Find the exact value of cos(11pi/12), showing steps, etc

[Snofrost]

I honestly don't understand these types of problems and hopefully one of you could help me! Here is the question I'm workin on right now:

"Find the exact value of cos(11pi/12). (show all steps and formulas used)"

I'm thinking it has something to do with special triangles...? Maybe adding or subtracting special triangle angles or something -- I gather from my notes -- but I just can't remember.

Please Help!!

Thanks... <3 Cherise <3

 

[pka]

We will use \(\displaystyle \L \,\left| {\cos \left( {\frac{\theta }{2}} \right)} \right|\, =\, \sqrt {\frac{{1\, +\, \cos (\theta )}}{2}}\).

Now \(\displaystyle \L \,\cos \left( \frac{11\pi }{6} \right)\, =\, \cos \left( {\frac{{ - \pi }}{6}} \right)\, =\, \frac{{\sqrt 3 }}{2}\).

To finish just note that \(\displaystyle \L \,\frac{{11\pi }}{{12}}\, =\, \frac{1}{2}\,\frac{{11\pi }}{6}\).

BUT take care \(\displaystyle \,\frac{{11\pi }}{{12}} \in II\).

 

[soroban]

Re: Help with Trig

Hello, Cherise!

There are a number of approaches.

Find the exact value of \(\displaystyle \cos\left(\frac{11\pi}{12}\right)\)

Your idea is correct . . . combining special angles.

You're expected to know the trig values for: \(\displaystyle \frac{\pi}{6},\:\frac{\pi}{4},\:\frac{\pi}{3},\:\cdots\)
. . and exercise your Imagination circuit.

How can we construct \(\displaystyle \frac{11\pi}{12}\) from known angles?

Since \(\displaystyle 11\:=\:3\,+\,8\), we have: \(\displaystyle \,\frac{11}{12}\:=\:\frac{3}{12}\,+\,\frac{8}{12}\:=\:\frac{1}{4}\,+\,\frac{2}{3}\)
. . Hence: \(\displaystyle \L\:\frac{11\pi}{12}\:=\:\frac{\pi}{4}\,+\,\frac{2\pi}{3}\;\) **

Compound-Angle Identity: \(\displaystyle \:\cos(A\,+\,B)\:=\:\cos(A)\cos(B)\,-\,\sin(A)\sin(B)\)


We have: \(\displaystyle \L\:\cos\left(\frac{\pi}{4}\,+\,\frac{2\pi}{3}\right) \;=\;\cos\left(\frac{\pi}{4}\right)\cos\left(\frac{2\pi}{3}\right) \,-\,\sin\left(\frac{\pi}{4}\right)\sin\left(\frac{2\pi}{3}\right)\)

. . . . . . . . . . . . . . . . . . . . \(\displaystyle \L= \;\left(\frac{\sqrt{2}}{2}\right)\left(-\frac{1}{2}\right)\,-\,\left(\frac{\sqrt{2}}{2}\right)\left(\frac{\sqrt{3}}{2}\right)\)

. . . . . . . . . . . . . . . . . . . . \(\displaystyle \L=\;\fbox{-\frac{\sqrt{2}\,+\,\sqrt{6}}{4}}\)

**
Since \(\displaystyle 11\:=\:2\,+\,9\;\;\Rightarrow\;\;\frac{11}{12}\:=\:\frac{2}{12}\,+\,\frac{9}{12} \:=\:\frac{1}{6}\,+\,\frac{3}{4}\),

. . we could use: \(\displaystyle \frac{11\pi}{12} \:=\:\frac{\pi}{6}\,+\,\frac{3\pi}{4}\)

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

pka's method is faster.

Double-angle Identity: \(\displaystyle \L\:\cos^2\theta \;=\;\frac{1\,+\,\cos2\theta}{2}\)

So we have: \(\displaystyle \L\:\cos^2\left(\frac{11\pi}{12}\right) \;=\;\frac{1\,+\,\cos\left(\frac{11\pi}{6}\right)}{2} \;=\; \frac{1\,+\,\frac{\sqrt{3}}{2}}{2}\;=\;\frac{2\,+\,\sqrt{3}}{4}\)

. . Hence: \(\displaystyle \L\:\cos\left(\frac{11\pi}{12}\right) \;=\;\pm\frac{\sqrt{2\,+\,\sqrt{3}}}{2}\)


Since \(\displaystyle \frac{11\pi}{12}\) is in Qudrant II, where \(\displaystyle \cos\theta\) is negative:

. . \(\displaystyle \L\cos\left(\frac{11\pi}{12}\right)\;=\;\fbox{-\frac{\sqrt{2\,+\,\sqrt{3}}}{2}}\)


And yes, the two answers are equal . . . Prove it yourself!

 

[Snofrost]

Oh wow thats why i couldn't get it! Thank you so much. I can see that pka's method is much faster but it confuses me a lot.

Thanks so much pka and soraban

Much appreciated!