Twierdzenie o sumie i różnicy funkcji trygonometrycznych

Dla dowolnego $x,y\in\mathbb{R}$ :

\begin{aligned} \sin x+\sin y&=2 \cdot \sin \frac{x+y}{2} \cdot \cos \frac{x-y}{2}\\ \sin x-\sin y&=2 \cdot \sin \frac{x-y}{2} \cdot \cos \frac{x+y}{2}\\ \cos x+\cos y&=2 \cdot \cos \frac{x+y}{2} \cdot \cos \frac{x-y}{2}\\ \cos x-\cos y&=-2 \cdot \sin \frac{x+y}{2} \cdot \cos \frac{x-y}{2}\\ \tg{x}+\tg{y}&=\frac{\sin{\left(x+y\right )}}{\cos{x}\cos{y}}\\ \tg{x}-\tg{y}&=\frac{\sin{\left (x-y\right )}}{\cos{x}\cos{y}}\\ \ctg{x}+\ctg{y}&=\frac{\sin{\left (y+x\right)}}{\sin{x}\sin{y}}\\ \ctg{x}-\ctg{y}&=\frac{\sin{\left(y-x\right )}}{\sin{x}\sin{y}} \end{aligned}

\begin{aligned} \sin x+\sin y&=2 \cdot \sin \frac{x+y}{2} \cdot \cos \frac{x-y}{2}\\ \sin x-\sin y&=2 \cdot \sin \frac{x-y}{2} \cdot \cos \frac{x+y}{2}\\ \cos x+\cos y&=2 \cdot \cos \frac{x+y}{2} \cdot \cos \frac{x-y}{2}\\ \cos x-\cos y&=-2 \cdot \sin \frac{x+y}{2} \cdot \cos \frac{x-y}{2}\\ \tg{x}+\tg{y}&=\frac{\sin{\left(x+y\right )}}{\cos{x}\cos{y}}\\ \tg{x}-\tg{y}&=\frac{\sin{\left (x-y\right )}}{\cos{x}\cos{y}}\\ \ctg{x}+\ctg{y}&=\frac{\sin{\left (y+x\right)}}{\sin{x}\sin{y}}\\ \ctg{x}-\ctg{y}&=\frac{\sin{\left(y-x\right )}}{\sin{x}\sin{y}} \end{aligned}

Twierdzenie o sumie i różnicy funkcji trygonometrycznych

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