logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
0 votes

Evaluate \(cos\bigg(cos^{-1}\frac{-\sqrt{3}}{2}+\frac{\pi}{6}\bigg)\)

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • \(-cos\theta=cos(\pi-\theta)\)
  • Principal interval of cos is \([0,\pi]\)
  • \(cos\large\frac{\pi}{6}=\large\frac{\sqrt{3}}{2}\)
  • \(cos\pi=-1\)
\(-\large\frac{\sqrt{3}}{2}=-cos\large\frac{\pi}{6}\)
By taking \(\theta=\large\frac{\pi}{6},\:-cos\large\frac{\pi}{6}=cos(\pi-\large\frac{\pi}{6})\)
\(cos^{-1}-\large\frac{\sqrt{3}}{2}=cos^{-1}(-cos\large\frac{\pi}{6})=cos^{-1}(cos(\pi-\large\frac{\pi}{6}))\)
\(=(\pi-\large\frac{\pi}{6})\)
 
The given expression becomes
\( cos\big(cos^{-1}(-\large\frac{\sqrt{3}}{2})+\large\frac{\pi}{6}\big)=cos( \pi-\large\frac{\pi}{6}+\large\frac{\pi}{6})\)
\( = cos\pi=-1\)

 

answered Mar 2, 2013 by thanvigandhi_1
edited Mar 19, 2013 by thanvigandhi_1
 
Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...