Question

Solve this $1 + \cos {\text{ 5}}{{\text{6}}^ \circ } + \cos {\text{ 5}}{{\text{8}}^ \circ } - \cos {\text{ 6}}{{\text{6}}^ \circ } = {\text{ }}m\cos {\text{ 2}}{{\text{8}}^ \circ }.{\text{ cos 2}}{{\text{9}}^ \circ }.\sin {\text{ 3}}{{\text{3}}^ \circ }$ . Find $m$.

Answer 1

Hint: To solve this we have to take L.H.S of the question and try to derive R.H.S terms by simplifying it. Here in this question, we have to use the concept and formulas of trigonometry and its identities and its functions. Thus, in this way we get our desired answer.

Complete step-by-step answer:
By rearranging the L.H.S. of given equation

$ = \left( {1 - \cos {\text{ 6}}{{\text{6}}^ \circ }} \right) + \left( {\cos {\text{ 5}}{{\text{6}}^ \circ } + \cos \;{\text{5}}{{\text{8}}^ \circ }} \right)$

Now, by using $\left( {1 - \cos 2\theta } \right) = 2{\sin ^2}\theta $ and$\cos x + \cos y = 2\cos \dfrac{{x + y}}{2}\cos \dfrac{{x - y}}{2}$

$

   = 2{\sin ^2}{33^ \circ } + \left( {2\cos {{57}^ \circ }{\text{.cos}}{{\text{1}}^ \circ }} \right) \\

   = 2{\sin ^2}{33^ \circ } + \left( {2\cos \left( {{{90}^ \circ } - {{33}^ \circ }} \right)\cos \left( {{{90}^ \circ } - {{89}^ \circ }} \right)} \right) \\

 $

Now using $\cos {90^ \circ } - \theta = \sin \theta $

$

   = 2{\sin ^2}{33^ \circ } + \left( {2\sin {{33}^ \circ }.\sin {{89}^ \circ }} \right) \\

   = 2\sin {33^ \circ }\left( {\sin {{33}^ \circ } + \sin {{89}^ \circ }} \right) \\

 $

Now using the formula $\sin x + \sin y = 2\sin \dfrac{{x + y}}{2}\cos \dfrac{{x + y}}{2}$

$

   = 2\sin {33^ \circ }\left( {2\sin {{61}^ \circ }.\cos {{28}^ \circ }} \right) \\

   = 4\sin {33^ \circ }.\sin {61^ \circ }.\cos {28^ \circ } \\

   = 4\sin {33^ \circ }.\sin \left( {{{90}^ \circ } - {{29}^ \circ }} \right).\cos {28^ \circ } \\

 $

Now, using $\sin \left( {{{90}^ \circ } - \theta } \right) = \cos \theta $

$ = 4\cos {28^ \circ }.\cos {29^ \circ }.\sin {33^ \circ }$

Therefore, the value of $m$ is $4$.

Note: Solving this question requires a stronghold on the topic of trigonometry and its concepts and many of the basic formulas that we use in this question. Here, we should follow the approach in which we solve the terms on the left side of the equation and try to derive the terms on the right side using them. This approach will give us our required value as a constant number is also derived with the terms on the right side. Thus, through this approach we can derive the required answer.