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I need to determine the exact value for cos 195°. The answer is \( {- \sqrt{3} + 1 \over \sqrt{3}+1}\) or \({\sqrt{3} - 2}\) in the textbook, what did I do wrong?

 

Thank you! :)

 Feb 21, 2019
 #1
avatar+21848 
+4

Determining the exact value of each trig expression.

 

Your calculation is correct.

\(\cos(195^\circ)=-\cos(15°) = -\dfrac14 *\sqrt{2}*(\sqrt{3}+1) \\ \tan(165^\circ)=-\tan(15°) = \sqrt{3} - 2\)

 

laugh

 Feb 21, 2019
 #2
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So is my answer plausible despite the answer in the textbook?

 

Also, I'm a little confused about the method you've used.

Guest Feb 21, 2019
 #4
avatar+21848 
+3

Hi Guest,


\(\begin{array}{|rcll|} \hline && \mathbf{\cos(195^\circ)} \\ &=& \cos(180^\circ +15^\circ)\\ &=& \underbrace{\cos(180^\circ)}_{=-1}\cos(15^\circ) - \underbrace{\sin(180^\circ)}_{=0}\sin(15^\circ)\\ &=& -\cos(15^\circ) - 0\cdot \sin(15^\circ)\\ &\mathbf{=}& \mathbf{-\cos(15^\circ)} \\\\ &=& -\cos(45^\circ - 30^\circ) \\ &=& -\Big( \underbrace{\cos(45^\circ)}_{=\dfrac{\sqrt{2}}{2}} \underbrace{\cos(30^\circ) }_{=\dfrac{\sqrt{3}}{2}} + \underbrace{\sin(45^\circ)}_{=\dfrac{\sqrt{2}}{2}} \underbrace{\sin(30^\circ)}_{=\dfrac{1}{2}}\Big) \\ &=& -\dfrac14 *\sqrt{2}*(\sqrt{3}+1) \\ \hline \end{array}\)

 

laugh

heureka  Feb 22, 2019
edited by heureka  Feb 22, 2019
 #3
avatar+98130 
+1

cos (195)  = 

 

cos (135 + 60) =

 

cos135cos60 - sin135sin60 = 

 

-√2/2 * 1/2  -  √2/2 * √3/2    =

 

-√2/4 - √6/4 =

 

- [ √2 + √6 ] / 4  =

 

-√2 [ 1 + √ 3 ] /  4

 

 

 

cool cool cool  

 Feb 21, 2019
edited by CPhill  Feb 21, 2019

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