Little Trigonometry

[Rabeeb]

If cos 3=$\sqrt p$,
Then cos 3.cos 6.cos 12$\cdots$cos 384=?
*All angles expressed in degrees

[Nadim Ul Abrar]

$\displaystyle \cos 3.\cos 6.\cos12 ... \cos 384$ $\displaystyle =\frac{2^7.\sin3.\cos 3.\cos 6.\cos12 ... \cos 384}{2^7.\sin3}=\frac{\sin 768}{2^7.\sin3}=\frac{\sin 48}{2^7.\sin3}$ $\displaystyle =\frac{\sin3+\cos3}{2^7 \sqrt {2}.\sin 3}=\frac{1}{2^7 \sqrt {2}}+\frac{\cot 3}{2^7 \sqrt {2}}$

[*Mahi*]

Use \sin, \cos and \tan instead of sin,cos or tan, this way they look more tidy.