Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

How to find $\arccos(\cos15\pi/11)$?

I'm completely lost. $15\pi/11$ isn't on the unit circle so how do I find the cosine of it?

share|improve this question
1  
Do you understand radians? Pi radians are 180 degrees, 2pi radian to a full circle. –  JoeTaxpayer Dec 6 '13 at 18:14
    
@JoeTaxpayer I mean that 15pi/11 isn't on the unit circle in the sense that it's not one of the generic ones like pi/3 or 3pi/2. So how do I find the cosine of something that's not one of the generic values on the unit circle? –  Anon Dec 6 '13 at 18:21
1  
Sorry. I understand now. Using a calculator isn't an option I suppose? A large enough well drawn circle will let you interpolate interim values. 1.3636 pi. –  JoeTaxpayer Dec 6 '13 at 18:24

1 Answer 1

As $\cos x=\cos A\implies x=2n\pi\pm A$ where $n$ is any integer

The general value of $$\arccos\left(\cos\frac{15\pi}{11}\right)=2n\pi\pm \frac{15\pi}{11}$$

Based on the definition of principal value inverse cosine ratio, $\displaystyle0\le \arccos x\le\pi$

Taking '+' sign, $\displaystyle0\le 2n\pi+\frac{15\pi}{11}\le\pi\implies 0\le 22n+15\le 11\implies0\le n<0$ which is impossible

Taking '-' sign, $\displaystyle0\le 2n\pi-\frac{15\pi}{11}\le\pi\implies 0\le 22n-15\le 11\implies 1\le n\le1\implies n=1$

So, the principal value is $\displaystyle 2\pi-\frac{15\pi}{11}=??$


Alternatively, as we know $\cos\left(\pi\pm y\right)=-\cos y$

$$\cos\left(\frac{15\pi}{11}\right)=\cos\left(\pi+\frac{4\pi}{11}\right)=-\cos\left(\frac{4\pi}{11}\right)=\cos\left(\pi-\frac{4\pi}{11}\right)=\cos\left(\frac{7\pi}{11}\right)$$

share|improve this answer
    
"15pi/11 isn't on the unit circle" - Do you think OP will find 7pi/11 any easier? –  JoeTaxpayer Dec 6 '13 at 18:16
    
@JoeTaxpayer, does not matter. I am supplying a generic solution –  lab bhattacharjee Dec 6 '13 at 18:18
    
@labbhattacharjee I don't quite understand your solution. I understand radians. –  Anon Dec 6 '13 at 18:20
    
@Anon, please find the first method –  lab bhattacharjee Dec 6 '13 at 18:27
    
@JoeTaxpayer, would you mind checking the edited answer? –  lab bhattacharjee Dec 6 '13 at 18:30

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.