Course: 14.14 使用总和和差异公式解决细数

Section outline

Select section General

Collapse Expand
General

Collapse all Expand all
- Select activity 新闻通告
  
  新闻通告 Forum
Select section 使用总和和差异公式解决Trig 公式

Collapse Expand
使用总和和差异公式解决Trig 公式
As Agent Trigonometry, you are given a piece of the puzzle: $\begin{align*}\sin(\frac{\pi}{2} - x) = -1\end{align*}$ . What is the value of $\begin{align*}x\end{align*}$ ?
::作为Trigonology探员,你被赋予了拼图的一部分:sin(2-x)1.x值是多少?

Solving Trigonometric Functions
::解决三角函数

W e can use the to solve trigonometric equations. For this concept, we will only find solutions in the interval $\begin{align*}0\le x <2\pi\end{align*}$ .
::我们可以用它来解决三角方程。对于这个概念, 我们只能在 0.x < 2 间距内找到解决方案。

Solve the function using the sum and difference formulas: $\begin{align*}\cos (x-\pi)=\frac{\sqrt{2}}{2}\end{align*}$
::使用总和和差数公式:cos(x)=22 来解析函数

Use the formula to simplify the left-hand side and then solve for $\begin{align*}x\end{align*}$ .
::使用公式简化左手侧,然后解析 x。

$\begin{aligned} \cos (x - π) & = \frac{\sqrt{2}}{2} \\ \cos x \cos π + \sin x \sin π & = \frac{\sqrt{2}}{2} \\ - \cos x & = \frac{\sqrt{2}}{2} \\ \cos x & = - \frac{\sqrt{2}}{2} \end{aligned}$
$\begin{align*}\cos (x-\pi) &=\frac{\sqrt{2}}{2} \\ \cos x \cos \pi +\sin x \sin \pi &=\frac{\sqrt{2}}{2} \\ -\cos x &=\frac{\sqrt{2}}{2}\\ \cos x &=-\frac{\sqrt{2}}{2}\end{align*}$
::COS(x) = 22cos xcos sin xsin 22-cosx=22cos x 22

The cosine negative in the $\begin{align*}2^{nd}\end{align*}$ and $\begin{align*}3^{rd}\end{align*}$ quadrants. $\begin{align*}x=\frac{3\pi}{4}\end{align*}$ and $\begin{align*}\frac{5\pi}{4}\end{align*}$ .
::第2和第3四舍方的余弦阴性。 x=34和54。

Solve the function using the sum and difference formulas: $\begin{align*}\sin \left(x+\frac{\pi}{4}\right)+1=\sin \left(\frac{\pi}{4}-x\right)\end{align*}$
::使用总和和差数公式来解析函数:sin(x4)+1=sin(4-x)

$\begin{aligned} \sin (x + \frac{π}{4}) + 1 & = \sin (\frac{π}{4} - x) \\ \sin x \cos \frac{π}{4} + \cos x \sin \frac{π}{4} + 1 & = \sin \frac{π}{4} \cos x - \cos \frac{π}{4} \sin x \\ \sin x \cdot \frac{\sqrt{2}}{2} + \cos x \cdot \frac{\sqrt{2}}{2} + 1 & = \frac{\sqrt{2}}{2} . \cos x - \frac{\sqrt{2}}{2} \cdot \sin x \\ \sqrt{2} \sin x & = - 1 \\ \sin x & = - \frac{1}{\sqrt{2}} = - \frac{\sqrt{2}}{2} \end{aligned}$
$\begin{align*}\sin \left(x+\frac{\pi}{4}\right)+1 &=\sin \left(\frac{\pi}{4}-x\right) \\ \sin x \cos \frac{\pi}{4}+\cos x \sin \frac{\pi}{4}+1 &=\sin \frac{\pi}{4}\cos x -\cos \frac{\pi}{4}\sin x \\ \sin x \cdot \frac{\sqrt{2}}{2}+\cos x \cdot \frac{\sqrt{2}}{2}+1 &=\frac{\sqrt{2}}{2}.\cos x -\frac{\sqrt{2}}{2} \cdot \sin x \\ \sqrt{2} \sin x &=-1 \\ \sin x &=-\frac{1}{\sqrt{2}}=-\frac{\sqrt{2}}{2}\end{align*}$
::=============================================================================================================================================================================================================================================================================================================================================================================================================================================================================================================================

In the interval, $\begin{align*}x=\frac{5\pi}{4}\end{align*}$ and $\begin{align*}\frac{7\pi}{4}\end{align*}$ .
::在间隔内, x=54 和 74。

Solve the function using the sum and difference formulas: $\begin{align*}2\sin \left(x+\frac{\pi}{3}\right)=\tan \frac{\pi}{3}\end{align*}$
::使用总和和差数公式解决函数 : 2sin(x3) =tan3

$\begin{aligned} 2 \sin (x + \frac{π}{3}) & = \tan \frac{π}{3} \\ 2 (\sin x \cos \frac{π}{3} + \cos x \sin \frac{π}{3}) & = \sqrt{3} \\ 2 \sin x \cdot \frac{1}{2} + 2 \cos x \cdot \frac{\sqrt{3}}{2} & = \sqrt{3} \\ \sin x + \sqrt{3} \cos x & = \sqrt{3} \\ \sin x & = \sqrt{3} (1 - \cos x) \\ \sin^{2} x & = 3 (1 - 2 \cos x + \cos^{2} x) square both sides \\ 1 - \cos^{2} x & = 3 - 6 \cos x + 3 \cos^{2} x substitute \sin^{2} x = 1 - \cos^{2} x \\ 0 & = 4 \cos^{2} x - 6 \cos x + 2 \\ 0 & = 2 \cos^{2} x - 3 \cos x + 1 \end{aligned}$
$\begin{align*}2\sin\left(x+\frac{\pi}{3}\right) &=\tan \frac{\pi}{3} \\ 2\left(\sin x \cos \frac{\pi}{3}+\cos x \sin \frac{\pi}{3}\right) &=\sqrt{3} \\ 2\sin x \cdot \frac{1}{2}+2\cos x \cdot \frac{\sqrt{3}}{2} &=\sqrt{3} \\ \sin x +\sqrt{3}\cos x &=\sqrt {3} \\ \sin x &=\sqrt{3}(1-\cos x) \\ \sin ^2x &=3(1-2\cos x+\cos ^2x) \qquad \text{square both sides} \\ 1-\cos ^2x &=3-6\cos x + 3\cos ^2 x \qquad \ \ \text{substitute} \ \sin ^2{x} = 1-\cos ^2x \\ 0 &=4\cos^2x-6\cos x +2 \\ 0 &=2\cos^2x-3\cos x +1\end{align*}$
::2sin(x3) =tan}32(sinxcosos) =32sinx12+2cos2x32=3sinx+3cososxx=3sin3sin2xx=3(1-cosx+cos2x) x=3(1-2cos2x=3-6cosx+3cos2}3+cos%3+3cos2x =2x 替代sin2xx_x0=4cos2}}x_6cosx=20cos2x_x3cos_x3xxxx+1

At this point, we can factor the equation to be $\begin{align*}(2\cos x -1)(\cos x -1)=0\end{align*}$ . $\begin{align*}\cos x =\frac{1}{2}\end{align*}$ , and 1, so $\begin{align*}x=0, \frac{\pi}{3}, \frac{5 \pi}{3}\end{align*}$ . Be careful with these answers. When we check these solutions it turns out that $\begin{align*}\frac{5\pi}{3}\end{align*}$ does not work.
::在这一点上,我们可以将方程乘以 (2cosx-1)(cosx-1) = 0. cosx=12, 和 1, 所以 x=0, 3- 5}3. 注意这些答案。当我们检查这些解决方案时, 结果发现 53 无效。

$\begin{aligned} 2 \sin (\frac{5 π}{3} + \frac{π}{3}) & = \tan \frac{π}{3} \\ 2 \sin 2 π & = \sqrt{3} \\ 0 & \neq \sqrt{3} \end{aligned}$
$\begin{align*}2\sin\left(\frac{5\pi}{3}+\frac{\pi}{3}\right) &=\tan \frac{\pi}{3} \\ 2\sin 2\pi &=\sqrt{3} \\ 0 &\ne \sqrt{3}\end{align*}$
::2sin(533333333333333333333333333333333333333333333333333333333333333333333333333

Therefore, $\begin{align*}\frac{5\pi}{3}\end{align*}$ is an extraneous solution.
::因此,5+3是一个不相干的解决办法。

Examples
::实例

Example 1
::例1

Earlier, you were asked to find the value of x from the equation $\begin{align*}\sin(\frac{\pi}{2} - x) = -1\end{align*}$ .
::早些时候,有人要求您从公式sin(% 2- x)\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ x的方程式中找到 x的值。

First, simplify the expression $\begin{align*}\sin(\frac{\pi}{2} - x)\end{align*}$ as:
::首先,将sin(%2-x)的表达式简化为:

$\begin{aligned} \sin (\frac{π}{2} - x) = \sin \frac{π}{2} \cos x - \cos \frac{π}{2} \sin x \\ = 1 \cdot \cos x - 0 \cdot \sin x \\ = c o s x \end{aligned}$
$\begin{align*}\sin(\frac{\pi}{2} - x)=\sin \frac{\pi}{2} \cos x - \cos \frac{\pi}{2} \sin x \\ &=1\cdot \cos x - 0\cdot \sin x \\ &=cos x\end{align*}$
:% 2 - x) =sin *% 2cos *x -cos *2sin *x=1cos *x -0sin*x=cosx

So what you're now looking for is the value of $\begin{align*}x\end{align*}$ where $\begin{align*}\cos x = -1\end{align*}$ .
::所以,现在你正在寻找的是 x 的值, 在哪里cosx1 。

The cosine of $\begin{align*}180^\circ\end{align*}$ is equal to $\begin{align*}-1\end{align*}$ .
::180的余弦等于-1。

Example 2
::例2

Solve in the interval $\begin{align*}0\le x<2\pi:\end{align*}$ $\begin{align*}\cos(2\pi - x)=\frac{1}{2}\end{align*}$
::在间隔 0x<2: cos( 2x) =12 中解决

$\begin{aligned} \cos (2 π - x) & = \frac{1}{2} \\ \cos 2 π \cos x + \sin 2 π \sin x & = \frac{1}{2} \\ \cos x & = \frac{1}{2} \\ x & = \frac{π}{3} a n d \frac{5 π}{3} \end{aligned}$
$\begin{align*}\cos (2\pi-x) &=\frac{1}{2} \\ \cos 2\pi \cos x +\sin 2 \pi \sin x &=\frac{1}{2} \\ \cos x &=\frac{1}{2} \\ x &=\frac{\pi}{3} \ and \ \frac{5\pi}{3}\end{align*}$
::coms( 2x) = 12cos @ 2cos @ x+sin @ 2sin @ x= 12cosx= 12x}3 和 53

Example 3
::例3

Solve in the interval $\begin{align*}0\le x<2\pi:\end{align*}$ $\begin{align*}\sin \left(\frac{\pi}{6}-x\right)+1 = \sin \left(x+\frac{\pi}{6}\right)\end{align*}$
::在间隔 0x<2: sin( 6- x) +1=sin( x6) 中解决

$\begin{aligned} \sin (\frac{π}{6} - x) + 1 & = \sin (x + \frac{π}{6}) \\ \sin \frac{π}{6} \cos x - \cos \frac{π}{6} \sin x + 1 & = \sin x \cos \frac{π}{6} + \cos x \sin \frac{π}{6} \\ \frac{1}{2} \cos x - \frac{\sqrt{3}}{2} \sin x + 1 & = \frac{\sqrt{3}}{2} \sin x + \frac{1}{2} \cos x \\ 1 & = \sqrt{3} \sin x \\ \frac{1}{\sqrt{3}} & = \sin x \\ x & = \sin^{- 1} (\frac{1}{\sqrt{3}}) = 0.6155 a n d 2.5261 rad \end{aligned}$
$\begin{align*}\sin \left(\frac{\pi}{6}-x\right)+1 &=\sin \left(x+\frac{\pi}{6}\right)\\ \sin \frac{\pi}{6}\cos x-\cos \frac{\pi}{6}\sin x +1 &=\sin x \cos \frac{\pi}{6}+\cos x \sin \frac{\pi}{6} \\ \frac{1}{2}\cos x - \frac{\sqrt{3}}{2}\sin x+1 &=\frac{\sqrt{3}}{2}\sin x +\frac{1}{2} \cos x \\ 1 &=\sqrt{3}\sin x \\ \frac{1}{\sqrt{3}} &=\sin x \\ x &=\sin ^{-1}\left(\frac{1}{\sqrt{3}}\right)=0.6155 \ and \ 2.5261 \ \text{rad}\end{align*}$
::\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ -\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\可以\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

Example 4
::例4

Solve in the interval $\begin{align*}0\le x<2\pi:\end{align*}$ $\begin{align*}\cos \left(\frac{\pi}{2}+x\right)=\tan \frac{\pi}{4}\end{align*}$
::在间隔 0x<2: cos( 2+x) =tan4 中解决

$\begin{aligned} \cos (\frac{π}{2} + x) & = \tan \frac{π}{4} \\ \cos \frac{π}{2} \cos x - \sin \frac{π}{2} \sin x & = 1 \\ - \sin x & = 1 \\ \sin x & = - 1 \\ x & = \frac{3 π}{2} \end{aligned}$
$\begin{align*}\cos \left(\frac{\pi}{2}+x\right) &=\tan \frac{\pi}{4} \\ \cos \frac{\pi}{2} \cos x - \sin \frac{\pi}{2} \sin x &=1 \\ -\sin x &=1 \\ \sin x &=-1 \\ x &=\frac{3\pi}{2}\end{align*}$
::COS(% 2+x) = tan4cos2cos2x-sin2sin2sinx=1-sin1x=1sin1sinxx1x=32

Review
::回顾

Solve the following trig equations in the interval $\begin{align*}0\le x < 2\pi\end{align*}$ .
::在 0. x < 2 间距内解决以下三角方程式。

$\begin{align*}\sin (x-\pi)=-\frac{\sqrt{2}}{2}\end{align*}$
:x)

$\begin{align*}\cos(2\pi +x)=-1\end{align*}$
::cos( 2x) 1

$\begin{align*}\tan \left(x+\frac{\pi}{4}\right)=1\end{align*}$
::tan( x4)=1

$\begin{align*}\sin \left(\frac{\pi}{2}-x\right)=\frac{1}{2}\end{align*}$
:%2-x)=12

$\begin{align*}\sin \left(x+\frac{3\pi}{4}\right)+\sin \left(x-\frac{3\pi}{4}\right)=1\end{align*}$
:x+34)+sin(x-34)=1

$\begin{align*}\sin \left(x+\frac{\pi}{6}\right)=-\sin \left(x-\frac{\pi}{6}\right)\end{align*}$
:x6) sin (x6)

$\begin{align*}\cos \left(x+\frac{\pi}{6}\right)=\cos \left(x-\frac{\pi}{6}\right)+1\end{align*}$
::cos(x6)=cos(x6)+1

$\begin{align*}\cos \left(x+\frac{\pi}{3}\right)+\cos \left(x-\frac{\pi}{3}\right)=1\end{align*}$
::cos(x3)+cos(x3)=1

$\begin{align*}\tan(x+\pi)+2\sin (x+\pi)=0\end{align*}$
::tan(x)+2sin(x)=0

$\begin{align*}\tan (x+\pi)+\cos \left(x+\frac{\pi}{2}\right)=0\end{align*}$
::tan(x)+cos(x2)=0

$\begin{align*}\tan \left(x+\frac{\pi}{4}\right)=\tan \left(x-\frac{\pi}{4}\right)\end{align*}$
::tan(x4) =tan(x4)

$\begin{align*}\sin \left(x-\frac{5\pi}{3}\right)-\sin \left(x-\frac{2\pi}{3}\right)=0\end{align*}$
:x-53)-(x-23)=0

$\begin{align*}4\sin (x+\pi)-2=2\cos\left(x+\frac{\pi}{2}\right)\end{align*}$
::4sin(x)-2=2cos(x2)

$\begin{align*}1+2\cos(x-\pi)+\cos x =0\end{align*}$
::1+2cos(x)+cosx=0

Real Life Application The height, $\begin{align*}h\end{align*}$ (in feet), of two people in different seats on a Ferris wheel can be modeled by $\begin{align*}h_1=50\cos 3t+46\end{align*}$ and $\begin{align*}h_2=50\cos 3\left(t-\frac{3\pi}{4}\right)+46\end{align*}$ where $\begin{align*}t\end{align*}$ is the time (in minutes). When are the two people at the same height?
::身高, h( 以脚为单位), 两人坐在Ferris轮的不同座位上, 可以用 h1 = 50cos3t+46 和 h2 = 50cos3 (t- 34)+46 来模拟。时间( 分钟) 。两个人何时在同一高度?

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

使用总和和差异公式解决Trig 公式

Solving Trigonometric Functions ::解决三角函数

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)

Solving Trigonometric Functions
::解决三角函数

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)