Solveeit Logo

Question

Question: If \( \cos \dfrac{\pi }{{33}}\cos \dfrac{{2\pi }}{{33}}\cos \dfrac{{4\pi }}{{33}}\cos \dfrac{{8\pi }...

If cosπ33cos2π33cos4π33cos8π33cos16π33cos=1m\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 =?

Explanation

Solution

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 2sinπ332sinπ33\dfrac{{2\sin \dfrac{\pi }{{33}}}}{{2\sin \dfrac{\pi }{{33}}}} then apply the formula sin2θ=2sinθcosθ\sin 2\theta = 2\sin \theta \cos \theta in the equation to find the value of m.

Complete step-by-step answer:
Let cosπ33cos2π33cos4π33cos8π33cos16π33cos\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 2sinπ332sinπ33\dfrac{{2\sin \dfrac{\pi }{{33}}}}{{2\sin \dfrac{\pi }{{33}}}}
\Rightarrow cosπ33cos2π33cos4π33cos8π33cos16π33cos\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 2sinπ332sinπ33\dfrac{{2\sin \dfrac{\pi }{{33}}}}{{2\sin \dfrac{\pi }{{33}}}} =sin2π332sinπ33cos2π33cos4π33cos8π33cos16π33cos\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 sin2θ=2sinθcosθ\sin 2\theta = 2\sin \theta \cos \theta
sin2π332sinπ33cos2π33cos4π33cos8π33cos16π33cos\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 24{2^4}
24sin2π3325sinπ33cos2π33cos4π33cos8π33cos16π33cos\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 sin2θ=2sinθcosθ\sin 2\theta = 2\sin \theta \cos \theta 4 times to simplify the equation into simplest form i.e.
\Rightarrow sin32π3332sinπ33\dfrac{{\sin \dfrac{{32\pi }}{{33}}}}{{32\sin \dfrac{\pi }{{33}}}} = sin(ππ33)32sinπ33\dfrac{{\sin (\pi - \dfrac{\pi }{{33}})}}{{32\sin \dfrac{\pi }{{33}}}}
\Rightarrow sinπ3332sinπ33=132\dfrac{{\sin \dfrac{\pi }{{33}}}}{{32\sin \dfrac{\pi }{{33}}}} = \dfrac{1}{{32}} By the formula of sin(πθ)=sinθ\sin (\pi - \theta ) = \sin \theta
So the value of 1m=132\dfrac{1}{m} = \dfrac{1}{{32}} .

Note: In the solution we have used the term trigonometric identities are equalities that involve trigonometric functions like sinθ\sin \theta , cosθ\cos \theta , tanθ\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.