Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Trignometry
0 votes

If $A+B+C=\pi$ then $\cos^2A+\cos^2B+\cos^2C$ is equal to

$\begin{array}{1 1}(a)\;1-\cos A\cos B\cos C&(b)\;1-2\sin A\sin B\sin C\\(c)\;1-\sin A\sin B\sin C&(d)\;1-2\cos A\cos B\cos C\end{array}$

Can you answer this question?

1 Answer

0 votes
Step 1:
We have $\cos^2A+\cos ^2B+\cos ^2C$
$\sin ^2B+\cos^2B=1$
$\Rightarrow \cos^2A+(1-\sin ^2B)+\cos ^2C$
$\Rightarrow (\cos ^2A-\sin^2B)+\cos^2C+1$
$\Rightarrow \cos(A+B)\cos(A-B)+\cos^2C+1$
Step 2:
The above equation can be written as
$\Rightarrow -\cos C\cos(A-B)+\cos^2C+1$
$\Rightarrow -\cos C[\cos (A-B)-\cos C]+1$
$\Rightarrow -\cos C[\cos(A-B)-\cos\{\pi-(A+B)\}]+1$
$\Rightarrow -\cos C[\cos(A-B)+\cos(A+B)]+1$
$\cos(A-B)+\cos(A+B)=2\cos A\cos B$
$\Rightarrow -\cos C(2\cos A\cos B)+1$
$\Rightarrow 1-2\cos A\cos B\cos C$
Hence (d) is the correct answer.
answered Oct 21, 2013 by sreemathi.v

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App