3}{5}$ e $\cos \beta=\frac{5}{13}, \operatorname{con} \frac{\pi}{2}<\alpha<\pi$ e $0<\beta<\frac{\pi}{2}$, calcola: $\cos (\alpha+\beta), \sin (\alpha-\beta)$, tan $2 \alpha$.
Sapendo che $\sin \alpha=\frac{3}{5}$ e $\cos \beta=\frac{5}{13}, \operatorname{con} \frac{\pi}{2}<\alpha<\pi$ e $0<\beta<\frac{\pi}{2}$, calcola: $\cos (\alpha+\beta), \sin (\alpha-\beta)$, tan $2 \alpha$.
SIN(α) = 3/5
COS(β) = 5/13
pi/2 < α < pi : angolo del 2° quadrante: seno positivo coseno negativo
0 < β < pi/2 : angolo del 1° quadrante: seno e coseno positivi
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COS(α + β) = COS(α)·COS(β) - SIN(α)·SIN(β)
COS(α) = - √(1 - (3/5)^2)= - 4/5
SIN(β) = √(1 - (5/13)^2) = 12/13
quindi:
COS(α + β) = (- 4/5)·(5/13) - 3/5·(12/13)------->COS(α + β) = - 56/65
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SIN(α - β) = SIN(α)·COS(β) - SIN(β)·COS(α)
SIN(α - β) = 3/5·(5/13) - 12/13·(- 4/5)------->SIN(α - β) = 63/65
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TAN(2·α) = SIN(2·α)/COS(2·α)
TAN(2·α) = 2·SIN(α)·COS(α)/(COS(α)^2 - SIN(α)^2)
TAN(2·α) = 2·(3/5)·(- 4/5)/((- 4/5)^2 - (3/5)^2)
TAN(2·α) = - 24/7