#3**+20 **

I doubt you will read this, but I'm actually from New Zealand and I'm in Auckland.

EighthMersennePrime Feb 15, 2017

#1**+10 **

Hi Eight MersenePrime,

I was thinking about your username... I looked up meridians to see where you might come from.

I am thinking that maybe you are from Strasbourg in France.... :)

I had not realized before that Bamako is further west than london - quite a lot further west !!

I love looking at maps :))

Now to answer your question :)

cos(19\pi/12)

\(cos(\frac{19\pi}{12})\\ \text{4th quadrant}\\ =+cos(\frac{(24-19)\pi}{12})\\ =+cos(\frac{5\pi}{12})\\ =+cos(\frac{3\pi}{12}+\frac{2\pi}{12})\\ =cos(\frac{\pi}{4}+\frac{\pi}{6})\\ =cos(\frac{\pi}{4})cos(\frac{\pi}{6})-sin(\frac{\pi}{4})sin(\frac{\pi}{6})\\ =cos(45^\circ)cos(30^\circ)-sin(45^\circ)sin(30^\circ)\\ =\frac{1}{\sqrt2}\times \frac{\sqrt3}{2}-\frac{1}{\sqrt2}\times \frac{1}{2}\\ =\frac{\sqrt3-1}{2\sqrt2}\\ =\frac{\sqrt3-1}{2\sqrt2}\times \frac{\sqrt2}{\sqrt2}\\ =\frac{\sqrt6-\sqrt2}{2*2}\\ =\frac{\sqrt6-\sqrt2}{4}\\ \)

.Melody Feb 14, 2017

#3**+20 **

Best Answer

I doubt you will read this, but I'm actually from New Zealand and I'm in Auckland.

EighthMersennePrime
Feb 15, 2017

#2**+15 **

**Random math IV**

**1.**

\(\begin{array}{|rcll|} \hline \cos(\frac{19}{12}\pi) &=& \cos(285^{\circ}) \\ &=& \cos(285^{\circ}-360^{\circ}) \\ &=& \cos(-75^{\circ}) \\ &=& \cos(15^{\circ}-90^{\circ}) \\ &=& \cos(-(90^{\circ}-15^{\circ})) \\ &=& \cos( 90^{\circ}-15^{\circ} ) \\ &=& \mathbf{\sin( 15^{\circ} )} \\\\ \mathbf{\cos(\frac{19}{12}\pi) } &\mathbf{=}& \mathbf{\sin( 15^{\circ} )}\\ \hline \end{array} \)

**2.**

\(\begin{array}{|lrcll|} \hline (1) & \cos(45^{\circ}-15^{\circ})= \cos(30^{\circ}) &=& \cos(45^{\circ})\cdot\cos(15^{\circ})+\sin(45^{\circ})\cdot \sin(15^{\circ})\\ (2) & \cos(45^{\circ}+15^{\circ})= \cos(60^{\circ}) &=& \cos(45^{\circ})\cdot\cos(15^{\circ})-\sin(45^{\circ})\cdot \sin(15^{\circ})\\ \\ \hline (1)-(2): & \cos(30^{\circ})-\cos(60^{\circ}) &=& 2\cdot\sin(45^{\circ})\cdot \sin(15^{\circ})\\\\ & \cos(30^{\circ}) &=& \frac{\sqrt{3}}{2} \\ & \cos(60^{\circ}) &=& \frac{1}{2} \\ & \sin(45^{\circ}) &=& \frac{\sqrt{2}}{2} \\ \\ & \frac{\sqrt{3}}{2}-\frac{1}{2} &=& 2\cdot \frac{\sqrt{2}}{2}\cdot \sin(15^{\circ})\\ & \frac{\sqrt{3}-1}{2} &=& \sqrt{2}\cdot \sin(15^{\circ})\\ & \frac{\sqrt{3}-1}{2\cdot \sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} &=& \sin(15^{\circ})\\ & \frac{\sqrt{2}\sqrt{3}-\sqrt{2}}{4} &=& \sin(15^{\circ})\\ & \frac{\sqrt{6}-\sqrt{2}}{4} &=& \sin(15^{\circ})\\\\ & \mathbf{\cos(\frac{19}{12}\pi) = \sin( 15^{\circ} ) }&\mathbf{=}& \mathbf{\frac{\sqrt{6}-\sqrt{2}}{4}} \\ \hline \end{array} \)

heureka Feb 14, 2017