三角公式 | JasonQian's Blog

三角函数

三角函数的基本关系

$$1. \ csc\alpha \ = \ \frac{1}{sinα}$$

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$$2. \ secα \ = \ \frac{1}{cosα}$$

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$$3.\ cot\alpha \ = \ \frac{1}{tan\alpha}$$

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$$4.\ tan\alpha \ = \ \frac{sin\alpha}{cos\alpha}$$

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$$5.\ cot\alpha \ = \ \frac{cos\alpha}{sin\alpha}$$

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$$6.\ sin^2{\alpha} \ + \ cos^2{\alpha} \ = \ 1$$

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$$7. \ tan^2{\alpha} \ + \ 1 \ = \ sec^2{\alpha}$$

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$$8.\ cot^2{\alpha} \ + \ 1 \ = \ csc^2{\alpha}$$

倍角公式

$$1. \ sin{2\alpha} \ = \ 2sin\alpha cos\alpha$$

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$$2.\ cos{2\alpha} \ = \ cos^2\alpha \ - \ sin^2\alpha \ = \ 1 \ - 2sin^2\alpha \ = \ 2cos^2\alpha \ - \ 1$$

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$$3. \ sin{3\alpha} \ = \ -4sin^3\alpha \ + \ 3sin\alpha$$

证明：

$sin{3\alpha}$

$= \ sin(\alpha \ + \ 2\alpha)$

$= \ sin\alpha cos{2\alpha} \ + \ sin{2\alpha} cos\alpha$

$= \ sin\alpha (1 \ - \ 2sin^2\alpha) \ + \ (2sin\alpha cos\alpha) cos\alpha$

$= \ sin\alpha \ - \ 2sin^3\alpha \ + \ 2sin\alpha cos^2\alpha$

$= \ sin\alpha \ - \ 2sin^3\alpha \ + \ sin\alpha(2cos^2\alpha)$

$= \ sin\alpha \ - \ 2sin^3\alpha \ + \ sin\alpha(2 \ - \ 2sin^2\alpha)$

$= \ -4sin^3\alpha \ + \ 3sin\alpha$

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$$4. \ cos3\alpha \ = \ 4cos^3\alpha \ - \ 3cos\alpha$$

证明：

$cos3\alpha$

$= \ cos(\alpha \ + \ 2\alpha)$

$= \ cos\alpha cos2\alpha - sin\alpha sin2\alpha$

$= \ cos\alpha(2cos^2\alpha \ - \ 1) \ - \ sin\alpha (2sin\alpha cos\alpha)$

$= \ 3cos^3\alpha \ - \ cos\alpha \ - \ 2sin^2\alpha cos\alpha$

$= \ 3cos^3\alpha \ - \ cos\alpha \ - \ (2 \ - \ cos^2\alpha) cos\alpha$

$= \ 4cos^4 \ - \ 3cos\alpha$

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$$5.\ tan{2\alpha} \ = \ \frac{2tan\alpha}{1 \ - \ tan^2\alpha}$$

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$$6. \ cos{2\alpha \ = \ \frac{1}{tan2\alpha} \ = \ \frac{1 \ - \ tan^2\alpha}{2tan\alpha}}$$

半角公式

$$1. \ sin^2{\frac{\alpha}{2}} \ = \ \frac{1}{2}(1 \ - \ cos\alpha)(降幂公式)$$

证明：

$利用了 1 \ - \ cos2\alpha \ = \ 2sin^2\alpha(降幂公式)$

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$$2.\ cos^2{\frac{\alpha}{2}} \ = \ \frac{1 \ + \ cos\alpha}{2}$$

证明：

$利用了2cos^2\alpha \ - \ 1 \ = \ cos2\alpha$

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$$3. \ sin\frac{\alpha}{2} \ = \ \pm \sqrt{\frac{1 \ - \ cos\alpha}{2}}$$

证明：

只是把1. 去掉平方了，但是注意要加$\pm$号

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$$4. \ cos\frac{\alpha}{2} \ = \ \pm \sqrt{\frac{1 \ + \ cos\alpha}{2}}$$

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$$5. \ tan\frac{\alpha}{2} \ = \ \pm \sqrt{\frac{1 \ - \ cos\alpha}{1 \ + \ cos\alpha}}$$

证明：

$tan\frac{\alpha}{2}$

$= \ \frac{sin\frac{\alpha}{2}}{cos\frac{\alpha}{2}}$

$= \ \frac{sin^2\frac{\alpha}{2}}{cos\frac{\alpha}{2} sin\frac{\alpha}{2}}$

$= \ \frac{2sin^2\frac{\alpha}{2}}{2cos\frac{\alpha}{2} sin\frac{\alpha}{2}}$

$= \ \frac{1 \ - \ cos\alpha}{sin\alpha}$

$= \ \frac{(1 \ - \ cos\alpha) \ * \ (1 \ + \ cos\alpha)}{sin\alpha \ * \ (1 \ + \ cos\alpha)}$

$= \ \frac{sin\alpha}{1 \ + \ cos\alpha}$

$= \ \pm\sqrt{\frac{sin^2\alpha}{(1 \ + \ cos\alpha)^2}}$

$= \ \pm\sqrt{\frac{1 \ - \ cos^2\alpha}{(1 \ + \ cos\alpha)^2}}$

$= \ \pm\sqrt{\frac{1 \ - \ cos\alpha}{1 \ + \ cos\alpha}}$

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$$6. \ cot\frac{\alpha}{2} \ = \ \sqrt{\frac{1 \ + \ cos\alpha}{1 \ - \ cos\alpha}}$$

证明：

同上，$ \ cot\alpha \ 就是 \ tan\alpha \ 的倒数$

和差公式

$$1. \ sin(\alpha \ \pm \ \beta) \ = \ sin\alpha cos\beta \ \pm \ cos\alpha sin\beta$$

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$$2. \ cos(\alpha \ \pm \ \beta) \ = \ cos\alpha cos\beta \ \mp \ sin\alpha sin\beta$$

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$$3. \ tan(\alpha \ + \ \beta) \ = \ \frac{tan\alpha \ \pm \ tan\beta}{1 \ \mp \ tan\alpha tan\beta}$$

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$$4. \ cot(\alpha \ + \ \beta) \ = \ \frac{cot\alpha cot\beta \ \mp \ 1}{cot\beta \ \pm \ cot\alpha}$$

积化和差和与和差化积公式

积化和差公式

$$1. \ sin\alpha cos\beta \ = \ \frac{1}{2}[sin(\alpha \ + \ \beta) \ + \ sin(\alpha \ - \ \beta)]$$

证明：

$\frac{1}{2}[sin(\alpha \ + \ \beta) \ + \ sin(\alpha \ - \ \beta)]$

$= \ \frac{1}{2}[sin\alpha cos\beta \ + \ cos\alpha sin\beta \ + \ sin\alpha cos\beta \ - \ cos\alpha sin\beta]$

$= \ \frac{1}{2}(sin\alpha cos\beta \ + \ cos\alpha sin\beta)$

$= \ sin\alpha cos\beta$

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$$2.\ cos\alpha sin\beta \ = \ \frac{1}{2}[sin(\alpha \ + \ \beta) \ - \ sin(\alpha \ - \ \beta)]$$

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$$3. \ cos\alpha cos\beta \ = \ \frac{1}{2}[cos(\alpha \ + \ \beta) \ + \ cos(\alpha \ - \ \beta)]$$

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$$4. \ sin\alpha sin\beta \ = \ \frac{1}{2}[cos(\alpha \ - \ \beta) \ - \ cos(\alpha \ + \ \beta)]$$

和差化积公式

$$1. \ sin\alpha \ + \ sin\beta \ = \ 2sin\frac{\alpha \ + \ \beta}{2} cos\frac{\alpha \ - \ \beta}{2}$$

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$$2. \ sin\alpha \ - \ sin\beta \ = \ 2sin\frac{\alpha \ - \ \beta}{2} cos\frac{\alpha \ + \ \beta}{2}$$

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$$3. \ cos\alpha \ + \ cos\beta \ = \ 2cos\frac{\alpha \ + \ \beta}{2} sin\frac{\alpha \ - \ \beta}{2}$$

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$$4. \ cos\alpha \ - \ cos\beta \ = \ {-}2sin\frac{\alpha \ + \ \beta}{2}sin\frac{\alpha \ - \ \beta}{2}$$

证明：

${-}2sin\frac{\alpha \ + \ \beta}{2}sin\frac{\alpha \ - \ \beta}{2}$

$= \ -2(sin\frac{\alpha}{2} cos\frac{\beta}{2} \ + \ cos\frac{\alpha}{2} sin\frac{\beta}{2})(sin\frac{\alpha}{2} cos\frac{\beta}{2} \ - \ cos\frac{\alpha}{2} sin\frac{\beta}{2})$

$= \ -2(sin^2\frac{\alpha}{2} cos^2\frac{\beta}{2} \ - \ sin\frac{\alpha}{2} cos\frac{\beta}{2}cos\frac{\alpha}{2} sin\frac{\beta}{2} \ + \ cos\frac{\alpha}{2} sin\frac{\beta}{2}sin\frac{\alpha}{2} cos\frac{\beta}{2} \ - \ cos^2\frac{\alpha}{2} sin^2\frac{\beta}{2})$

$= \ -2(sin^2\frac{\alpha}{2} cos^2\frac{\beta}{2} \ - \ cos^2\frac{\alpha}{2} sin^2\frac{\beta}{2})$

$= \ -2[(\frac{1 \ - \ cos\alpha}{2})(\frac{1 \ + \ cos\beta}{2}) \ - \ (\frac{1 \ + \ cos\alpha}{2})(\frac{1 \ - \ cos\beta}{2})]$

$= \ -\frac{(1 \ + \ cos\beta \ - \ cos\alpha \ - \ cos\alpha cos\beta) \ - \ (1 \ - \ cos\beta \ + \ cos\alpha \ - \ cos\alpha cos\beta)}{2}$

$= \ -\frac{2cos\beta \ - \ 2cos\alpha}{2}$

$= \ cos\alpha \ - \ cos\beta$