3.10 双角特性

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  双角度特征

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  Finding the values for trig functions is pretty familiar to you by now. The trig functions of some particular angles may even seem obvious, since you've worked with them so many times. In some cases, you might be able to use this knowledge to your benefit to make calculating the values of some trig equations easier. For example, if someone asked you to evaluate
  ::查找 trig 函数的值对于您来说已经相当熟悉了。 某些特定角度的 trig 函数可能看起来更显而易见, 因为您已经与它们合作了这么多次 。 在某些情况下, 您也许能够利用这些知识来帮助您, 从而更容易计算一些 trig 方程的值 。 例如, 如果有人要求您评估的话 。

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  $\begin{array}{r}\cos {120}^{\circ }\end{array}$
  ::120

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  without consulting a table of trig values, could you do it?
  ::无需咨询三重值的表格,您可以做吗?

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  You might notice right away that this is equal to four times $\begin{array}{r}{30}^{\circ }\end{array}$ . Can this help you? Read this lessons, and at its conclusion you'll know how to use certain formulas to simplify multiples of familiar angles to solve problems.
  ::您可能会立刻注意到, 这等于 4 乘以 30 。 这能帮助您吗 ? 读读这个教训, 并在它结束时, 您将会知道如何使用某些公式来简化多个熟悉角度来解决问题 。

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  Double Angle Identities
  ::双角度特征

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  Here we'll start with the for sine, cosine, and tangent. We can use these identities to help derive a new formula for when we are given a trig function that has twice a given angle as the argument.
  ::在这里,我们将从正弦、正弦和正弦开始。 我们可以使用这些身份来帮助我们找到一个新的公式, 当我们得到一个三重函数时, 它具有两倍的 特定角度作为论据。

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  For example, $\begin{array}{r}\sin (2\theta )\end{array}$ . This way, if we are given $\begin{array}{r}\theta \end{array}$ and are asked to find $\begin{array}{r}\sin (2\theta )\end{array}$ , we can use our new double angle identity to help simplify the problem. Let's start with the derivation of the double angle identities.
  ::例如, sin( 2) 。 这样, 如果我们被给予 并被要求找到 sin( 2) , 我们可以使用我们新的双角度身份来帮助简化问题。 让我们从双重角度身份的衍生开始 。

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  One of the formulas for calculating the sum of two angles is:
  ::计算两个角度之和的公式之一是:

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  $$\begin{array}{r}\sin (\alpha +\beta )=\sin \alpha \cos \beta +\cos \alpha \sin \beta \end{array}$$

  
  :)=sincoscoscossinsin

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  If $\begin{array}{r}\alpha \end{array}$ and $\begin{array}{r}\beta \end{array}$ are both the same angle in the above formula, then
  ::如果α和β在上述公式中是同一角度,那么

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  $$\begin{array}{rl}\sin (\alpha +\alpha )& =\sin \alpha \cos \alpha +\cos \alpha \sin \alpha \\ \sin 2\alpha & =2\sin \alpha \cos \alpha \end{array}$$

  
  ::-=YTET -伊甸园字幕组=-伊甸园字幕组=-伊甸园字幕组=-伊甸园字幕组=-伊甸园字幕组=-伊甸园字幕组=-伊甸园字幕组=-伊甸园字幕组=-伊甸园字幕组=-伊甸园字幕组=-伊甸园字幕组=-伊甸园字幕组=-伊甸园字幕组=-伊甸园字幕组=-伊甸园字幕组=-伊甸园字幕组=-伊甸园字幕组=- 翻译:

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  This is the double angle formula for the sine function. The same procedure can be used in the sum formula for cosine, start with the sum angle formula:
  ::这是正弦函数的双角公式。 comsine 的和公式中可以使用相同的程序, 从和角公式开始 :

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  $$\begin{array}{r}\cos (\alpha +\beta )=\cos \alpha \cos \beta -\sin \alpha \sin \beta \end{array}$$

  
  :)=coscossinsinsin

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  If $\begin{array}{r}\alpha \end{array}$ and $\begin{array}{r}\beta \end{array}$ are both the same angle in the above formula, then
  ::如果α和β在上述公式中是同一角度,那么

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  $$\begin{array}{rl}\cos (\alpha +\alpha )& =\cos \alpha \cos \alpha -\sin \alpha \sin \alpha \\ \cos 2\alpha & ={\cos }^2\alpha -{\sin }^2\alpha \end{array}$$

  
  :)=coscossinsinsinsincos2cos2sin2sin2}

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  This is one of the double angle formulas for the cosine function. Two more formulas can be derived by using the Pythagorean Identity, $\begin{array}{r}{\sin }^2\alpha +{\cos }^2\alpha =1\end{array}$ .
  ::这是余弦函数的双角公式之一。 使用 Pythagoren 身份, sin2 cos2\\\\\\\\\\\\\可以得出另外两个公式 。

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  $\begin{array}{r}{\sin }^2\alpha =1-{\cos }^2\alpha \end{array}$ and likewise $\begin{array}{r}{\cos }^2\alpha =1-{\sin }^2\alpha \end{array}$
  ::和同样的原因2 -1 -sin2\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\可以\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

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  $$\begin{array}{rlrl}\text{Using}{\sin }^2\alpha & =1-{\cos }^2\alpha :& & \text{Using}{\cos }^2\alpha =1-{\sin }^2\alpha :\\ \cos 2\alpha & ={\cos }^2\alpha -{\sin }^2\alpha & & \cos 2\alpha ={\cos }^2\alpha -{\sin }^2\alpha \\ & ={\cos }^2\alpha -(1-{\cos }^2\alpha )& & =(1-{\sin }^2\alpha )-{\sin }^2\alpha \\ & ={\cos }^2\alpha -1+{\cos }^2\alpha & & =1-{\sin }^2\alpha -{\sin }^2\alpha \\ & =2{\cos }^2\alpha -1& & =1-2{\sin }^2\alpha \end{array}$$

  
  ::使用 sin2\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\2\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\2\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

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  Therefore, the double angle formulas for $\begin{array}{r}\cos 2\alpha \end{array}$ are:
  ::因此,COs2α的双角公式是:

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  $$\begin{array}{rl}\cos 2\alpha & ={\cos }^2\alpha -{\sin }^2\alpha \\ \cos 2\alpha & =2{\cos }^2\alpha -1\\ \cos 2\alpha & =1-2{\sin }^2\alpha \end{array}$$

  
  ::222222222212212122222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222

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  Finally, we can calculate the double angle formula for tangent, using the tangent sum formula :
  ::最后,我们可以用相切总和公式计算正切值的双角公式:

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  $$\begin{array}{r}\tan (\alpha +\beta )=\frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }\end{array}$$

  
  ::~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

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  If $\begin{array}{r}\alpha \end{array}$ and $\begin{array}{r}\beta \end{array}$ are both the same angle in the above formula, then
  ::如果α和β在上述公式中是同一角度,那么

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  $$\begin{array}{rl}\tan (\alpha +\alpha )& =\frac{\tan \alpha +\tan \alpha }{1-\tan \alpha \tan \alpha }\\ \tan 2\alpha & =\frac{2\tan \alpha }{1-{\tan }^2\alpha }\end{array}$$

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

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  We can use these formulas to help simplify calculations of trig functions of certain arguments.
  ::我们可以使用这些公式来帮助简化某些论点的三角函数的计算。

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  Let's look at a few problems involving double angle identities. 
  ::让我们来看看几个问题 涉及双重角度的身份。

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  1. If $\begin{array}{r}\sin a=\frac{5}{13}\end{array}$ and $\begin{array}{r}a\end{array}$ is in Quadrant II, find $\begin{array}{r}\sin 2a\end{array}$ , $\begin{array}{r}\cos 2a\end{array}$ , and $\begin{array}{r}\tan 2a\end{array}$ .
  ::1. 如果sina=513和a在Quadrant II中,发现 sina, cos2a和tan2a。

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  To use $\begin{array}{r}\sin 2a=2\sin a\cos a\end{array}$ , the value of $\begin{array}{r}\cos a\end{array}$ must be found first.
  ::要使用 sin2a=2sinacosa, 必须先找到 cosa 的值 。

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  $$\begin{array}{rl}& ={\cos }^{2}a+{\sin }^{2}a=1\\ & ={\cos }^{2}a+{\left(\frac{5}{13}\right)}^2=1\\ & ={\cos }^{2}a+\frac{25}{169}=1\\ & ={\cos }^{2}a=\frac{144}{169},\cos a=\pm \frac{12}{13}\end{array}$$

  .
  ::=cos2a+sin2a=1=cos2a+(513)2=1=cos2a+25169=1=cos2a=144169,cosáa=1213。

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  However since $\begin{array}{r}a\end{array}$ is in Quadrant II, $\begin{array}{r}\cos a\end{array}$ is negative or $\begin{array}{r}\cos a=-\frac{12}{13}\end{array}$ .
  ::然而,由于a在Quadrant II,cosa为负或cosa 1213。

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  $$\begin{array}{r}\sin 2a=2\sin a\cos a=2\left(\frac{5}{13}\right)\times \left(-\frac{12}{13}\right)=\sin 2a=-\frac{120}{169}\end{array}$$

  
  ::-=============================================================================================================================================================================================================================================================================================================================================================================================================================================================================================================================

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  For $\begin{array}{r}\cos 2a\end{array}$ , use $\begin{array}{r}\cos (2a)={\cos }^{2}a-{\sin }^{2}a\end{array}$
  ::对于 cos2a, 使用 cos( 2a) = cos2a- sin2a

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  $$\begin{array}{rl}\cos (2a)& ={\left(-\frac{12}{13}\right)}^2-{\left(\frac{5}{13}\right)}^2\text{or}\frac{144-25}{169}\\ \cos (2a)& =\frac{119}{169}\end{array}$$

  
  :2a) = (- 1213) 2-( 513) 2 或 144- 25169cos * (2a) = 119169

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  For $\begin{array}{r}\tan 2a\end{array}$ , use $\begin{array}{r}\tan 2a=\frac{2\tan a}{1-{\tan }^{2}a}\end{array}$ . From above, $\begin{array}{r}\tan a=\frac{\frac{5}{13}}{-\frac{12}{13}}=-\frac{5}{12}\end{array}$ .
  ::对于 tan2a, 使用 tan2a = 2tana1 - tan2a 。 上面, tana = 513 - 1213 512 。

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  $$\begin{array}{r}\tan (2a)=\frac{2\cdot \frac{-5}{12}}{1-{\left(\frac{-5}{12}\right)}^2}=\frac{\frac{-5}{6}}{1-\frac{25}{144}}=\frac{\frac{-5}{6}}{\frac{119}{144}}=-\frac{5}{6}\cdot \frac{144}{119}=-\frac{120}{119}\end{array}$$

  
  :2a)=25121-(-512)2561-25144}*5611914}}*56}}*56}}}}}*56}}}1}1}1}1}1}1}1}1}1}1}1}1}1}1}1}1}1}1}1}1}1}1}1}1}1}1}1}1}1}1}1}1}1}1}

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  2. Find $\begin{array}{r}\cos 4\theta \end{array}$ .
  ::2. 寻找cos4。

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  Think of $\begin{array}{r}\cos 4\theta \end{array}$ as $\begin{array}{r}\cos (2\theta +2\theta )\end{array}$ .
  ::将Cos4 视为(222) 。

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  $$\begin{array}{r}\cos 4\theta =\cos (2\theta +2\theta )=\cos 2\theta \cos 2\theta -\sin 2\theta \sin 2\theta ={\cos }^{2}2\theta -{\sin }^{2}2\theta \end{array}$$

  
  ::============================================================================================================================================================ ================================================================================================================================================================================================================================================================================================================================================================

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  Now, use the double angle formulas for both . For cosine, you can pick which formula you would like to use. In general, because we are proving a cosine identity, stay with cosine.
  ::现在,对两者都使用双角度公式。 对于余弦, 您可以选择您想要使用的公式。 一般来说, 因为我们正在证明一种余弦特性, 请与余弦保持在一起 。

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  $$\begin{array}{rl}& =(2{\cos }^2\theta -1{)}^2-(2\sin \theta \cos \theta {)}^2\\ & =4{\cos }^4\theta -4{\cos }^2\theta +1-4{\sin }^2\theta {\cos }^2\theta \\ & =4{\cos }^4\theta -4{\cos }^2\theta +1-4(1-{\cos }^2\theta ){\cos }^2\theta \\ & =4{\cos }^4\theta -4{\cos }^2\theta +1-4{\cos }^2\theta +4{\cos }^4\theta \\ & =8{\cos }^4\theta -8{\cos }^2\theta +1\end{array}$$

  
  ::= (2cos21)2-2-2=4cos44444421-4sin22244cos244cos2444444cos24444448cos21

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  3. Solve the trigonometric equation $\begin{array}{r}\sin 2x=\sin x\end{array}$ such that $\begin{array}{r}(-\pi \le x<\pi )\end{array}$
  ::3. 解决三重方程sin2x=sinx等值 (x)

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  Using the sine double angle formula:
  ::使用正弦双角公式 :

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  $$\begin{array}{rl}& \sin 2x=\sin x\\ & 2\sin x\cos x=\sin x\\ & 2\sin x\cos x-\sin x=0\\ & \sin x(2\cos x-1)=0\\ & \downarrow \searrow \\ & 2\cos x-1=0\\ & 2\cos x=1\\ & \sin x=0\\ & x=0,-\pi \cos x=\frac{1}{2}\\ & x=\frac{\pi }{3},-\frac{\pi }{3}\end{array}$$

  
  ::=0 =0 =0 =0 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =

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  Examples
  ::实例

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  Example 1
  ::例1

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  Earlier, you were asked to solve  $\begin{array}{r}\cos {120}^{\circ }\end{array}$ .
  ::早些时候,你被要求解决Cos120。

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  You can simplify this into a familiar angle:
  ::您可以将其简化为熟悉的角度 :

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  $\begin{array}{r}\cos (2\times {60}^{\circ })\end{array}$
  ::COs( 2x60)

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  And then apply the double angle identity:
  ::然后应用双角身份 :

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  $$\begin{array}{rl}\cos (2\times {60}^{\circ })& =2{\cos }^2{60}^{\circ }-1\\ & =(2)(\cos {60}^{\circ })(\cos {60}^{\circ })-1\\ & =(2)(\frac{1}{2})(\frac{1}{2})-1\\ & =-\frac{1}{2}\end{array}$$

  
  :2×60)=2cos2601=(2)(cos60)(cos60)-1=(2)(2)(12)-1=(2)(12)-112

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  Example 2
  ::例2

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  If $\begin{array}{r}\sin x=\frac{4}{5}\end{array}$ and $\begin{array}{r}x\end{array}$ is in Quad II, find the exact values of $\begin{array}{r}\cos 2x,\sin 2x\end{array}$ and $\begin{array}{r}\tan 2x\end{array}$
  ::如果 sinx=45 和 x 在 Quad II 中, 请找到 cos2x, sin2x 和 tan2x 的准确值 。

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   If $\begin{array}{r}\sin x=\frac{4}{5}\end{array}$ and in Quadrant II, then cosine and tangent are negative. Also, by the , the third side is $\begin{array}{r}3(b=\sqrt{5^2-4^2})\end{array}$ . So, $\begin{array}{r}\cos x=-\frac{3}{5}\end{array}$ and $\begin{array}{r}\tan x=-\frac{4}{3}\end{array}$ . Using this, we can find $\begin{array}{r}\sin 2x,\cos 2x\end{array}$ , and $\begin{array}{r}\tan 2x\end{array}$ .
  ::如果 sinx=45 和在 Quadrant II 中, 余弦和正切为负值。 另外, 在 3 (b) = 52 - 42。 所以, cosx =35 和 tan x {43。 使用它, 我们可以找到 sin2x, cos2x 和 tan2x 。

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  $$\begin{array}{rlrlrl}& & & \cos 2x=1-{\sin }^{2}x& & \tan 2x=\frac{2\tan x}{1-{\tan }^{2}x}\\ & & & =1-2\cdot {\left(\frac{4}{5}\right)}^2& & =\frac{2\cdot -\frac{4}{3}}{1-{\left(-\frac{4}{3}\right)}^2}\\ & \sin 2x=2\sin x\cos x& & =1-2\cdot \frac{16}{25}& & =\frac{-\frac{8}{3}}{1-\frac{16}{9}}=-\frac{8}{3}÷-\frac{7}{9}\\ & =2\cdot \frac{4}{5}\cdot -\frac{3}{5}& & =1-\frac{32}{25}& & =-\frac{8}{3}\cdot -\frac{9}{7}\\ & =-\frac{24}{25}& & =-\frac{7}{25}& & =\frac{24}{7}\end{array}$$

  
  ::=22x=1-sin22x tan2x=2tan2x2x1-tan2x=1-2(452)2=2431-(43-2)2x=2sin2x=2sin2xxx=1-22x1625}831_______83__79=245}35=1-3225 83*97 2425725=247

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  Example 3
  ::例3

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  Find the exact value of $\begin{array}{r}{\cos }^2{15}^{\circ }-{\sin }^2{15}^{\circ }\end{array}$
  ::查找 com2 @% 15 @sin2_ 15 @ 15_ 准确值

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  This is one of the forms for $\begin{array}{r}\cos 2x\end{array}$ .
  ::这是 Cos2x 的表格之一 。

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  $$\begin{array}{rl}{\cos }^2{15}^{\circ }-{\sin }^2{15}^{\circ }& =\cos ({15}^{\circ }\cdot 2)\\ & =\cos {30}^{\circ }\\ & =\frac{\sqrt{3}}{2}\end{array}$$

  
  ::-=======================================================================================================================================================================================================================================================

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  Example 4
  ::例4

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  Verify the identity: $\begin{array}{r}\cos 3\theta =4{\cos }^3\theta -3\cos \theta \end{array}$
  ::验证身份: cos34cos333cos}

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  Step 1: Use the cosine sum formula
  ::第1步:使用余弦和公式

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  $$\begin{array}{rl}\cos 3\theta & =4{\cos }^3\theta -3\cos \theta \\ \cos (2\theta +\theta )& =\cos 2\theta \cos \theta -\sin 2\theta \sin \theta \end{array}$$

  
  ::=============================================================================================================================================================================================================================================================================================================================================================================================================================================================================================================================

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  Step 2: Use double angle formulas for $\begin{array}{r}\cos 2\theta \end{array}$ and $\begin{array}{r}\sin 2\theta \end{array}$
  ::第2步:对 cos2和 sin2使用双角公式

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  $$\begin{array}{r}=(2{\cos }^2\theta -1)\cos \theta -(2\sin \theta \cos \theta )\sin \theta \end{array}$$

  
  ::= (2cos21)cos(2sincos)sin(2sincos)sin()

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  Step 3: Distribute and simplify.
  ::第3步:分配和简化。

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  $$\begin{array}{rl}& =2{\cos }^3\theta -\cos \theta -2{\sin }^2\theta \cos \theta \\ & =-\cos \theta (-2{\cos }^2\theta +2{\sin }^2\theta +1)\\ & =-\cos \theta [-2{\cos }^2\theta +2(1-{\cos }^2\theta )+1]& & \to \text{Substitute}1-{\cos }^2\theta \text{for}{\sin }^2\theta \\ & =-\cos \theta [-2{\cos }^2\theta +2-2{\cos }^2\theta +1]\\ & =-\cos \theta (-4{\cos }^2\theta +3)\\ & =4{\cos }^3\theta -3\cos \theta \end{array}$$

  
  ::=2cos32222222221222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222233333333333334c2322222222

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  Review
  ::回顾

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  Simplify each expression so that it is in terms of $\begin{array}{r}\sin (x)\end{array}$ and $\begin{array}{r}\cos (x)\end{array}$ .
  ::简化每个表达式, 使其以sin( x) 和cos( x) 表示 。

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  1. $\begin{array}{r}\sin 2x+\cos x\end{array}$
     ::sin *% 2x+cos *x
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  2. $\begin{array}{r}\sin 2x+\cos 2x\end{array}$
     ::sin =% 2x+cos=%2x
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  3. $\begin{array}{r}\sin 3x+\cos 2x\end{array}$
     ::sin 3x+cos%2x sin 3x+cos%2x
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  4. $\begin{array}{r}\sin 2x+\cos 3x\end{array}$
     ::sin =% 2x+cos=%3x sin =% 2x+cos=% 3x

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  Solve each equation on the interval $\begin{array}{r}[0,2\pi )\end{array}$ .
  ::在间隔[0,2]中解决每个方程。

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  5. $\begin{array}{r}\sin (2x)=2\sin (x)\end{array}$
     :xx)=2sin *(x)
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  6. $\begin{array}{r}\cos (2x)=\sin (x)\end{array}$
     ::COs( 2x) =sin( x)
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  7. $\begin{array}{r}\sin (2x)-\tan (x)=0\end{array}$
     :x) =0 (% 2x)
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  8. $\begin{array}{r}{\cos }^2(x)+\cos (x)=\cos (2x)\end{array}$
     ::cos2(x)+cos(x)=cos(2x)
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  9. $\begin{array}{r}\cos (2x)=\cos (x)\end{array}$
     ::cos( 2x) =cos( x)

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  Simplify each expression so that only one calculation would be needed in order to evaluate.
  ::简化每个表达式,以便评价只需要一个计算法。

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  10. $\begin{array}{r}2{\cos }^2({15}^{\circ })-1\end{array}$
      ::2cos2(15)-1
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  11. $\begin{array}{r}2\sin ({25}^{\circ })\cos ({25}^{\circ })\end{array}$
      ::2sin(25)cos(25)
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  12. $\begin{array}{r}1-2{\sin }^2({35}^{\circ })\end{array}$
      ::1-2sin2__________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
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  13. $\begin{array}{r}{\cos }^2({60}^{\circ })-{\sin }^2({60}^{\circ })\end{array}$
      ::coms2(60)-sin2(60)
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  14. $\begin{array}{r}2\sin ({125}^{\circ })\cos ({125}^{\circ })\end{array}$
      ::2sin(125)cos(125)
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  15. $\begin{array}{r}1-2{\sin }^2({32}^{\circ })\end{array}$
      ::1-2sin2(32)

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  Review (Answers)
  ::回顾(答复)

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  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 。