Course: 3.9 总计和差异公式的应用

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总计和差异公式的应用
You are quite likely familiar with the values of trig functions for a variety of angles. Angles such as $\begin{align*}30^\circ\end{align*}$ , $\begin{align*}60^\circ\end{align*}$ , and $\begin{align*}90^\circ\end{align*}$ are common. However, if you were asked to find the value of a trig function for a more rarely used angle, could you do so? Or what if you were asked to find the value of a trig function for a sum of angles? For example, if you were asked to find $\begin{align*}\sin \left( \frac{3\pi}{2} + \frac{\pi}{4} \right)\end{align*}$ could you?
::您非常熟悉各种角度的三角函数值。 30 、 60 和 90 等角度是常见的。但是, 如果您被要求为更少使用的角度找到三角函数值, 您可以这样做吗 ? 或者如果您被要求为角度总和找到三角函数值呢 ? 例如, 如果您被要求找到罪( 3 ) 2 4 , 您可以吗 ?

Read on, and in this section, you'll get practice with simplifying trig functions of angles using the .
::阅读,在本节, 你会得到实践简化三角函数角度使用。

Sum and Difference Formulas
::总和和差额和差额公式

Quite frequently one of the main obstacles to solving a problem in trigonometry is the inability to transform the problem into a form that makes it easier to solve. Sum and difference formulas can be very valuable in helping with this.
::解决三角测量问题的主要障碍之一经常是无法将问题转化为较容易解决的形式。求和差异公式对于帮助解决这一问题非常有价值。

Here we'll get some extra practice putting the sum and difference formulas to good use. If you haven't gone through them yet, you might want to review the sections on the Sum and Difference Formulas for sine, cosine, and tangent.
::在这里,我们将得到一些额外的实践, 使总和和差值公式得到很好的使用。如果您还没有通过它们, 您可能想要查看关于正弦、顺弦和正弦的和差异公式的总计和差异公式的章节。

Solve using the Sum Formula
::使用总和公式解决

Verify the identity $\begin{align*}\frac{\cos (x-y)}{\sin x \sin y} = \cot x \cot y + 1\end{align*}$
::校验身份( x-y) sin *xsin *y=cot *xcot+1

$\begin{aligned} \cot x \cot y + 1 & = \frac{\cos (x - y)}{\sin x \sin y} \\ = \frac{\cos x \cos y}{\sin x \sin y} + \frac{\sin x \sin y}{\sin x \sin y} & Expand using the cosine difference formula . \\ = \frac{\cos x \cos y}{\sin x \sin y} + 1 \\ \cot x \cot y + 1 & = \cot x \cot y + 1 & cotangent equals cosine over sine \end{aligned}$
$\begin{align*}\cot x \cot y + 1 & = \frac{\cos(x-y)}{\sin x \sin y} \\ & = \frac{\cos x \cos y}{\sin x \sin y} + \frac{\sin x \sin y}{\sin x \sin y} && \text{Expand using the cosine difference formula}. \\ & = \frac{\cos x \cos y}{\sin x \sin y} + 1 \\ \cot x \cot y +1 & = \cot x \cot y +1 && \text{cotangent equals cosine over sine}\end{align*}$
::COTxcoty+1=cos(x-y)sinxsinxsinxsinxsiny+sinxsinxsinxsiny Expand 使用共弦公式。 =cosxxcossinxsiny+1cotxcot}}+1=cotxcoty+1=cotxcoty+1 cotangent 等于正弦共弦值

Solve using the Difference Formula
::使用差异公式解决

Solve $\begin{align*}3 \sin (x-\pi)=3\end{align*}$ in the interval $\begin{align*}[0, 2\pi)\end{align*}$ .
::在间隔 [0,2] 中, Solve 3sin(x)=3。

First, get $\begin{align*}\sin(x - \pi)\end{align*}$ by itself, by dividing both sides by $\begin{align*}3\end{align*}$ .
::首先,通过将双方除以3 来获得罪(x) 本身,通过将双方除以3 来获得罪(x) 。

$\begin{aligned} \frac{3 \sin (x - π)}{3} & = \frac{3}{3} \\ \sin (x - π) & = 1 \end{aligned}$
$\begin{align*}\frac{3 \sin (x- \pi)}{3} & = \frac{3}{3} \\ \sin (x - \pi) & = 1\end{align*}$
::3sin(x)3=33sin(x)=1

Now, expand the left side using the sine difference formula .
::现在,用正弦差公式扩大左侧。

$\begin{aligned} \sin x \cos π - \cos x \sin π & = 1 \\ \sin x (- 1) - \cos x (0) & = 1 \\ - \sin x & = 1 \\ \sin x & = - 1 \end{aligned}$
$\begin{align*}\sin x \cos \pi - \cos x \sin \pi & = 1 \\ \sin x (-1) - \cos x (0) & = 1 \\ - \sin x & = 1 \\ \sin x & = - 1 \end{align*}$
::=1 -sinx=1 -sinx=1xx=1xx=1xxx=1xxxxxx#1xxx#1xxx_1xxxx_1xxxx_1xxxx_1xxxxx_1xxxx_1xxx_1xxx_1xx_1xx_1xxx_1xx_1xx_1xxx_1xx_1xx_1xxx_1xxx_1xx_1xxxx_1_1xxx_1xx_1xx_1xxxxxx_1_1xxxxxx_1xxxxxxxx_1_1_xxxxxxxx_xxx_xxxxx_1_xx_x_x_x_x_x_

The $\begin{align*}\sin x = -1\end{align*}$ when $\begin{align*}x\end{align*}$ is $\begin{align*}\frac{3\pi}{2}\end{align*}$ .
::x 是 3Q2 时的 sinx1 。

Solve using the Sum Formula
::使用总和公式解决

Find all the solutions for $\begin{align*}2 \cos^2 \left (x+ \frac{\pi}{2} \right ) = 1\end{align*}$ in the interval $\begin{align*}[0, 2\pi)\end{align*}$ .
::在间隔 [0,2] 中查找 2cos2(x2) =1 的所有解决方案。

Get the $\begin{align*}\cos^2 \left (x+ \frac{\pi}{2} \right )\end{align*}$ by itself and then take the square root.
::自行获取 Cos2(x2) , 然后取平方根。

$\begin{aligned} 2 \cos^{2} (x + \frac{π}{2}) & = 1 \\ \cos^{2} (x + \frac{π}{2}) & = \frac{1}{2} \\ \cos (x + \frac{π}{2}) & = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} \end{aligned}$
$\begin{align*}2 \cos^2 \left (x+ \frac{\pi}{2} \right ) & = 1 \\ \cos^2 \left (x+ \frac{\pi}{2} \right ) & = \frac{1}{2} \\ \cos \left (x+ \frac{\pi}{2} \right ) & = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}\end{align*}$
::2cos2(x2)=1cos2(x2)=12cos*(x2)=12=22

Now, use the cosine sum formula to expand and solve.
::现在,使用余弦总和公式来扩大和解决。

$\begin{aligned} \cos x \cos \frac{π}{2} - \sin x \sin \frac{π}{2} & = \frac{\sqrt{2}}{2} \\ \cos x (0) - \sin x (1) & = \frac{\sqrt{2}}{2} \\ - \sin x & = \frac{\sqrt{2}}{2} \\ \sin x & = - \frac{\sqrt{2}}{2} \end{aligned}$
$\begin{align*}\cos x \cos \frac{\pi}{2} - \sin x \sin \frac{\pi}{2} & = \frac{\sqrt{2}}{2} \\ \cos x(0) - \sin x (1) & = \frac{\sqrt{2}}{2} \\ - \sin x & = \frac{\sqrt{2}}{2} \\ \sin x & = - \frac{\sqrt{2}}{2} \end{align*}$
::COSxcos2-sinxsin2=22cosx(0)-sinx(1)=22-sinx=22sinx=22sin}xx=22sin}xx=22sinxxx22

The $\begin{align*}\sin x = - \frac{\sqrt{2}}{2}\end{align*}$ is in Quadrants III and IV, so $\begin{align*}x = \frac{5 \pi}{4}\end{align*}$ and $\begin{align*}\frac{7 \pi}{4}\end{align*}$ .
::x22 是在二次和第四次二次曲线中, 所以 x=54 和 74 。

Examples
::实例

Example 1
::例1

Earlier, you were asked to find $\begin{align*}\sin \left( \frac{3\pi}{2} + \frac{\pi}{4} \right)\end{align*}$ , use the sine sum formula :
::早些时候,你被要求找到sin(324),使用正数公式:

$\begin{aligned} \sin (a + b) = \sin (a) \cos (b) + \cos (a) \sin (b) \\ \sin (\frac{3 π}{2} + \frac{π}{4}) = \sin (\frac{3 π}{2}) \times \cos (\frac{π}{4}) + \cos (\frac{3 π}{2}) \times \sin (\frac{π}{4}) \\ = (- 1) (\frac{\sqrt{2}}{2}) + (0) (\frac{\sqrt{2}}{2}) \\ = - \frac{\sqrt{2}}{2} \end{aligned}$
$\begin{align*} \sin (a + b) = \sin(a)\cos(b) + \cos(a)\sin(b)\\ \sin \left( \frac{3\pi}{2} + \frac{\pi}{4} \right) = \sin \left( \frac{3\pi}{2} \right) \times \cos \left( \frac{\pi}{4} \right) + \cos \left( \frac{3\pi}{2} \right) \times \sin \left( \frac{\pi}{4} \right)\\ = (-1)\left( \frac{\sqrt{2}}{2} \right) + (0)\left( \frac{\sqrt{2}}{2} \right)\\ = -\frac{\sqrt{2}}{2}\\ \end{align*}$
:a+b)=sin(a)cos(b)+cos(a)sin(b)sin(324)=sin(324)xxcos(4)+cos(32)xsin(4)=(-1)(22)+(0)(22)____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

Example 2
::例2

Find all solutions to $\begin{align*}2 \cos^2 \left (x+ \frac{\pi}{2} \right ) = 1\end{align*}$ , when $\begin{align*}x\end{align*}$ is between $\begin{align*}[0, 2\pi)\end{align*}$ .
::当 x 介于 [ 0, 2] 之间时, 查找 2cos2 {( x2) =1 的所有解决方案。

To find all the solutions, between $\begin{align*}[0, 2\pi)\end{align*}$ , we need to expand using the sum formula and isolate the $\begin{align*}\cos x\end{align*}$ .
::为了找到所有的解决办法,在[0,2]之间,我们需要使用总和公式进行扩展,并分离COsx。

$\begin{aligned} 2 \cos^{2} (x + \frac{π}{2}) & = 1 \\ \cos^{2} (x + \frac{π}{2}) & = \frac{1}{2} \\ \cos (x + \frac{π}{2}) & = \pm \sqrt{\frac{1}{2}} = \pm \frac{\sqrt{2}}{2} \\ \cos x \cos \frac{π}{2} - \sin x \sin \frac{π}{2} & = \pm \frac{\sqrt{2}}{2} \\ \cos x \cdot 0 - \sin x \cdot 1 & = \pm \frac{\sqrt{2}}{2} \\ - \sin x & = \pm \frac{\sqrt{2}}{2} \\ \sin x & = \pm \frac{\sqrt{2}}{2} \end{aligned}$
$\begin{align*}2 \cos^2 \left (x + \frac{\pi}{2} \right ) & = 1 \\ \cos^2 \left (x + \frac{\pi}{2} \right ) & = \frac{1}{2} \\ \cos \left (x + \frac{\pi}{2} \right ) & = \pm\sqrt{\frac{1}{2}} = \pm\frac{\sqrt{2}}{2} \\ \cos x \cos \frac{\pi}{2} - \sin x \sin \frac{\pi}{2} & = \pm\frac{\sqrt{2}}{2} \\ \cos x \cdot 0 - \sin x \cdot 1 & = \pm\frac{\sqrt{2}}{2} \\ - \sin x & = \pm\frac{\sqrt{2}}{2} \\ \sin x & = \pm\frac{\sqrt{2}}{2} \end{align*}$
::2cos2(x2) =1cos2(x2) =12cos (x2) 1222cos xcos2-sin 222cosxin_xin_222_sin_x1}22-sin_xí_x_x_x_x_x_x_x_x_x_x__sin_2_sin_x_x__2_sin_x__x_x_2___sin_x_x_x_x__x___sin_x___x_x_x_x_____x____x______x_______x____x______x_____________x_______________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

This is true when $\begin{align*}x = \frac{\pi}{4}, \frac{3 \pi}{4}, \frac{5 \pi}{4}\end{align*}$ , or $\begin{align*}\frac{7 \pi}{4}\end{align*}$
::当 x+4,34,54,或 74 时,情况就是这样。

Example 3
::例3

Solve for all values of $\begin{align*}x\end{align*}$ between $\begin{align*}[0, 2\pi)\end{align*}$ for $\begin{align*}2 \tan^2 \left (x+ \frac{\pi}{6} \right ) + 1 = 7\end{align*}$ .
::2tan2(x6)+1=7, 在 [0,2] 之间解决所有 x 值之间的所有值。

First, solve for $\begin{align*}\tan (x+ \frac{\pi}{6})\end{align*}$ .
::首先, Ten(x6) 的解答。

$\begin{aligned} 2 \tan^{2} (x + \frac{π}{6}) + 1 & = 7 \\ 2 \tan^{2} (x + \frac{π}{6}) & = 6 \\ \tan^{2} (x + \frac{π}{6}) & = 3 \\ \tan (x + \frac{π}{6}) & = \pm \sqrt{3} \end{aligned}$
$\begin{align*}2 \tan^2 \left (x+ \frac{\pi}{6} \right ) +1 & = 7 \\ 2 \tan^2 \left (x+ \frac{\pi}{6} \right ) & = 6 \\ \tan^2 \left (x+ \frac{\pi}{6} \right ) & = 3 \\ \tan \left (x+ \frac{\pi}{6} \right ) & = \pm\sqrt{3}\end{align*}$
::2tan2(x6)+1=72tan2(x6)=6tan2(x6)=3tan(x6)3

Now, use the tangent sum formula to expand for when $\begin{align*}\tan (x+ \frac{\pi}{6}) = \sqrt{3}\end{align*}$ .
::现在, 使用正切总和公式来扩展当 tan( x6) = 3 时的公式。

$\begin{aligned} \frac{\tan x + \tan \frac{π}{6}}{1 - \tan x \tan \frac{π}{6}} & = \sqrt{3} \\ \tan x + \tan \frac{π}{6} & = \sqrt{3} (1 - \tan x \tan \frac{π}{6}) \\ \tan x + \frac{\sqrt{3}}{3} & = \sqrt{3} - \sqrt{3} \tan x \cdot \frac{\sqrt{3}}{3} \\ \tan x + \frac{\sqrt{3}}{3} & = \sqrt{3} - \tan x \\ 2 \tan x & = \frac{2 \sqrt{3}}{3} \\ \tan x & = \frac{\sqrt{3}}{3} \end{aligned}$
$\begin{align*}\frac{\tan x + \tan \frac{\pi}{6}}{1 - \tan x \tan \frac{\pi}{6}} & = \sqrt{3} \\ \tan x + \tan \frac{\pi}{6} & = \sqrt{3} \left (1 - \tan x \tan \frac{\pi}{6} \right ) \\ \tan x + \frac{\sqrt{3}}{3} & = \sqrt{3} - \sqrt{3} \tan x \cdot \frac{\sqrt{3}}{3} \\ \tan x + \frac{\sqrt{3}}{3} & = \sqrt{3} - \tan x \\ 2 \tan x & = \frac{2 \sqrt{3}}{3} \\ \tan x & = \frac{\sqrt{3}}{3}\end{align*}$
::\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ -\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ -\\\\\\\\\\\\\\\ -\\\\\\\\\\33\\\\\33\\\\\\\\\\\\\\\\33。

This is true when $\begin{align*}x = \frac{\pi}{6}\end{align*}$ or $\begin{align*}\frac{7 \pi}{6}\end{align*}$ .
::当 x6 或 76 或 76 时,情况就是这样。

If the tangent sum formula to expand for when $\begin{align*}\tan (x+ \frac{\pi}{6}) = -\sqrt{3}\end{align*}$ , we get no solution as shown.
::如果在 tan(x6)3 时要扩展的正切和公式, 我们没有得到显示的解决方案。

$\begin{aligned} \frac{\tan x + \tan \frac{π}{6}}{1 - \tan x \tan \frac{π}{6}} & = - \sqrt{3} \\ \tan x + \tan \frac{π}{6} & = - \sqrt{3} (1 - \tan x \tan \frac{π}{6}) \\ \tan x + \frac{\sqrt{3}}{3} & = - \sqrt{3} + \sqrt{3} \tan x \cdot \frac{\sqrt{3}}{3} \\ \tan x + \frac{\sqrt{3}}{3} & = - \sqrt{3} + \tan x \\ \frac{\sqrt{3}}{3} & = - \sqrt{3} \end{aligned}$
$\begin{align*}\frac{\tan x + \tan \frac{\pi}{6}}{1 - \tan x \tan \frac{\pi}{6}} & = -\sqrt{3} \\ \tan x + \tan \frac{\pi}{6} & = -\sqrt{3} \left (1 - \tan x \tan \frac{\pi}{6} \right ) \\ \tan x + \frac{\sqrt{3}}{3} & = -\sqrt{3} + \sqrt{3} \tan x \cdot \frac{\sqrt{3}}{3} \\ \tan x + \frac{\sqrt{3}}{3} & = -\sqrt{3} + \tan x \\ \frac{\sqrt{3}}{3} & = -\sqrt{3}\\\end{align*}$
::{\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... {\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...

Therefore, the tangent sum formula cannot be used in this case. However, since we know that $\begin{align*}\tan(x+\frac{\pi}{6}) = -\sqrt{3}\end{align*}$ when $\begin{align*}x+\frac{\pi}{6} = \frac{5\pi}{6}\end{align*}$ or $\begin{align*}\frac{11\pi}{6}\end{align*}$ , we can solve for $\begin{align*}x\end{align*}$ as follows.
::因此,在此情况下不能使用相切的和数公式。但是, 因为我们知道当 x 6= 5 6 或 11 6 时, tan (x 6) 3 我们可以解决 x 如下。

$\begin{aligned} x + \frac{π}{6} = \frac{5 π}{6} \\ x = \frac{4 π}{6} \\ x = \frac{2 π}{3} \\ x + \frac{π}{6} = \frac{11 π}{6} \\ x = \frac{10 π}{6} \\ x = \frac{5 π}{3} \end{aligned}$
$\begin{align*}x+\frac{\pi}{6}=\frac{5\pi}{6} \\ x = \frac{4\pi}{6} \\ x = \frac{2\pi}{3} \\ \\ x+\frac{\pi}{6}=\frac{11\pi}{6} \\ x = \frac{10\pi}{6} \\ x = \frac{5\pi}{3}\end{align*}$
::x6=56x=46x=23x6=116x=106x=53

Therefore, all of the solutions are $\begin{align*}x=\frac{\pi}{6}, \frac{2\pi}{3}, \frac{7 \pi}{6}, \frac{5\pi}{3}\end{align*}$
::因此,所有解决方案都是x6,23,76,53

Example 4
::例4

Find all solutions to $\begin{align*}\sin \left (x+ \frac{\pi}{6} \right ) = \sin \left (x- \frac{\pi}{4} \right )\end{align*}$ , when $\begin{align*}x\end{align*}$ is between $\begin{align*}[0, 2\pi)\end{align*}$ .
::当 x 介于 [ 0. 2] 之间时, 找到所有 sin 的解决方案。

To solve, expand each side:
::要解决, 扩大每侧 :

$\begin{aligned} \sin (x + \frac{π}{6}) & = \sin x \cos \frac{π}{6} + \cos x \sin \frac{π}{6} = \frac{\sqrt{3}}{2} \sin x + \frac{1}{2} \cos x \\ \sin (x - \frac{π}{4}) & = \sin x \cos \frac{π}{4} - \cos x \sin \frac{π}{4} = \frac{\sqrt{2}}{2} \sin x - \frac{\sqrt{2}}{2} \cos x \end{aligned}$
$\begin{align*}\sin \left (x + \frac{\pi}{6} \right ) & = \sin x \cos \frac{\pi}{6} + \cos x \sin \frac{\pi}{6} = \frac{\sqrt{3}}{2} \sin x + \frac{1}{2} \cos x \\ \sin \left (x - \frac{\pi}{4} \right ) & = \sin x \cos \frac{\pi}{4} - \cos x \sin \frac{\pi}{4} = \frac{\sqrt{2}}{2} \sin x - \frac{\sqrt{2}}{2} \cos x\end{align*}$
::\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

Set the two sides equal to each other:
::设两对为等:

$\begin{aligned} \frac{\sqrt{3}}{2} \sin x + \frac{1}{2} \cos x & = \frac{\sqrt{2}}{2} \sin x - \frac{\sqrt{2}}{2} \cos x \\ \sqrt{3} \sin x + \cos x & = \sqrt{2} \sin x - \sqrt{2} \cos x \\ \sqrt{3} \sin x - \sqrt{2} \sin x & = - \cos x - \sqrt{2} \cos x \\ \sin x (\sqrt{3} - \sqrt{2}) & = \cos x (- 1 - \sqrt{2}) \\ \frac{\sin x}{\cos x} & = \frac{- 1 - \sqrt{2}}{\sqrt{3} - \sqrt{2}} \\ \tan x & = \frac{- 1 - \sqrt{2}}{\sqrt{3} - \sqrt{2}} \cdot \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} + \sqrt{2}} \\ = \frac{- \sqrt{3} - \sqrt{2} + \sqrt{6} - 2}{3 - 2} \\ = - 2 + \sqrt{6} - \sqrt{3} - \sqrt{2} \end{aligned}$
$\begin{align*}\frac{\sqrt{3}}{2} \sin x + \frac{1}{2} \cos x & = \frac{\sqrt{2}}{2} \sin x - \frac{\sqrt{2}}{2} \cos x \\ \sqrt{3} \sin x + \cos x & = \sqrt{2} \sin x - \sqrt{2} \cos x \\ \sqrt{3} \sin x - \sqrt{2} \sin x & = - \cos x - \sqrt{2} \cos x \\ \sin x \left (\sqrt{3} - \sqrt{2} \right ) & = \cos x \left (-1 - \sqrt{2} \right ) \\ \frac{\sin x}{\cos x} & = \frac{-1- \sqrt{2}}{\sqrt{3} - \sqrt{2}} \\ \tan x & = \frac{-1- \sqrt{2}}{\sqrt{3} - \sqrt{2}} \cdot \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} + \sqrt{2}} \\ & = \frac{- \sqrt{3} - \sqrt{2} + \sqrt{6} - 2}{3-2} \\ & = -2 + \sqrt{6} - \sqrt{3} - \sqrt{2}\end{align*}$
::32sinx+12cosx=22sinx22cosxxxx3sinxxxxxxxx=2sinx2cosx3sin}x2sinxxxxxxxxxxxxxxx(3-2)xxxx(3-2)=cosxxx(-1-2)xxxxxxx*1-2322-2tanx1-233+23+2}}3-2+623 -2}+2

As a decimal, this is $\begin{align*}-2.69677\end{align*}$ , so $\begin{align*}\tan^{-1}(-2.69677) = x, x = 290.35^\circ\end{align*}$ and $\begin{align*}110.35^\circ\end{align*}$ .
::小数点为-2.69677, 所以tan- 1(-2.69677)=x,x=290.35和110.35。

Review
::回顾

Prove each identity.
::证明每个身份

$\begin{align*}\cos(3x)+\cos(x)=2\cos(2x)\cos(x)\end{align*}$
::cos( 3x) +cos( x) = 2cos( 2x) cos( x)

$\begin{align*}\cos(3x)=\cos^3(x)-3\sin^2(x)\cos(x)\end{align*}$
::cos( 3x) =cos3( x)- 3sin2( x) cos( x)

$\begin{align*}\sin(3x)=3\cos^2(x)\sin(x)-\sin^3(x)\end{align*}$
:x) -sin3 *(x)=3cos2 *(x)sin *(x)_sin3*(x)

$\begin{align*}\sin(4x)+\sin(2x)=2\sin(3x)\cos(x)\end{align*}$
:4x)+sin(2x)=2sin(3x)cos(x)

$\begin{align*}\tan(5x)\tan(3x)=\frac{\tan^2(4x)-\tan^2(x)}{1-\tan^2(4x)\tan^2(x)}\end{align*}$
::tan( 5x) tan( 3x) =tan2( 4x) -tan2( (x) 1) -tan2( 4x) tan2( 4x) tan2( x)

$\begin{align*}\cos((\frac{\pi}{2}-x)-y)=\sin(x+y)\end{align*}$
::COs( (%2- x)-y) =sin( x+y)

Use sum and difference formulas to help you graph each function.
::使用和差公式来帮助您绘制每个函数的图形。

$\begin{align*}y=\cos(3)\cos(x)+\sin(3)\sin(x)\end{align*}$
::y=cos(3)cos(x)+sin(3)sin(x)

$\begin{align*}y=\cos(x)\cos(\frac{\pi}{2})+\sin(x)\sin(\frac{\pi}{2})\end{align*}$
::y=cos(x)cos(2)+sin(x)sin(2)

$\begin{align*}y=\sin(x)\cos(\frac{\pi}{2})+\cos(x)\sin(\frac{\pi}{2})\end{align*}$
::y=sin(x)cos(2)+cos(x)sin(2)

$\begin{align*}y = \sin(x)\cos\left(\frac{3\pi}{2}\right) - \cos(x)\sin\left(\frac{3\pi}{2}\right)\end{align*}$
::y=sin(x)cos(32)-cos(x)sin(32)

$\begin{align*}y=\cos(4x)\cos(2x)-\sin(4x)\sin(2x)\end{align*}$
::y=cos(4x)cos(2x)-sin(4x)sin(2x)

$\begin{align*}y=\cos(x)\cos(x)-\sin(x)\sin(x)\end{align*}$
::y=cos *(x)cos *(x)-sin *(x)sin *(x)sin *(x)

Solve each equation on the interval $\begin{align*}[0,2\pi)\end{align*}$ . Put your answers in radians.
::在间隔 [0,2] 上解决每个方程。请将答案以弧度表示。

$\begin{align*}2\sin(x-\frac{\pi}{2})=1\end{align*}$
::2sin(x2)=1

$\begin{align*}4\cos(x-\pi)=4\end{align*}$
::4cos(x)=4

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

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

总计和差异公式的应用

Sum and Difference Formulas ::总和和差额和差额公式

Solve using the Sum Formula ::使用总和公式解决

Solve using the Difference Formula ::使用差异公式解决

Solve using the Sum Formula ::使用总和公式解决

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)

Sum and Difference Formulas
::总和和差额和差额公式

Solve using the Sum Formula
::使用总和公式解决

Solve using the Difference Formula
::使用差异公式解决

Solve using the Sum Formula
::使用总和公式解决

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)