节: 使用双角和半角公式解析 Trig 公式 | 14.17 用双角和半角公式解析三角方程

章节大纲

Trig Riddle: I am an angle x such that $\begin{aligned} 0 \leq x < 2 π \end{aligned}$ . I satisfy the equation $\begin{aligned} \sin 2 x - \sin x = 0 \end{aligned}$ . What angle am I?
::Trig Riddle: 我是一个角度x 0x<2Q。我符合 sin2x-sinx=0的方程式。我是什么角度?

Solve Trigonometric Equations
::解析三角等量等量

We can use the half and double angle formulas to solve trigonometric equations.
::我们可以用半角度和双角度公式解决三角方程

Let's solve the following trigonometric equations.
::让我们解决以下三角方程。

Solve $\begin{aligned} \tan 2 x + \tan x = 0 \end{aligned}$ when $\begin{aligned} 0 \leq x < 2 π \end{aligned}$ .
::0x<2 时, 解决 tan2x+tanx=0 。

Change $\begin{aligned} \tan 2 x \end{aligned}$ and simplify.
::更改 tan2x 并简化。

$\begin{aligned} \tan 2 x + \tan x & = 0 \\ \frac{2 \tan x}{1 - \tan^{2} x} + \tan x & = 0 \\ 2 \tan x + \tan x (1 - \tan^{2} x) & = 0 \to Multiply everything by 1 - \tan^{2} x to eliminate denominator. \\ 2 \tan x + \tan x - \tan^{3} x & = 0 \\ 3 \tan x - \tan^{3} x & = 0 \\ \tan x (3 - \tan^{2} x) & = 0 \end{aligned}$

::2x+tanx=02tanx1-tan2x+tanx=02tanx+tanx(1-tan2x)=0*1-tan2x消除负分母2tanx+tanx-tan3}x=03tanx -tan3x=0tanx(3-tan2x)=0

Set each factor equal to zero and solve.
::设定每个系数等于零并解析。

$\begin{aligned} 3 - \tan^{2} x = 0 \\ - \tan^{2} x = - 3 \\ \tan x = 0 a n d \tan^{2} x = 3 \\ x = 0 a n d π \tan x = \pm \sqrt{3} \\ x = \frac{π}{3}, \frac{2 π}{3}, \frac{4 π}{3}, \frac{5 π}{3} \end{aligned}$

::3 - tan2x=0 - tan2x _ 3x=0 和 tan2x=3x=0 和 tanx3 x 3,23,3,4,43,53

Solve $\begin{aligned} 2 \cos \frac{x}{2} + 1 = 0 \end{aligned}$ when $\begin{aligned} 0 \leq x < 2 π \end{aligned}$ .
::0x<2 时, 溶解 2cosx2+1=0 。

In this case, you do not have to use the half-angle formula. Solve for $\begin{aligned} \frac{x}{2} \end{aligned}$ .
::在此情况下, 您不必使用半角公式。解决 x2 。

$\begin{aligned} 2 \cos \frac{x}{2} + 1 & = 0 \\ 2 \cos \frac{x}{2} & = - 1 \\ \cos \frac{x}{2} & = - \frac{1}{2} \end{aligned}$

::2cos_x2+1=02cos_x2}%1cos_x2}%2#12}=02cos_x2=12

Now, let’s find $\begin{aligned} \cos a = - \frac{1}{2} \end{aligned}$ and then solve for $\begin{aligned} x \end{aligned}$ by dividing by 2.
::现在,让我们找到cosaa12,然后用除以2来解析 x 。

$\begin{aligned} \frac{x}{2} & = \frac{2 π}{3}, \frac{4 π}{3} \\ = \frac{4 π}{3}, \frac{8 π}{3} \end{aligned}$

::x2 = 23,43 = 43,83

Now, the second solution is not in our range, so the only solution is $\begin{aligned} x = \frac{4 π}{3} \end{aligned}$ .
::现在,第二个解决方案不在我们的范围之内, 所以唯一的解决方案是 x=43。

Solve $\begin{aligned} 4 \sin x \cos x = \sqrt{3} \end{aligned}$ for $\begin{aligned} 0 \leq x < 2 π \end{aligned}$ .
::0x<2=3 溶解 4sinxcosx=3。

Pull a 2 out of the left-hand side and use the $\begin{aligned} \sin 2 x \end{aligned}$ formula.
::从左手边拉出一个2 并使用 sin2x 公式。

$\begin{aligned} 4 \sin x \cos x & = \sqrt{3} \\ 2 \cdot 2 \sin x \cos x & = \sqrt{3} \\ 2 \cdot \sin 2 x & = \sqrt{3} \\ \sin 2 x & = \frac{\sqrt{3}}{2} \\ 2 x & = \frac{π}{3}, \frac{5 π}{3}, \frac{7 π}{3}, \frac{11 π}{3} \\ x & = \frac{π}{6}, \frac{5 π}{6}, \frac{7 π}{6}, \frac{11 π}{6} \end{aligned}$

::{\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}

Examples
::实例

Example 1
::例1

Earlier, you were asked to find the angle x, where $\begin{aligned} 0 \leq x < 2 π \end{aligned}$ , such that x satisfies the equation $\begin{aligned} \sin 2 x - \sin x = 0 \end{aligned}$ .
::早些时候, 您被要求找到角度 x, 位置为 0x < 2, 这样 x 就能满足 sin_ 2x- sinx=0 的方程式。

Use the double angle formula and simplify.
::使用双角度公式并简化。

$\begin{aligned} \sin 2 x - \sin x = 0 \\ 2 \sin x \cos x - \sin x = 0 \\ \sin x (2 \cos x - 1) = 0 \\ \sin x = 0 O R \cos x = \frac{1}{2} \end{aligned}$

::sin *% 2x -sin *x=02sin *xcos *x-sin*x=0sin*xx*x( 2cos *x- 1)=0sin*x=0ORcos*x=12

Under the constraint $\begin{aligned} 0 \leq x < 2 π \end{aligned}$ , $\begin{aligned} \sin x = 0 \end{aligned}$ when $\begin{aligned} x = 0 \end{aligned}$ or when $\begin{aligned} x = π \end{aligned}$ . Under this same constraint, $\begin{aligned} \cos x = \frac{1}{2} \end{aligned}$ when $\begin{aligned} x = \frac{π}{3} \end{aligned}$ or when $\begin{aligned} x = \frac{5 π}{3} \end{aligned}$ .
::在限制下 0x < 2, 当 x=0 或 x 时sinx=0。在相同的限制下, x3 或 x=53 时cosx=12。

Example 2
::例2

Solve the following equation for $\begin{aligned} 0 \leq x < 2 π \end{aligned}$ .
::解决 0x<2 的以下方程式。

$\begin{aligned} \sin \frac{x}{2} = - 1 \end{aligned}$
::一九九九二一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一

$\begin{aligned} \sin \frac{x}{2} & = - 1 \\ \frac{x}{2} & = \frac{3 π}{2} \\ x & = 3 π \end{aligned}$

::___1x2=32x=33x

From this we can see that there are no solutions within our interval.
::从这一点可以看出,我们之间的间隔内没有解决办法。

Example 3
::例3

Solve the following equation for $\begin{aligned} 0 \leq x < 2 π \end{aligned}$ .
::解决 0x<2 的以下方程式。

$\begin{aligned} \cos 2 x - \cos x = 0 \end{aligned}$
::COs% 2x- cos_x=0

$\begin{aligned} \cos 2 x - \cos x & = 0 \\ 2 \cos^{2} x - \cos x - 1 & = 0 \\ (2 \cos x + 1) (\cos x - 1) & = 0 \end{aligned}$

::cos2x-coosx=02cos2x-cosx_1=0(2cos_x+1)(cos_x-1)=0

Set each factor equal to zero and solve.
::设定每个系数等于零并解析。

$\begin{aligned} 2 \cos x + 1 & = 0 \\ 2 \cos x & = - 1 \cos x - 1 = 0 \\ \cos x & = - \frac{1}{2} a n d \cos x = 1 \\ x & = \frac{2 π}{3}, \frac{4 π}{3} x = 0, 2 π \end{aligned}$

::2cos*x+1=02cos*x*1cos*x*1x1=0cos*x**12andcos*x=1x=1x=2}3,4}3x=0,2}

Review
::回顾

Solve the following equations for $\begin{aligned} 0 \leq x < 2 π \end{aligned}$ .
::解决 0x<2 的以下方程式。

$\begin{aligned} \cos x - \cos \frac{1}{2} x = 0 \end{aligned}$
::cosx-cos12x=0

$\begin{aligned} \sin 2 x \cos x = \sin x \end{aligned}$
::{\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}为什么? {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}为什么?

$\begin{aligned} \cos 3 x - \cos^{3} x = 3 \sin^{2} x \cos x \end{aligned}$
::COs% 3x- cos3x=3sin2xcosxxxx

$\begin{aligned} \tan 2 x - \tan x = 0 \end{aligned}$
::tan2x-tanx=0

$\begin{aligned} \cos 2 x - \cos x = 0 \end{aligned}$
::COs% 2x- cos_x=0

$\begin{aligned} 2 \cos^{2} \frac{x}{2} = 1 \end{aligned}$
::2cos2x2=1

$\begin{aligned} \tan \frac{x}{2} = 4 \end{aligned}$
::tanx2=4

$\begin{aligned} \cos \frac{x}{2} = 1 + \cos x \end{aligned}$
::COsx2=1+cosx

$\begin{aligned} \sin 2 x + \sin x = 0 \end{aligned}$
::sin *% 2x+sin*x=0

$\begin{aligned} \cos^{2} x - \cos 2 x = 0 \end{aligned}$
::cos2_x-cos%2x=0

$\begin{aligned} \frac{\cos 2 x}{\cos^{2} x} = 1 \end{aligned}$
::COs =% 2xcos2}% xx=1

$\begin{aligned} \cos 2 x - 1 = \sin^{2} x \end{aligned}$
::COs%% 2x- 1 =sin2 x

$\begin{aligned} \cos 2 x = \cos x \end{aligned}$
::COs=2x=cosx

$\begin{aligned} \sin 2 x - \cos 2 x = 1 \end{aligned}$
::sin *% 2x - cos *2x=1

$\begin{aligned} \sin^{2} x - 2 = \cos 2 x \end{aligned}$
::sin2\\ x- 2 =cos% 2x =cos2x

$\begin{aligned} \cot x + \tan x = 2 \csc 2 x \end{aligned}$
::comtx+tanx=2csc=2x

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。

章节大纲

Solve Trigonometric Equations ::解析三角等量等量

Examples ::实例

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

Review ::回顾

Review (Answers) ::回顾(答复)

Solve Trigonometric Equations
::解析三角等量等量

Examples
::实例

Example 1
::例1

Example 2
::例2

Example 3
::例3

Review
::回顾

Review (Answers)
::回顾(答复)