Question

The value of $\cos {12}^{\circ }\cos {24}^{\circ }\cos {36}^{\circ }\cos {48}^{\circ }\cos {72}^{\circ }\cos {84}^{\circ }$ is

• A. $\frac{1}{64}$
• B. $\frac{1}{32}$
• C. $\frac{1}{16}$
• D. $\frac{1}{128}$

Answer 1

The correct option is A $\frac{1}{64}$

Using identity $\{\because 2\sin \theta \cos \theta =\sin 2\theta \}$
$a=\cos {12}^{}\cos {24}^{}\cos {36}^{}\cos {48}^{}\cos {72}^{}\cos {84}^{}$
$\Rightarrow a=\frac{\left(2\sin {12}^{}\cos {12}^{}\right)\cos {24}^{}\cos {36}^{}\cos {48}^{}\cos {72}^{}\cos {84}^{}}{2\sin {12}^{}}$
$\Rightarrow a=\frac{2\sin {24}^{}\cos {24}^{}\cos {36}^{}\cos {48}^{}\cos {72}^{}\cos {84}^{}}{4\sin {12}^{}}$
$\Rightarrow a=\frac{-2\sin {48}^{}\cos {48}^{}\cos {36}^{}\cos {72}^{}\cos ({180}^{}-{84}^{})}{8\sin {12}^{}}$
$\Rightarrow a=\frac{-2\sin {96}^{}\cos {96}^{}\cos {36}^{}\cos {72}^{}}{16\sin {12}^{}}$
$\Rightarrow a=\frac{-\sin {192}^{}\cos {36}^{}\cos {72}^{}}{16\sin {12}^{}}=\frac{-\sin ({180}^{}+{12}^{})\cos {36}^{}\cos {72}^{}}{16\sin {12}^{}}$
$\Rightarrow a=\frac{\sin {12}^{}\left(2\sin {36}^{}\cos {36}^{}\right)\cos {72}^{}}{32\sin {12}^{}\sin {36}^{}}=\frac{2\sin {72}^{}\cos {72}^{}}{64\sin {36}^{}}$
$\Rightarrow a=\frac{\sin {144}^{}}{64\sin {36}^{}}=\frac{\sin ({180}^{}-{144}^{})}{64\sin {36}^{}}=\frac{1}{64}$
Hence, option 'A' is correct.