Course: 3.13 Sine 和 Consine 产品公式总和

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Sine 和 Consine 产品公式总和
Can you solve problems that involve the sum of sines or cosines? For example, consider the equation:
::您能否解决涉及正弦或余弦之和的问题?例如,考虑公式:

$\begin{align*}\cos 10t + \cos 3t\end{align*}$
::cos% 10t+cos% 3t

You could just compute each expression separately and add their values at the end. However, there is an easier way to do this. You can simplify the equation first, and then solve.
::您可以单独计算每个表达式, 并在结尾处添加它们的值。但是, 这样做比较容易。您可以先简化方程, 然后解析。

Sine and Cosine Sum to Product Formulas
::Sine 和 Coine 与产品公式之和

In some problems, the product of two trigonometric functions is more conveniently found by the sum of two trigonometric functions by use of identities.
::在某些问题上,两种三角函数的产物更方便地通过使用身份的两种三角函数之和找到。

Here is an example:
::以下是一个例子:

$\begin{aligned} \sin α + \sin β = 2 \sin \frac{α + β}{2} \times \cos \frac{α - β}{2} \end{aligned}$
$\begin{align*}\sin \alpha + \sin \beta = 2 \sin \frac{\alpha + \beta}{2} \times \cos \frac{\alpha - \beta}{2}\end{align*}$
::{\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}不! {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}不! {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不

This can be verified by using the :
::可以通过使用 :

$\begin{aligned} 2 \sin \frac{α + β}{2} \cos \frac{α - β}{2} \\ = 2 [\begin{matrix} \sin (\frac{α}{2} + \frac{β}{2}) \cos (\frac{α}{2} - \frac{β}{2}) \end{matrix}] \\ = 2 [\begin{matrix} (\sin \frac{α}{2} \cos \frac{β}{2} + \cos \frac{α}{2} \sin \frac{β}{2}) (\cos \frac{α}{2} \cos \frac{β}{2} + \sin \frac{α}{2} \sin \frac{β}{2}) \end{matrix}] \\ = 2 [\begin{matrix} \sin \frac{α}{2} \cos \frac{α}{2} \cos^{2} \frac{β}{2} + \sin^{2} \frac{α}{2} \sin \frac{β}{2} \cos \frac{β}{2} + \sin \frac{β}{2} \cos^{2} \frac{α}{2} \cos \frac{β}{2} + \sin \frac{α}{2} \sin^{2} \frac{β}{2} \cos \frac{α}{2} \end{matrix}] \\ = 2 [\begin{matrix} \sin \frac{α}{2} \cos \frac{α}{2} (\sin^{2} \frac{β}{2} + \cos^{2} \frac{β}{2}) + \sin \frac{β}{2} \cos \frac{β}{2} (\sin^{2} \frac{α}{2} + \cos^{2} \frac{α}{2}) \end{matrix}] \\ = 2 [\begin{matrix} \sin \frac{α}{2} \cos \frac{α}{2} + \sin \frac{β}{2} \cos \frac{β}{2} \end{matrix}] \\ = 2 \sin \frac{α}{2} \cos \frac{α}{2} + 2 \sin \frac{β}{2} \cos \frac{β}{2} \\ = \sin (2 \cdot \frac{α}{2}) + \sin (2 \cdot \frac{β}{2}) \\ = \sin α + \sin β \end{aligned}$
$\begin{align*}& 2 \sin \frac{\alpha + \beta}{2} \cos \frac{\alpha - \beta}{2}\\ &= 2 \begin{bmatrix} \sin \left( \frac{\alpha}{2} + \frac{\beta}{2} \right) \cos \left(\frac{\alpha}{2} - \frac{\beta}{2} \right) \end{bmatrix} \\ &= 2 \begin{bmatrix} \left( \sin \frac{\alpha}{2} \cos \frac{\beta}{2} + \cos \frac{\alpha}{2} \sin \frac{\beta}{2} \left) \right( \cos \frac{\alpha}{2} \cos \frac{\beta}{2} + \sin \frac{\alpha}{2} \sin \frac{\beta}{2} \right ) \end{bmatrix}\\ &= 2 \begin{bmatrix} \sin \frac{\alpha}{2} \cos \frac{\alpha}{2} \cos ^2 \frac{\beta}{2} + \sin ^2 \frac{\alpha}{2} \sin \frac{\beta}{2} \cos \frac{\beta}{2} + \sin \frac{\beta}{2} \cos ^2 \frac{\alpha}{2} \cos \frac{\beta}{2} + \sin \frac{\alpha}{2} \sin ^2 \frac{\beta}{2} \cos \frac{\alpha}{2} \end{bmatrix}\\ &= 2 \begin{bmatrix} \sin \frac{\alpha}{2} \cos \frac{\alpha}{2} \left( \sin^2 \frac{\beta}{2} + \cos^2 \frac{\beta}{2} \right) + \sin \frac{\beta}{2} \cos \frac{\beta}{2} \left( \sin^2 \frac{\alpha}{2} + \cos^2 \frac{\alpha}{2} \right) \end{bmatrix}\\ &= 2 \begin{bmatrix} \sin \frac{\alpha}{2} \cos \frac{\alpha}{2} + \sin \frac{\beta}{2} \cos \frac{\beta}{2} \end{bmatrix}\\ &= 2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2} + 2 \sin \frac{\beta}{2} \cos \frac{\beta}{2}\\ &= \sin \left( 2 \cdot \frac{\alpha}{2} \right) + \sin \left( 2 \cdot \frac{\beta}{2} \right)\\ &= \sin \alpha + \sin \beta\end{align*}$
::222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222212222222222222222222222222222222211222221111111

The following variations can be derived similarly:
::以下差异可以得出类似的变化:

$\begin{aligned} \sin α - \sin β & = 2 \sin \frac{α - β}{2} \times \cos \frac{α + β}{2} \\ \cos α + \cos β & = 2 \cos \frac{α + β}{2} \times \cos \frac{α - β}{2} \\ \cos α - \cos β & = - 2 \sin \frac{α + β}{2} \times \sin \frac{α - β}{2} \end{aligned}$
$\begin{align*} \sin \alpha - \sin \beta &= 2 \sin \frac{\alpha - \beta}{2} \times \cos \frac{\alpha + \beta}{2}\\ \cos \alpha + \cos \beta &= 2 \cos \frac{\alpha + \beta}{2} \times \cos \frac{\alpha - \beta}{2}\\ \cos \alpha - \cos \beta &= -2 \sin \frac{\alpha + \beta}{2} \times \sin \frac{\alpha - \beta}{2}\\\end{align*}$
::2xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

Here are some problems using this type of transformation from a sum of terms to a product of terms.
::使用这种从条件总和转换成条件产物的转变,这里有一些问题。

1. Change $\begin{align*}\sin 5x - \sin 9x\end{align*}$ into a product.
::1. 将sin_5x-sin_sin_9x改变为产品。

Use the formula $\begin{align*}\sin \alpha - \sin \beta = 2 \sin \frac{\alpha - \beta}{2} \times \cos \frac{\alpha + \beta}{2}.\end{align*}$
::使用公式sin%2sin%2xcos%2。

$\begin{aligned} \sin 5 x - \sin 9 x & = 2 \sin \frac{5 x - 9 x}{2} \cos \frac{5 x + 9 x}{2} \\ = 2 \sin (- 2 x) \cos 7 x \\ = - 2 \sin 2 x \cos 7 x \end{aligned}$
$\begin{align*} \sin 5x - \sin 9x &= 2 \sin \frac{5x - 9x}{2} \cos \frac{5x + 9x}{2}\\ &= 2 \sin (-2x) \cos 7x\\ &= -2 \sin 2x \cos 7x\end{align*}$
::5x-sin%9x=2sin @5x-9x2cos @5x+9x2=2sin @(-2x)cos @7x%2sin @2xcosin=2xcos@2xcos=7xxx

2. Change $\begin{align*}\cos(-3x) + \cos 8x\end{align*}$ into a product.
::2. 将cos(-3x)+cos8x改变为产品。

Use the formula $\begin{align*} \cos \alpha + \cos \beta = 2 \cos \frac{\alpha + \beta}{2} \times \cos \frac{\alpha - \beta}{2}\end{align*}$
::使用公式 cosçç2cos2xcos2

$\begin{aligned} \cos (- 3 x) + \cos (8 x) & = 2 \cos \frac{- 3 x + 8 x}{2} \cos \frac{- 3 x - 8 x}{2} \\ = 2 \cos (2.5 x) \cos (- 5.5 x) \\ = 2 \cos (2.5 x) \cos (5.5 x) \end{aligned}$
$\begin{align*}\cos (-3x) + \cos (8x) &= 2 \cos \frac{-3x + 8x}{2} \cos \frac {-3x - 8x}{2}\\ &= 2 \cos (2.5x) \cos (-5.5x)\\ &= 2 \cos (2.5x) \cos (5.5x)\end{align*}$
:- 3x) +cos = 2cos = 3x + 8x2cos = 3x + 8x2x2= 2cos = (2.5x) = (2.5x) cos = (-5.5x) = 2cos = (2.5x) cos = (5.5x)

3. Change $\begin{align*}2 \sin 7x \cos 4x\end{align*}$ to a sum.
::3. 将2sin7xcos4x改为总和。

This is the reverse of what was done in the previous two examples. Looking at the four formulas above, take the one that has as a product, $\begin{align*}\sin \alpha + \sin \beta = 2 \sin \frac{\alpha + \beta}{2} \times \cos \frac{\alpha - \beta}{2}.\end{align*}$ Therefore, $\begin{align*}7x = \frac{\alpha + \beta}{2}\end{align*}$ and $\begin{align*}4x = \frac{\alpha - \beta}{2}\end{align*}$ .
::这是前两个例子的反差。看一下上面的四种公式, 取一个作为产品的公式, sin2sin2xxcos2。因此, 7x2 和 4x2 。

$\begin{aligned} 7 x & = \frac{α + β}{2} & 4 x = \frac{α - β}{2} \\ and \\ 14 x & = α + β & 8 x = α - β \\ α = 14 x - β & 8 x = [14 x - β] - β \\ so \\ - 6 x = - 2 β \\ 3 x = β \\ α = 14 x - 3 x \\ α = 11 x \end{aligned}$
$\begin{align*}7x & = \frac{\alpha + \beta}{2} &&&& 4x = \frac{\alpha - \beta}{2} \\ &&& \text{and} \\ 14x & = \alpha+\beta &&&& 8x= \alpha - \beta \\ &\alpha = 14x - \beta &&&& 8x=[14x-\beta]-\beta \\ &&& \text{so} \\ &&&&&-6x = -2\beta\\ &&&&&3x=\beta\\ \alpha=14x-3x\\ \alpha=11x\end{align*}$
::7x24x2和14x8x14x8x=[14x]so-6x23x}14x3x_1x11x

So, this translates to $\begin{align*}\sin(11x) + \sin(3x)\end{align*}$ . A shortcut for this problem, would be to notice that the sum of $\begin{align*}7x\end{align*}$ and $\begin{align*}4x\end{align*}$ is $\begin{align*}11x\end{align*}$ and the difference is $\begin{align*}3x\end{align*}$ .
::因此,这相当于sin( 11x) +sin( 3x) 。这个问题的一个捷径是注意 7x 和 4x 的总和是 11x , 差额是 3x 。

Examples
::实例

Example 1
::例1

Earlier, you were asked to solve
::早些时候,有人要求你解决

$\begin{align*}\cos 10t + \cos 3t\end{align*}$
::cos% 10t+cos% 3t

You can easily transform this equation into a product of two trig functions using:
::您可以很容易地将这个方程式转换成两个三角函数的产物,使用:

$\begin{align*}\cos \alpha + \cos \beta = 2 \cos \frac{\alpha + \beta}{2} \times \cos \frac{\alpha - \beta}{2}\end{align*}$
::cos=2cos=2cos=2xcos=2 COs=2

Substituting the known quantities:
::替代已知数量:

$\begin{align*}\cos 10t + \cos 3t = 2 \cos \frac{13t}{2} \times \cos \frac{7t}{2} = 2\cos(6.5t) \cos(3.5t)\end{align*}$
::cos10t+cos3t=2cos13t2×cos7t2=2cos(6.5t)cos(3.5t)

Example 2
::例2

Express the sum as a product: $\begin{align*}\sin 9x + \sin 5x\end{align*}$
::以产品表示总和:sin=9x+sin=5x

Using the sum-to-product formula:
::使用总和对产品公式:

$\begin{aligned} \sin 9 x + \sin 5 x \\ 2 (\sin (\frac{9 x + 5 x}{2}) \cos (\frac{9 x - 5 x}{2})) \\ 2 \sin 7 x \cos 2 x \end{aligned}$
$\begin{align*}& \sin 9x + \sin 5x\\ & 2 \left(\sin \left(\frac{9x + 5x}{2} \right) \cos \left(\frac{9x - 5x}{2} \right) \right)\\ & 2 \sin 7x \cos 2x\end{align*}$
:9x+5x2)2sin\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

Example 3
::例3

Express the difference as a product: $\begin{align*}\cos 4y - \cos 3y \end{align*}$
::以产品表示差异:cos4y-cos3y

Using the difference-to-product formula:
::使用差异对产品公式:

$\begin{aligned} \cos 4 y - \cos 3 y \\ - 2 \sin (\frac{4 y + 3 y}{2}) \sin (\frac{4 y - 3 y}{2}) \\ - 2 \sin \frac{7 y}{2} \sin \frac{y}{2} \end{aligned}$
$\begin{align*}& \cos 4y - \cos 3y\\ & -2 \sin \left(\frac{4y + 3y}{2} \right) \sin \left(\frac{4y - 3y}{2} \right)\\ & -2 \sin \frac{7y}{2} \sin \frac{y}{2}\end{align*}$
::============================================================================================

Example 4
::例4

Verify the identity (using sum-to-product formula): $\begin{align*}\frac{\cos 3a - \cos 5a}{\sin 3a - \sin 5a} = - \tan 4a\end{align*}$
::验证身份( 使用总和到产品公式): cos @ 3a- cos @ 5asin_ 3a- sin @ 5a_ tan @ 4a

Using the difference-to-product formulas:
::使用差异对产品公式:

$\begin{aligned} \frac{\cos 3 a - \cos 5 a}{\sin 3 a - \sin 5 a} = - \tan 4 a \\ \frac{- 2 \sin (\frac{3 a + 5 a}{2}) \sin (\frac{3 a - 5 a}{2})}{2 \sin (\frac{3 a - 5 a}{2}) \cos (\frac{3 a + 5 a}{2})} \\ - \frac{\sin 4 a}{\cos 4 a} \\ - \tan 4 a \end{aligned}$
$\begin{align*}& \qquad \qquad \frac{\cos 3a - \cos 5a}{\sin 3a - \sin 5a} = - \tan 4a\\ & \frac{-2 \sin \left (\frac{3a + 5a}{2} \right ) \sin \left (\frac {3a - 5a}{2} \right )}{2 \sin \left (\frac{3a - 5a}{2} \right ) \cos \left(\frac{3a + 5a}{2} \right)}\\ & \qquad \qquad \qquad \ \ - \frac{\sin 4a}{\cos 4a}\\ & \qquad \qquad \qquad \ \ - \tan 4a\end{align*}$
::=====================================================================================================================================================~=======================================================================================================================================================================================================================================================================================================================================================================

Review
::回顾

Change each sum or difference into a product.
::将每个总和或差异转化为产品。

$\begin{align*}\sin 3x + \sin 2x\end{align*}$
::3x+sin#%2x sin *%3x+sin%2x

$\begin{align*}\cos 2x + \cos 5x\end{align*}$
::COs=2x+COs=5x COs=2x+COs=5x

$\begin{align*}\sin (-x) - \sin 4x\end{align*}$
:-x)-(sin) 4x

$\begin{align*}\cos 12x + \cos 3x\end{align*}$
::COs% 12x+COs%3x

$\begin{align*}\sin 8x - \sin 4x\end{align*}$
::8x-sin4x

$\begin{align*}\sin x + \sin \frac{1}{2}x\end{align*}$
::sinx+sin+sin=12x

$\begin{align*}\cos 3x - \cos (-3x)\end{align*}$
::cos3x-cos(- 3x)

Change each product into a sum or difference.
::将每种产品变成一个总和或差异。

$\begin{align*}-2\sin 3.5x \sin 2.5x\end{align*}$
::-2sin=3.5xsin=2.5xxx=2.5xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

$\begin{align*}2\cos 3.5x \sin 0.5x\end{align*}$
::2cos=3.5xsin=0.5x

$\begin{align*}2\cos 3.5x \cos 5.5x\end{align*}$
::2cos=3.5xcos=5.5x

$\begin{align*}2\sin 6x \cos 2x\end{align*}$
::2sin 6xxcos 6xxcos 2x

$\begin{align*}-2\sin 3x \sin x\end{align*}$
::~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

$\begin{align*}2\sin 4x \cos x\end{align*}$
::2sin 4xcosxxxx 2sin 4xcosxxxxx

Show that $\begin{align*}\cos\frac{A+B}{2}\cos\frac{A-B}{2}=\frac{1}{2}(\cos A + \cos B)\end{align*}$ .
::显示 cosA+B2cosA-B2=12(cosA+cosB)。

Let $\begin{align*}u=\frac{A+B}{2}\end{align*}$ and $\begin{align*}v=\frac{A-B}{2}\end{align*}$ . Show that $\begin{align*}\cos u\cos v =\frac{1}{2}(\cos (u+v)+\cos(u-v)).\end{align*}$
::Letu = A+B2 和 v= A-B2. 显示 cosucosv= 12 (cos(u+v)+cos(u-v))。

Review (Answers)
::回顾(答复)

Click to see the answer key or go to the Table of Contents and click on the Answer Key under the 'Other Versions' option.
::单击可查看答题键, 或转到目录中, 单击“ 其他版本” 选项下的答题键。

Section outline

General

Sine 和 Consine 产品公式总和

Sine and Cosine Sum to Product Formulas ::Sine 和 Coine 与产品公式之和

Examples ::实例

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

Example 4 ::例4

Review ::回顾

Review (Answers) ::回顾(答复)

Sine and Cosine Sum to Product Formulas
::Sine 和 Coine 与产品公式之和

Examples
::实例

Example 1
::例1

Example 2
::例2

Example 3
::例3

Example 4
::例4

Review
::回顾

Review (Answers)
::回顾(答复)