总和公式

The sum and difference formulas state that

$\ begin {Aligned} \ sin（a + b）＆= \ sin a \ cos b + \ cos a \ sin b \ \ sin（a -b）＆= \ sin a a \ cos b - cos b- \ cos a \ cos a \ sin b \ sin b \ sin结束{对齐}$

然后

$\ begin {Aligned} \ cos（a + b）＆= \ cos a \ cos b - \ sin a \ sin a \ sin b \ \ cos（a -b）＆= \ cos a \ cos a \ cos b + sin a \ sin a \ sin b。\ end {Aligned}$

得出总和公式

$\ begin {Aligned} \ sin（a+b）＆= \ sin a \ cos b+\ cos a \ sin b \ \ cos（a+b）b。\ end {Aligned}$

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ByEuler's formulawe know that

$e^{i \ theta} = \ cos \ theta + i \ sin \ theta。$

如果 $\ theta = a + b$, then

$e^{i（a+b）} = \ cos（a+b）+i \ sin（a+b）。\ qquad（1）$

我们也从指数的代数属性中知道

$e^{i（a+b）} = e^{ia} e^{ib}。$

所以，

$\begin{aligned} e^{ia}e^{ib} &= (\cos a + i\sin a)(\cos b + i\sin b)\\ &=\cos a \cos b + i\sin a \cos b + i\sin b\cos a + i^2\sin a \sin b \\ &= \cos a \cos b - \sin a \sin b + i(\sin a \cos b + \sin b \cos a). \end{aligned}$

现在，来自 $(1)$we have

$\begin{aligned} e^{i(a+b)} &= \cos(a+b) + i\sin(a+b)\\ &=\cos a \cos b - \sin a \sin b + i(\sin a \cos b + \sin b \cos a). \end{aligned}$

但是由于 $\ cos$和 $\罪$are real-valued functions, it must be true that

$\begin{aligned} \cos(a+b) &= \cos a \cos b - \sin a \sin b \\ i\sin(a+b) &= i(\sin a \cos b + \sin b \cos a), \end{aligned}$

这意味着

$\begin{aligned} \cos(a+b) &= \cos a\cos b - \sin a \sin b \\ \sin(a+b) &= \sin a \cos b + \sin b \cos a. \ _\square \end{aligned}$

在图中，让点 $A$旋转到要点 $B$和 $C,$让角度 $\α$和 $\ beta$be defined as follows:

$\ Angle AOB = \ Alpha，\ Quad \ Angle Boc = \ beta。$

Also, let both $\ edline {cd}$和 $\overline{FG}$垂直于 $\ edline {oa}，$然后让 $E$be a point on $\ edline {cd}$这样 $\lvert \overline{ED}\rvert=\lvert\overline{FG}\rvert.$

然后是余弦的公式 $\ cos（\ alpha+\ beta），$which is $\frac{\lvert \overline{OD}\rvert}{\lvert \overline{OC}\rvert} ,$can be obtained as follows:

$\begin{aligned} \cos (\alpha + \beta) &= \frac{\lvert \overline{OD}\rvert}{\lvert \overline{OC}\rvert} \\ &= \frac{\lvert \overline{OG}\rvert}{\lvert \overline{OC}\rvert} - \frac{\lvert \overline{EF}\rvert}{\lvert \overline{OC}\rvert} \\ &= \frac{\lvert \overline{OG}\rvert}{\lvert \overline{OF}\rvert} \cdot \frac{\lvert \overline{OF}\rvert}{\lvert \overline{OC}\rvert} - \frac{\lvert \overline{EF}\rvert}{\lvert \overline{CF}\rvert} \cdot \frac{\lvert \overline{CF}\rvert}{\lvert \overline{OC}\rvert} \qquad \left(\text{since } \lvert \overline{OD}\rvert = \lvert \overline{OG}\rvert-\lvert \overline{EF}\rvert\right)\\ &= \cos \alpha \cdot \cos \beta - \sin \alpha \cdot \sin \beta. \end{aligned}$

余弦差异公式可以通过替换从余弦 - 和公式获得 $\ beta$和 $- \ beta，$并使用 $\ cos（ - \ beta）= \ cos \ beta$和 $\罪(-\beta) = -\sin \beta:$

$\begin{aligned} \cos(\alpha + \beta) &= \cos \alpha \cdot \cos \beta - \sin \alpha \cdot \sin \beta \\ \Rightarrow \cos(\alpha - \beta) &= \cos \alpha \cdot \cos (-\beta) - \sin \alpha \cdot \sin (-\beta) \\ &= \cos \alpha \cdot \cos \beta + \sin \alpha \cdot \sin \beta. \end{aligned}$

In summary, we have the following two formulas of cosine-sum and cosine-difference:

Cosine-sum formula: $\ cos(\alpha + \beta)= \cos \alpha \cdot \cos \beta - \sin \alpha \cdot \sin \beta ,$

余弦差异公式: $\ cos（\ alpha- \ beta）= \ cos \ alpha \ cdot \ cos \ cos \ beta + \ sin \ alpha \ alpha \ cdot \ sin \ beta。$

What is $\ cos 75^\ circ？$

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从余弦和公式中，我们有

$\ begin {Aligned} \ cos 75^\ circ＆= \ cos（45^\ circ + 30^\ circ）\\＆= \ cos 45^\ circ \ circ \ cdot \ cos 30^\ circ- \ circ- \ sin 45^\ sin 45^\ sincirc \ cdot \ sin 30^\ circ \\＆= \ frac {\ sqrt {2}}} {2} {2} \ cdot \ frac \ frac {\ sqrt {3}}} {2} {2} {2} - \ frac {\ frac {\ sqrt {2}}}2} \ cdot \ frac {1} {2} \\＆= \ frac {\ sqrt {6}}} {4} {4} - \ frac {\ sqrt {\ sqrt {2}}}} {4} {4}{6} - \ sqrt {2}} {4}。$

What is $\ cos15^\circ ?$

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From cosine-difference formula, we have

$\ begin {Aligned} \ cos 15^\ circ＆= \ cos（45^\ circ -circ -30^\ circ）\\＆= \ cos 45^\ circ \ circ \ cdot \ cos 30^\ circ + circ + circ + sin 45^\ sin 45^\ sincirc \ cdot \ sin 30^\ circ \\＆= \ frac {\ sqrt {2}}} {2} {2} \ cdot \ frac \ frac {\ sqrt {3}}} {2} {2} + \ frac {\ frac {\ sqrt {\ sqrt {2}}}2} \ cdot \ frac {1} {2} \\＆= \ frac {\ sqrt {\ sqrt {6}}} {4} + \ frac {\ sqrt {\ sqrt {2}}}} {4} {4}{6}+\ sqrt {2}} {4}。$

What is $\ cos 105^\ circ？$

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从余弦和公式， $\ cos 105^\ circ$是

$\begin{aligned} \cos 105^\circ &= \cos (60^\circ + 45^\circ) \\ &= \cos 60^\circ \cdot \cos 45^\circ - \sin 60^\circ \cdot \sin 45^\circ \\ &= \frac{1}{2} \cdot \frac{\sqrt{2}}{2} - \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2} \\ &= \frac{\sqrt{2}}{4} - \frac{\sqrt{6}}{4} \\ &= \frac{\sqrt{2}-\sqrt{6}}{4}.\ _\square \end{aligned}$

简化

$\ cos140^\circ \cdot \cos 50^\circ + \sin 140^\circ \cdot \sin 50^\circ .$

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From cosine-difference formula, we have

$\ begin {Aligned} \ cos 140^\ circ \ cdot \ cos 50^\ circ + \ sin 140^\ circ \ cdot \ sin 50^\ circ＆circ＆= \ cos（140^\ circ-50^\ circ）\＆= \ cos 90^\ circ \\ =0。\ _ \ square \ end {aligned}$

如果 $\ sin \ alpha = \ frac {13} {14}$和 $\罪\ beta= \frac{11}{14}$为了 $0 <\ alpha <\ frac {\ pi} {2}$和 $0 <\ beta <\ frac {\ pi} {2}，$what is $\α+ \beta ?$

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自从 $\罪^{2} x + \cos^{2}x =1,$

$\ begin {Aligned} \ cos \ alpha＆= \ sqrt {1- \ sin^{2} \ alpha} = \ sqrt {1- \ frac {13^2} {14^2} {14^2}}sqrt {3}} {14}，\\ cos \ beta＆= \ sqrt {1- \ sin^{2} \ beta} = \ sqrt {1- \ frac {11^2} {14^2} {14^2}} {14^2}}}= \ frac {5 \ sqrt {3}} {14}。\ end {Aligned}$

Thus, from cosine-sum formula, we have

$\ begin {aligned} \ cos（\ alpha + \ beta）＆= \ cos \ alpha \ cdot \ cos \ cos \ beta- \ sin \ alpha \ alpha \ cdot \ sin \ sin \ beta \ beta \\＆= \ frac {3 \ sqrt {3 \ sqrt {3} {3} {3} {3} {3}} {14} \ times \ frac {5 \ sqrt {3}} {14} {14} - \ frac {13} {14} {14} \ times \ times \ frac {11} {14} {14} {14} \\＆= - = - \ frac {1} {1} {1} {2} {2}。\ end {Aligned}$

因此，从那以后 $0 <\ alpha + \ beta <\ pi，$we can get $\α+ \beta$如下：

$\begin{aligned} \cos (\alpha + \beta) &= -\frac{1}{2} \\ \Rightarrow \alpha + \beta &= \frac{2}{3} \pi. \ _\square \end{aligned}$

The tangent sum and difference formulas are

$\begin{aligned} \tan(A+B) &= \dfrac{\tan A + \tan B}{1 - \tan A \tan B} \\\\ \tan(A-B) &= \dfrac{\tan A - \tan B}{1 + \tan A \tan B}. \end{aligned}$

得出切线和公式。

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We know that

$\begin{aligned} \sin(a+b) &= \sin a \cos b + \cos a \sin b &\qquad (1) \\ \cos(a+b) &= \cos a \cos b - \sin a \sin b. &\qquad (2) \end{aligned}$

Dividing $(1)$by $(2)$gives

$\dfrac{\sin(a+b)}{\cos(a+b)} = \dfrac{ \sin a \cos b + \cos a \sin b}{\cos a \cos b - \sin a \sin b}.$

Dividing the right side of this by $\ cos a \ cos b$gives

$\ tan（a + b）= \ dfrac {\ dfrac {\ sin a \ cos b} {\ cos a \ cos b} + \ dfrac {\ cos a \ sin \ sin b} {\ cos a \ cos a \ cos b}}} {\ dfrac {\ cos a \ cos b} {\ cos a \ cos b} - \ dfrac {\ sin a \ sin a \ sin b} {\ cos a \ cos a \ cos b}} = \ dfrac {\ tan a + \ tan a + \ tan a + \ tan b}{1- \ tan a \ tan b}，$

这是和形式ula. $_\正方形$

Derive the tangent difference formula.

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We know that

$\ tan（-a）= - \ tan a。$

Putting $b = -b$在切线和公式中，我们得到

$\begin{aligned} \tan\big(a+(-b)\big) &= \dfrac{\tan(a) + \tan(-b)}{1 - \tan a \tan(-b)} \\ \tan(a-b) &= \dfrac{\tan a - \tan b}{1 + \tan a \tan b}. \ _\square \end{aligned}$

These identities are useful for finding the values of tangents of angles which are not known.

找到价值 $\tan 75^{\circ}$。

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We want to break $75^{\ circ}$分为两个角度，我们知道的切线。一个明显的一对是 $（30^{\ circ}，45^{\ circ}）$。

Using the tangent sum formula, we have

$\ begin {aligned} \ tan 75^{\ circ} = \ tan（30^{\ circ} + 45^{\ circ}）＆= \ dfrac {\ tan 30^{\ circ}\ circ}}} {1 - \ tan 30^{\ circ} \ tan 45^{\ circ}}}} \\\\\\\\\\\\\\\\\\\\\\\\\\ frac {\ frac {1} {\ sqrt {\ sqrt {3}}}}}} + 1} {1} {1} {1} {1} {1- \ frac {1} {\ sqrt {3}}} \ cdot 1} \ \\\＆= \ frac {\ frac {\ frac {1 + \ sqrt {3}}} {\ sqrt {\ sqrt {3}}}}}}}}frac {\ sqrt {3} - 1} {\ sqrt {3}} \ \ \} \ \} \\\\\\\\\\\\\\\\\\\\\\\\\\ \ frac {\ sqrt {3} +1} {\ sqrt3-1}。\ _ \ Square \ End {Aligned}$

证明

$\ sin（18^\ circ）= \ frac14 \ big（\ sqrt5-1 \ big）。$

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