Question

If $ \cos \dfrac{\pi }{{33}}\cos \dfrac{{2\pi }}{{33}}\cos \dfrac{{4\pi }}{{33}}\cos \dfrac{{8\pi }}{{33}}\cos \dfrac{{16\pi }}{{33}}\cos = \dfrac{1}{m} $ , then m =?

Answer 1

Hint: Generally these types of question can be simplified easily but since in this question it is difficult to simplify it so here we can multiply the equation by \[\dfrac{{2\sin \dfrac{\pi }{{33}}}}{{2\sin \dfrac{\pi }{{33}}}}\] then apply the formula \[\sin 2\theta = 2\sin \theta \cos \theta \] in the equation to find the value of m.

Complete step-by-step answer:
Let \[\cos \dfrac{\pi }{{33}}\cos \dfrac{{2\pi }}{{33}}\cos \dfrac{{4\pi }}{{33}}\cos \dfrac{{8\pi }}{{33}}\cos \dfrac{{16\pi }}{{33}}\cos \] take as equation 1
Now let multiply equation 1 by \[\dfrac{{2\sin \dfrac{\pi }{{33}}}}{{2\sin \dfrac{\pi }{{33}}}}\]
 $ \Rightarrow $ \[\cos \dfrac{\pi }{{33}}\cos \dfrac{{2\pi }}{{33}}\cos \dfrac{{4\pi }}{{33}}\cos \dfrac{{8\pi }}{{33}}\cos \dfrac{{16\pi }}{{33}}\cos \] \[ \times \] \[\dfrac{{2\sin \dfrac{\pi }{{33}}}}{{2\sin \dfrac{\pi }{{33}}}}\] =\[\dfrac{{\sin \dfrac{{2\pi }}{{33}}}}{{2\sin \dfrac{\pi }{{33}}}}\cos \dfrac{{2\pi }}{{33}}\cos \dfrac{{4\pi }}{{33}}\cos \dfrac{{8\pi }}{{33}}\cos \dfrac{{16\pi }}{{33}}\cos \] by the formula \[\sin 2\theta = 2\sin \theta \cos \theta \]
\[\dfrac{{\sin \dfrac{{2\pi }}{{33}}}}{{2\sin \dfrac{\pi }{{33}}}}\cos \dfrac{{2\pi }}{{33}}\cos \dfrac{{4\pi }}{{33}}\cos \dfrac{{8\pi }}{{33}}\cos \dfrac{{16\pi }}{{33}}\cos \] (Equation 2)
Now multiplying and dividing the equation 2 by $ {2^4} $
\[\dfrac{{{2^4}\sin \dfrac{{2\pi }}{{33}}}}{{{2^5}\sin \dfrac{\pi }{{33}}}}\cos \dfrac{{2\pi }}{{33}}\cos \dfrac{{4\pi }}{{33}}\cos \dfrac{{8\pi }}{{33}}\cos \dfrac{{16\pi }}{{33}}\cos \] (Equation 3)
Now applying the formula \[\sin 2\theta = 2\sin \theta \cos \theta \] 4 times to simplify the equation into simplest form i.e.
 $ \Rightarrow $ $ \dfrac{{\sin \dfrac{{32\pi }}{{33}}}}{{32\sin \dfrac{\pi }{{33}}}} $ = $ \dfrac{{\sin (\pi - \dfrac{\pi }{{33}})}}{{32\sin \dfrac{\pi }{{33}}}} $
 $ \Rightarrow $ \[\dfrac{{\sin \dfrac{\pi }{{33}}}}{{32\sin \dfrac{\pi }{{33}}}} = \dfrac{1}{{32}}\] By the formula of $ \sin (\pi - \theta ) = \sin \theta $
So the value of $ \dfrac{1}{m} = \dfrac{1}{{32}} $ .

Note: In the solution we have used the term trigonometric identities are equalities that involve trigonometric functions like $ \sin \theta $ , $ \cos \theta $ , $ \tan \theta $ , etc. If we explain this term as geometrically, these are identities involving certain functions of one or more angles and they are distinct from triangle identities, which are identities potentially involving angles but also involving side lengths or other lengths of a triangle. There is an important application i.e. the integration of non-trigonometric functions: a common technique involves first using the substitution rule with a trigonometric function, and then simplifying the resulting integral with a trigonometric identity.