Question

In $\Delta ABC,$ with angles $A,B$ and$C,\cos \left(\frac{B+2C+3A}{2}\right)+\cos \left(\frac{A-B}{2}\right)=$

• A. $-1$
• B. $0$
• C. $1$
• D. $2$

Answer 1

The correct option is B $0$

Sum of Interior angles of at Triangle is $={180}^o$

So, $A+B+C={180}^o\Rightarrow C={180}^o-A-B$

$\cos \left(\frac{B+2C+3A}{2}\right)+\cos \left(\frac{A-B}{2}\right)=\cos \left(\frac{B+2({180}^o-A-B)+3A}{2}\right)+\cos \left(\frac{A-B}{2}\right)=\cos \left(\frac{{360}^o+A-B}{2}\right)+\cos \left(\frac{A-B}{2}\right)=-\cos \left(\frac{A-B}{2}\right)+\cos \left(\frac{A-B}{2}\right)=0$

Therefore, Answer is $0$