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Trigonometrijos formulės

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$$\sin ^2 \alpha +\cos ^2 \alpha = 1$$ $$\text{tg} \alpha=\frac{\sin\alpha}{\cos\alpha}, \ \text{ctg}\alpha=\frac{\cos\alpha}{\sin\alpha}$$ $$\text{tg}\alpha\cdot \text{ctg}\alpha=1$$
$$\sin({2\alpha})=2\sin{\alpha}\cos{\alpha}$$ $$\cos({2\alpha})=\cos^2\alpha-\sin^2\alpha$$ $$\text{tg}({2\alpha})=\frac{2\text{tg}\alpha}{1-\text{tg}^2\alpha}, \ \text{ctg}({2\alpha})=\frac{\text{ctg}^2\alpha-1}{2\text{ctg}\alpha}$$
$$\sin3\alpha=-4\sin^{3}\alpha+3\sin\alpha$$ $$\cos3\alpha=4\cos^{3}\alpha-3\cos\alpha$$
$$1+\cos\alpha=2\cos^{2}\frac{\alpha}{2}$$ $$1-\cos\alpha=2\sin^{2}\frac{\alpha}{2}$$ $$1+\text{tg}^{2}\alpha=\frac{1}{\cos^{2}\alpha}, \ 1+\text{ctg}^{2}\alpha=\frac{1}{\sin^{2}\alpha}$$
$$\text{tg}\alpha \pm \text{tg}\beta=\frac{\sin(\alpha \pm \beta)}{\cos\alpha\cos\beta}, \ \text{ctg}\alpha \pm \text{ctg}\beta=\frac{\sin(\beta \pm \alpha)}{\sin\alpha\sin\beta}$$ $$\sin\alpha \pm \sin\beta=2\sin\frac{\alpha \pm \beta}{2}\cos\frac{\alpha \mp \beta}{2}$$ $$\cos\alpha+\cos\beta=2\cos\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}$$ $$\cos\alpha-\cos\beta=2\sin\frac{\alpha+\beta}{2}\sin\frac{\beta-\alpha}{2}$$
$$\text{tg}(\alpha \pm \beta)=\frac{\text{tg}\alpha \pm \text{tg}\beta}{1 \mp \text{tg}\alpha \text{tg}\beta}, \ \text{ctg}(\alpha \pm \beta)=\frac{\text{ctg}\alpha \text{ctg}\beta\mp 1}{\text{ctg}\beta \pm \text{tg}\alpha}$$ $$\sin(\alpha \pm \beta)=\sin\alpha\cos\beta \pm \cos\alpha\sin\beta$$ $$\cos(\alpha \pm \beta)=\cos\alpha\cos\beta \mp \sin\alpha\sin\beta$$
$$\sin\alpha\cos\beta=\frac{1}{2}(\sin(\alpha-\beta)+\sin(\alpha+\beta))$$ $$\cos\alpha\cos\beta=\frac{1}{2}(\cos(\alpha-\beta)+\cos(\alpha+\beta))$$ $$\sin\alpha\sin\beta=\frac{1}{2}(\cos(\alpha-\beta)-\cos(\alpha+\beta))$$

Paskutinį kartą atnaujinta 2017-11-07

0

Pusės kampo formulės:
sin(α/2) = √((1-cosα)/2)
cox(α/2) = √((1+cosα)/2)
tg(α/2) = sinα/(1+cosα) = (1-cosα)/sinα

0

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