We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
95
4
avatar

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+22182 
+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
avatar
0

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+22182 
+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+100539 
+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

23 Online Users

avatar