Hello,

Can someone walk me through how to get -cos(5pi/6) + cos(0) as an exact number without decimals?

Thanks!

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- Sep 15th 2008, 10:01 PM2clientssimple approximation but need exact answer
Hello,

Can someone walk me through how to get -cos(5pi/6) + cos(0) as an exact number without decimals?

Thanks! - Sep 15th 2008, 10:14 PMJameson
cos(0) = 1.

$\displaystyle \frac{5 \pi}{6}$, convert this to degrees to see if it's a special angle we know about. I get that this is 150 deg, which is in the second quadrant and has an angle of 30 deg (I forgot the terminology for this angle but you get what I mean hopefully). But, a 30-60-90 triangle is one we know about! Try drawing the angle and working out the value. - Sep 15th 2008, 10:38 PM2clients
Okay great, I see where you're going. I forgot about the 30-60-90 thing - where it corresponds to n across from the 30, 2n across from the 90, and nsqrt3 across from the 60. However, Im still stuck and would love any more explanation.

Thanks! - Sep 15th 2008, 10:48 PMJameson
Ok, I can't draw well so I'll do my best.

This triangle has a horizontal leg of -2, a vertical leg of 1, and a hypotenuse of $\displaystyle \sqrt{3}$ Simple way to see this is the smallest leg is always across the smallest angle, thus the vertical leg across the 30 deg angle is 1. So draw that triangle to see it visually. Now if you take the cosine of the 30 deg, you should get adjacent/hyp. or $\displaystyle \frac{-2}{\sqrt{3}}$. Make sense? - Sep 16th 2008, 09:19 AMMoo
Hello,

Note that $\displaystyle \frac{5 \pi}{6}=\frac{6 \pi- \pi}{6}=\pi-\frac \pi 6$

Using the rule $\displaystyle \cos(a-b)=\cos(a)\cos(b)+\sin(a)\sin(b)$, we can prove that $\displaystyle \cos\left(\pi-\frac \pi 6\right)=-\cos\left(\frac \pi 6\right)$

And $\displaystyle \cos \left(\frac \pi 6\right)=\frac{\sqrt{3}}{2}$