Course: 14.16 使用双角和半角公式简化 Trig 表达式

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使用双和半角度公式简化 Trig 表达式
As Agent Trigonometry, you are given the following cryptic clue. How could you simplify this clue?
::作为Trigonology探员,你得到了以下隐秘的线索, 你怎么能简化这个线索?

$\begin{align*}\frac{\tan 2x}{\frac{tan x}{1 + \tan x}}\end{align*}$
::tan2xtanx1+tanx

Simplifying Trigonometric Expressions
::简化三角数表达式

We can also use the double-angle and half-angle formulas to simplify trigonometric expressions.
::我们还可以使用双角和半角公式来简化三角表达式。

Let's simplify $\begin{align*}\frac{\cos 2x}{\sin x \cos x}\end{align*}$ .
::让我们简化cos2xsinxcosxxxxxx。

Use $\begin{align*}\cos 2a=\cos^2a-\sin^2a\end{align*}$ and then factor.
::使用 cos2a =cos2çaa-sin2a 和乘数。

$\begin{aligned} \frac{\cos 2 x}{\sin x \cos x} & = \frac{\cos^{2} x - \sin^{2} x}{\sin x + \cos x} \\ = \frac{(\cos x - \sin x) (\cos x + \sin x)}{\sin x + \cos x} \\ = \cos x - \sin x \end{aligned}$
$\begin{align*}\frac{\cos 2x}{\sin x \cos x}&=\frac{\cos^2x- \sin^2x}{\sin x+ \cos x} \\ &=\frac{\left(\cos x- \sin x\right) \cancel{\left(\cos x + \sin x\right)}}{\cancel{\sin x+ \cos x}} \\ &=\cos x- \sin x\end{align*}$
::cos2x辛2xxxxxxin2xxxinxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxsxxxxxxxxxxxxxxxxxxxxxxxxxxxxx-sinxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

Now, let's find the formula for $\begin{align*}\sin 3x\end{align*}$ .
::现在,让我们找到sin_3x的配方。

You will need to use the sum formula and the double-angle formula. $\begin{align*}\sin 3x=\sin(2x+x)\end{align*}$
::您需要使用总和公式和双角公式。 sin3x=sin( 2x+x)

$\begin{aligned} \sin 3 x & = \sin (2 x + x) \\ = \sin 2 x \cos x + \cos 2 x \sin x \\ = 2 \sin x \cos x \cos x + \sin x (2 \cos^{2} x - 1) \\ = 2 \sin x \cos^{2} x + 2 \sin x \cos^{2} x - \sin x \\ = 4 \sin x \cos^{2} x - \sin x \\ = \sin x (4 \cos^{2} x - 1) \end{aligned}$
$\begin{align*}\sin 3x&=\sin (2x+x) \\ &=\sin 2x \cos x + \cos 2x \sin x \\ &=2 \sin x \cos x \cos x+ \sin x(2 \cos^2x-1) \\ &=2 \sin x \cos^2x+2 \sin x \cos^2 x- \sin x \\ &=4 \sin x \cos^2x- \sin x \\ &=\sin x(4 \cos^2x-1)\end{align*}$
::xx- 1= 2sinxxxcos2xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxsin%2xxsinxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx2xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx1

Finally, let's verify the identity $\begin{align*}\cos x+2 \sin^2 \frac{x}{2}=1\end{align*}$ .
::最后,让我们验证一下身份 cosx+2sin2x2=1。

Simplify the left-hand side use the half-angle formula.
::使用半角公式简化左手侧。

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

Examples
::实例

Example 1
::例1

Earlier, you were asked to simplify $\begin{align*}\frac{\tan 2x}{\frac{tan x}{1 + \tan x}}\end{align*}$ .
::早些时候,你被要求简化 tan2xtanx1+tanx。

Use $\begin{align*}\tan 2a=\frac {2\tan a}{1 - \tan^2 a}\end{align*}$ and then factor.
::使用 tan2a = 2tana1 - tan2a 和后因子。

$\begin{aligned} \frac{\tan 2 x}{\frac{t a n x}{1 + \tan x}} = \frac{2 \tan x}{1 - \tan^{2} x} \cdot \frac{1 + \tan x}{t a n x} \\ = \frac{2 \tan x}{(1 + \tan x) (1 - \tan x)} \cdot \frac{1 + \tan x}{t a n x} = \frac{2}{1 - \tan x} \end{aligned}$
$\begin{align*}\frac{\tan 2x}{\frac{tan x}{1 + \tan x}}=\frac{2\tan x}{1-\tan^2 x}\cdot \frac{1 + \tan x}{tan x} \\ =\frac{2\tan x}{(1 + \tan x)(1 - \tan x)}\cdot \frac{1 + \tan x}{tan x} =\frac{2}{1-\tan x}\end{align*}$
::~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Example 2
::例2

Simplify $\begin{align*}\frac{\sin 2x}{\sin x}\end{align*}$ .
::简化 sin=2xsin=xxxx

$\begin{align*}\frac{\sin 2x}{\sin x}=\frac{2 \sin x \cos x}{\sin x}=2 \cos x\end{align*}$
::2xsinx=2sinxcosxsinxx=2cosxx

Example 3
::例3

Verify $\begin{align*}\cos x+2 \cos^2 \frac{x}{2}=1+ 2 \cos x\end{align*}$ .
::校验 cosx+2cos2x2=1+2cosx。

$\begin{aligned} \cos x + 2 \cos^{2} \frac{x}{2} & = 1 + 2 \cos x \\ \cos x + 2 {\sqrt{\frac{1 + \cos x}{2}}}^{2} & = \\ \cos x + 1 + \cos x & = \\ 1 + 2 \cos x & = \end{aligned}$
$\begin{align*}\cos x+2 \cos^2 \frac{x}{2}&=1+2 \cos x \\ \cos x+2 \sqrt{\frac{1+ \cos x}{2}}^2&= \\ \cos x+1 + \cos x&= \\ 1+2 \cos x&=\end{align*}$
::cosx+2cos2x2=1+2cosxcos*x+21+cos*x22=cosx+1+cos*x=1+2cos*x=1

Review
::回顾

Simplify the following expressions.
::简化下列表达式。

$\begin{align*}\sqrt{2+2 \cos x} \left(\cos \frac{x}{2}\right)\end{align*}$
::2+2cosx(cosx2)

$\begin{align*}\frac{\cos 2x}{\cos^2x}\end{align*}$
::2xcos2xxxx

$\begin{align*}\tan 2x(1+ \tan x)\end{align*}$
::tan2x(1+tanx)

$\begin{align*}\cos 2x- 3 \sin^2x\end{align*}$
::COs% 2x- 3sin2xx

$\begin{align*}\frac{1+\cos 2x}{\cot x}\end{align*}$
::1+cos2xcotxx

$\begin{align*}(1+ \cos x)^2 \tan \frac{x}{2}\end{align*}$
: 1+cosx) 2tanx2

Verify the following identities.
::验证以下身份。

$\begin{align*}\cot \frac{x}{2}=\frac{\sin x}{1- \cos x}\end{align*}$
::comtx2=sinx1-cosx

$\begin{align*}\frac{\sin x}{1+ \cos x}=\frac{1- \cos x}{\sin x}\end{align*}$
::ciux1+cosx=1 -cosxinxx

$\begin{align*}\frac{\sin 2x}{1+ \cos 2x}= \tan x\end{align*}$
::sin *% 2x1+cos *2x=tan*x

$\begin{align*}(\sin x+ \cos x)^2=1+ \sin 2x\end{align*}$
:sinx+cosx) 2=1+sin2x

$\begin{align*}\sin x \tan \frac{x}{2}+2 \cos x=2 \cos^2 \frac{x}{2}\end{align*}$
::=2cos2x2x2xx=2cos2}x2x2

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

$\begin{align*}\cos 3x=4 \cos^3x-3 \cos x\end{align*}$
::COs*%3x=4cos3}*x_3cos*x

$\begin{align*}\cos 3x=\cos^3x-3 \sin^2x \cos x\end{align*}$
::COs=3x=cos3xx-3sin2xxcosxxxxxxxxx

$\begin{align*}\sin 2x-\tan x=\tan x \cos 2x\end{align*}$
::2x-tanx=tanxcos2x

$\begin{align*}\cos^4x-\sin^4x=\cos 2x\end{align*}$
::cos4x-sin4x=cos%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。

Section outline

General

使用双和半角度公式简化 Trig 表达式

Simplifying Trigonometric Expressions ::简化三角数表达式

Examples ::实例

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

Review ::回顾

Review (Answers) ::回顾(答复)

Simplifying Trigonometric Expressions
::简化三角数表达式

Examples
::实例

Example 1
::例1

Example 2
::例2

Example 3
::例3

Review
::回顾

Review (Answers)
::回顾(答复)