Question

$cos\frac{\pi }{12}+cos\frac{3\pi }{12}+cos\frac{9\pi }{12}+cos\frac{11\pi }{12}=$

• A. $cos2\pi$
• B. $cos\frac{5\pi }{12}$
• C. $cos\frac{6\pi }{12}$
• D. None

Answer 1

Answer: (C)

$cos\frac{6\pi }{12}$

$\cos \frac{\pi }{12}+\cos \frac{3\pi }{12}+\cos \frac{9\pi }{12}+\cos \frac{11\pi }{12}$

Using $\cos C+\cos D=\frac{C+D}{2}.\cos \frac{C-D}{2}$

$+\cos \frac{(\frac{\pi }{12}+\frac{3\pi }{12})}{2}.\cos \frac{(\frac{\pi }{12}-\frac{3\pi }{12})}{2}+2\cos \left(\frac{20\pi }{2.12}\right)+\cos \left(\frac{2\pi }{2.12}\right)$

$=2\cos \left(\frac{2\pi }{12}\right).\cos \left(\frac{\pi }{12}\right)+2\cos \left(\frac{10\pi }{12}\right).\cos \left(\frac{\pi }{12}\right)$

$=2\cos \left(\frac{\pi }{12}\right)[\cos \left(\frac{\pi }{6}\right)+\cos \left(\frac{5\pi }{6}\right)]$

$2\cos \left(\frac{\pi }{12}\right)[2\cos \left(\frac{6\pi }{6.2}\right).\cos \left(\frac{4\pi }{6.2}\right)]$

$2\cos \left(\frac{\pi }{12}\right)[2\cos \left(\frac{\pi }{2}\right).\cos \left(\frac{2\pi }{6}\right)]$

$=0$ or $\cos \left(\frac{5\pi }{12}\right)$