Sum and Difference Formulas

The sum and difference formulas state that

\[开始\{对齐}\罪(a + b) & = \ \因为b + \因为罪a \sin b \\ \sin(a-b) &= \sin a \cos b - \cos a \sin b \end{align}\]

and that

\[\begin{align} \cos(a+b) &= \cos a \cos b − \sin a \sin b \\ \cos(a-b) &= \cos a \cos b + \sin a \sin b. \end{align}\]

Derive the sum formulas

\[开始\{对齐}\罪(a + b) & = \ \因为b + \因为罪a \sin b \\ \cos(a+b) &= \cos a \cos b − \sin a \sin b.\end{align}\]

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ByEuler's formulawe know that

\[ e^{i\theta} = \cos \theta + i\sin \theta. \]

If \(\theta = a + b \), then

\[ e^{i(a+b)} = \cos(a+b) + i\sin(a+b). \qquad (1)\]

我们也知道从世博会的代数性质nentials that

\[ e^{i(a+b)} = e^{ia}e^{ib}. \]

Therefore,

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

Now, from \((1)\) we have

\[\begin{align} e^{i(a+b)} &= \cos(a+b) + i\sin(a+b)\\ &=\cos a \cos b - \sin a \sin b + i(\sin a \cos b + \sin b \cos a). \end{align}\]

But since \( \cos \) and \( \sin \) are real-valued functions, it must be true that

\[\begin{align} \cos(a+b) &= \cos a \cos b - \sin a \sin b \\ i\sin(a+b) &= i(\sin a \cos b + \sin b \cos a), \end{align} \]

which implies

\[\begin{align} \cos(a+b) &= \cos a\cos b - \sin a \sin b \\ \sin(a+b) &= \sin a \cos b + \sin b \cos a. \ _\square \end{align}\]

In the diagram, let point \(A\) revolve to points \(B\) and \(C,\) and let the angles \(\alpha\) and \(\beta\) be defined as follows:

\[ \angle AOB = \alpha, \quad \angle BOC = \beta. \]

Also, let both \(\overline{CD}\) and \(\overline{FG}\) be perpendicular to \(\overline{OA},\) and let \(E\) be a point on \(\overline{CD}\) such that \(\lvert \overline{ED}\rvert=\lvert\overline{FG}\rvert.\)

Then the formula for cosine-sum \( \cos (\alpha+\beta) ,\) which is \( \frac{\lvert \overline{OD}\rvert}{\lvert \overline{OC}\rvert} ,\) can be obtained as follows:

\[开始\{对齐}\ cos(\ \α+β)& = \压裂{\ lvert \overline{OD}\rvert}{\lvert \overline{OC}\rvert} \\ &= \frac{\lvert \overline{OG}\rvert}{\lvert \overline{OC}\rvert} - \frac{\lvert \overline{EF}\rvert}{\lvert \overline{OC}\rvert} \\ &= \frac{\lvert \overline{OG}\rvert}{\lvert \overline{OF}\rvert} \cdot \frac{\lvert \overline{OF}\rvert}{\lvert \overline{OC}\rvert} - \frac{\lvert \overline{EF}\rvert}{\lvert \overline{CF}\rvert} \cdot \frac{\lvert \overline{CF}\rvert}{\lvert \overline{OC}\rvert} \qquad \left(\text{since } \lvert \overline{OD}\rvert = \lvert \overline{OG}\rvert-\lvert \overline{EF}\rvert\right)\\ &= \cos \alpha \cdot \cos \beta - \sin \alpha \cdot \sin \beta. \end{align} \]

The cosine-difference formula can be obtained from the cosine-sum formula by replacing \(\beta\) with \( - \beta,\) and using \(\cos( -\beta) = \cos \beta\) and \(\sin(-\beta) = -\sin \beta:\)

\[ \begin{align} \cos(\alpha + \beta) &= \cos \alpha \cdot \cos \beta - \sin \alpha \cdot \sin \beta \\ \Rightarrow \cos(\alpha - \beta) &= \cos \alpha \cdot \cos (-\beta) - \sin \alpha \cdot \sin (-\beta) \\ &= \cos \alpha \cdot \cos \beta + \sin \alpha \cdot \sin \beta. \end{align} \]

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

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

Cosine-difference formula: \[\cos(\alpha - \beta) = \cos \alpha \cdot \cos \beta + \sin \alpha \cdot \sin \beta .\]

What is \( \cos 75^\circ ?\)

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From cosine-sum formula, we have

\[ \begin{align} \cos 75^\circ &= \cos (45^\circ + 30^\circ) \\ &= \cos 45^\circ \cdot \cos 30^\circ - \sin 45^\circ \cdot \sin 30^\circ \\ &= \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2} \cdot \frac{1}{2} \\ &= \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4} \\ &= \frac{\sqrt{6}-\sqrt{2}}{4}.\ _\square \end{align} \]

What is \( \cos 15^\circ ?\)

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From cosine-difference formula, we have

\[ \begin{align} \cos 15^\circ &= \cos (45^\circ - 30^\circ) \\ &= \cos 45^\circ \cdot \cos 30^\circ + \sin 45^\circ \cdot \sin 30^\circ \\ &= \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \cdot \frac{1}{2} \\ &= \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} \\ &= \frac{\sqrt{6}+\sqrt{2}}{4}.\ _\square \end{align} \]

What is \( \cos 105^\circ ?\)

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From cosine-sum formula, \( \cos 105^\circ \) is

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

Simplify

\[ \cos 140^\circ \cdot \cos 50^\circ + \sin 140^\circ \cdot \sin 50^\circ .\]

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From cosine-difference formula, we have

\[ \begin{align} \cos 140^\circ \cdot \cos 50^\circ + \sin 140^\circ \cdot \sin 50^\circ &= \cos(140^\circ-50^\circ) \\ &= \cos 90^\circ \\ &= 0. \ _\square \end{align} \]

If \( \sin \alpha = \frac{13}{14}\) and \(\sin \beta = \frac{11}{14} \) for \( 0 < \alpha < \frac{\pi}{2}\) and \(0< \beta < \frac{\pi}{2} ,\) what is \( \alpha + \beta ?\)

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Since \( \sin^{2} x + \cos^{2}x =1,\)

\[ \begin{align} \cos \alpha &= \sqrt{1-\sin^{2}\alpha} = \sqrt{1-\frac{13^2}{14^2}} = \frac{3\sqrt{3}}{14}, \\ \cos \beta &= \sqrt{1-\sin^{2}\beta} = \sqrt{1-\frac{11^2}{14^2}} = \frac{5\sqrt{3}}{14}. \end{align} \]

Thus, from cosine-sum formula, we have

\[ \begin{align} \cos (\alpha + \beta) &= \cos \alpha \cdot \cos \beta - \sin \alpha \cdot \sin \beta \\ &= \frac{3\sqrt{3}}{14} \times \frac{5\sqrt{3}}{14} - \frac{13}{14} \times \frac{11}{14} \\ &= -\frac{1}{2}. \end{align} \]

Hence, since \( 0< \alpha + \beta < \pi ,\) we can get \( \alpha + \beta \) as follows:

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

The tangent sum and difference formulas are

\[\begin{align} \tan(A+B) &= \dfrac{\tan A + \tan B}{1 - \tan A \tan B} \\\\ \tan(A-B) &= \dfrac{\tan A - \tan B}{1 + \tan A \tan B}. \end{align} \]

Derive the tangent sum formula.

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We know that

\[开始\{对齐}\罪(a + b) & = \ \因为b + \因为罪a \sin b &\qquad (1) \\ \cos(a+b) &= \cos a \cos b − \sin a \sin b. &\qquad (2) \end{align}\]

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

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

which is the sum formula. \(_\square\)

Derive the tangent difference formula.

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We know that

\[ \tan(-a) = -\tan a .\]

Putting \( b = -b \) in the tangent sum formula, we get

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

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

Find the value of \( \tan 75^{\circ} \).

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We want to break \( 75^{\circ} \) into two angles whose tangents we know. One obvious pair is \( (30^{\circ}, 45^{\circ}) \).

Using the tangent sum formula, we have

\[\begin{align} \tan 75^{\circ} = \tan( 30^{\circ} + 45^{\circ}) &= \dfrac{\tan 30^{\circ} + \tan 45^{\circ}}{1 - \tan 30^{\circ} \tan 45^{\circ}} \\\\ &= \frac{ \frac{1}{\sqrt{3}} + 1}{1 - \frac{1}{\sqrt{3}}\cdot 1} \\\\ &= \frac{\frac{1 + \sqrt{3}}{\sqrt{3}}}{\ \ \ \frac{\sqrt{3} - 1}{\sqrt{3}}\ \ \ } \\\\ &= \frac{\sqrt{3}+1}{\sqrt3 - 1}. \ _\square \end{align} \]

Prove that

\[\sin(18^\circ) = \frac14\big(\sqrt5-1\big).\]

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