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! :)

Guest Feb 21, 2019

#1**+4 **

**Determining the exact value of each trig expression.**

Your calculation is correct.

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

heureka Feb 21, 2019

#2**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**+3 **

Hi Guest,

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

heureka
Feb 22, 2019