Trigonometrie plana si sferica

Trigonometrie plana si sferica

Formule trigonometrice

Formulele functiilor trigonometrice ale sumei si diferentei:

@d\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta@d

@d\cos(\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta@d

@d\sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta@d

@d\sin(\alpha-\beta)=\sin\alpha\cos\beta-\cos\alpha\sin\beta@d

@d\operatorname{tg}(\alpha+\beta)=\frac{\operatorname{tg}\alpha+\operatorname{tg}\beta}{1-\operatorname{tg}\alpha\operatorname{tg}\beta}@d

@d\operatorname{tg}(\alpha-\beta)=\frac{\operatorname{tg}\alpha-\operatorname{tg}\beta}{1+\operatorname{tg}\alpha\operatorname{tg}\beta}@d

@d\operatorname{ctg}(\alpha+\beta)=\frac{\operatorname{ctg}\alpha\operatorname{ctg}\beta-1}{\operatorname{ctg}\alpha+\operatorname{ctg}\beta}@d

@d\operatorname{ctg}(\alpha-\beta)=\frac{\operatorname{ctg}\alpha\operatorname{ctg}\beta+1}{\operatorname{ctg}\beta-\operatorname{ctg}\alpha}@d

Consecinte ale formulelor pentru suma:

@d\sin2x=2\sin x\cos x@d

@d\cos2x=\cos^2x-\sin^2x=2\cos^2x-1=1-2\sin^2x@d

@d\operatorname{tg}2x=\frac{2\operatorname{tg}x}{1-\operatorname{tg}^2x},\operatorname{ctg}2x=\frac{\operatorname{ctg}^2x-1}{2\operatorname{ctg} x}@d

@d\sin3x=3\sin x-4\sin^3x, \cos3x=4\cos^3x-3\cos x@d

@d\operatorname{tg}3x=\frac{3\operatorname{tg} x-\operatorname{tg}^3x}{1-3\operatorname{tg}^2x}, \operatorname{ctg}3x=\frac{\operatorname{ctg}^3x-3\operatorname{ctg} x}{3\operatorname{ctg}^2x-1}@d

Din formulele pentru $$\cos2x$$ obtinem:

@d\cos^2x=\frac{1+\cos2x}{2}, \sin^2x=\frac{1-\cos2x}{2}@d

Inlocuind pe $$x$$ cu $$\frac{x}{2}$$ gasim:

@d\cos^2\frac{x}{2}=\frac{1+\cos x}{2}, \sin^2\frac{x}{2}=\frac{1-\cos x}{2}@d

@d\cos x=\frac{1-\operatorname{tg}^2\frac{x}{2}}{1+\operatorname{tg}^2\frac{x}{2}}, \sin x=\frac{2\operatorname{tg}\frac{x}{2}}{1+\operatorname{tg}^2\frac{x}{2}}@d

Adunand si scazand formulele pentru suma si diferenta se obtine:

@d\cos(\alpha+\beta)+\cos(\alpha-\beta)=2\cos\alpha\cos\beta@d

@d\cos(\alpha+\beta)-\cos(\alpha-\beta)=-2\sin\alpha\sin\beta@d

@d\sin(\alpha+\beta)+\sin(\alpha-\beta)=2\sin\alpha\cos\beta@d

@d\sin(\alpha+\beta)-\sin(\alpha-\beta)=2\cos\alpha\sin\beta@d

Notam $$\alpha+\beta=x, \alpha-\beta=y$$. Atunci $$\alpha=\frac{x+y}{2}, \beta=\frac{x-y}{2}$$ si avem

@d\cos x+\cos y=2\cos\frac{x+y}{2}\cos\frac{x-y}{2}@d

@d\cos x-\cos y=-2\sin\frac{x+y}{2}\sin\frac{x-y}{2}@d

@d\sin x+\sin y=2\sin\frac{x+y}{2}\cos\frac{x-y}{2}@d

@d\sin x-\sin y=2\sin\frac{x-y}{2}\cos\frac{x+y}{2}@d

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