Question

The value of sin6°sin42°sin 66°sin 102° is equal to k then 1/2k​

Answer 1

SOLUTION

GIVEN

$\displaystyle \sf{ \sin {6}^{ \circ} \sin {42}^{ \circ} \sin {66}^{ \circ} \sin {102}^{ \circ} = k}$

TO DETERMINE

The value of

$\displaystyle \sf{ \frac{1}{2k} }$

EVALUATION

Here it is given that

$\displaystyle \sf{ \sin {6}^{ \circ} \sin {42}^{ \circ} \sin {66}^{ \circ} \sin {102}^{ \circ} = k}$

Now

$\displaystyle \sf{ \sin {6}^{ \circ} \sin {42}^{ \circ} \sin {66}^{ \circ} \sin {102}^{ \circ} }$

$\displaystyle \sf{ = \frac{1}{4} \times 2 \sin {6}^{ \circ} \sin {66}^{ \circ} \times 2 \sin {42}^{ \circ} \sin {102}^{ \circ} }$

$\displaystyle \sf{ = \frac{1}{4} \times (\cos {60}^{ \circ} - \cos {72}^{ \circ} ) (\cos {60}^{ \circ} - \cos {144}^{ \circ} ) }$

$\displaystyle \sf{ = \frac{1}{4} \times \bigg( \frac{1}{2} - \frac{ \sqrt{5} - 1}{4} \bigg) \bigg( \frac{1}{2} + \cos {36}^{ \circ} \bigg) }$

$\displaystyle \sf{ = \frac{1}{4} \times \bigg( \frac{1}{2} - \frac{ \sqrt{5} - 1}{4} \bigg) \bigg( \frac{1}{2} + \frac{ \sqrt{5} + 1}{4} \bigg)}$

$\displaystyle \sf{ = \frac{1}{4} \times \bigg( \frac{ 3 - \sqrt{5} }{4} \bigg) \bigg( \frac{ 3 + \sqrt{5} }{4} \bigg)}$

$\displaystyle \sf{ = \frac{1}{4} \times \bigg( \frac{ {3}^{2} - {( \sqrt{5} )}^{2} }{4 \times 4} \bigg) }$

$\displaystyle \sf{ = \frac{1}{4} \times \bigg( \frac{ 9 - 5 }{16} \bigg) }$

$\displaystyle \sf{ = \frac{1}{4} \times \bigg( \frac{ 4 }{16} \bigg) }$

$\displaystyle \sf{ = \frac{1}{16} }$

This gives $\displaystyle \sf{ k= \frac{1}{16} }$

Thus we have

$\displaystyle \sf{ \frac{1}{2k} }$

$\displaystyle \sf{ = \frac{16}{2} }$

$\displaystyle \sf{ = 8}$

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Answer 2

$\underline{\textbf{Given:}}$

$\mathsf{sin\,6^\circ\;sin\,42^\circ\;sin\,66^\circ\;sin\,102^\circ=k}$

$\underline{\textbf{To find:}}$

$\textsf{The value of}$

$\mathsf{\dfrac{1}{2k}}$

$\underline{\textbf{Solution:}}$

$\underline{\textsf{Formula used:}}$

$\boxed{\mathsf{sin(60^\circ-A)\;sinA\;sin(60^\circ+A)=\dfrac{1}{4}sin\,3A}}$

$\mathsf{Consider,}$

$\mathsf{sin\,6^\circ\;sin\,42^\circ\;sin\,66^\circ\;sin\,102^\circ=k}$

$\mathsf{(sin\,6^\circ\;sin\,66^\circ)(sin\,42^\circ\;sin\,102^\circ)=k}$

$\textsf{This can be written as,}$

$\mathsf{\left(\dfrac{sin\,54\;sin\,6^\circ\;sin\,66^\circ}{sin\,54}\right)\left(\dfrac{sin\,54\;sin\,42^\circ\;sin\,102^\circ}{sin\,18}\right)=k}$

$\mathsf{\left(\dfrac{sin\,54^\circ\;sin\,6^\circ\;sin\,66^\circ}{sin\,54^\circ}\right)\left(\dfrac{sin\,18^\circ\;sin\,42^\circ\;sin\,102^\circ}{sin\,18^\circ}\right)=k}$

$\mathsf{\left(\dfrac{sin(60^\circ-6^\circ)\;sin\,6^\circ\;sin(60^\circ+6^\circ)}{sin\,54^\circ}\right)\left(\dfrac{sin(60^\circ-42^\circ)\;sin\,42^\circ\;sin(60^\circ+42^\circ)}{sin\,18^\circ}\right)=k}$

$\mathsf{Using\;the\;above\;formula,}$

$\mathsf{\left(\dfrac{1}{4}sin\,3(6^\circ){\times}\dfrac{1}{sin\,54^\circ}\right)\left(\dfrac{1}{4}sin\,3(42^\circ){\times}\dfrac{1}{sin\,18^\circ}\right)=k}$

$\mathsf{\left(\dfrac{1}{4}sin\,18^\circ{\times}\dfrac{1}{sin\,54^\circ}\right)\left(\dfrac{1}{4}sin\,126^\circ{\times}\dfrac{1}{sin\,18^\circ}\right)=k}$

$\mathsf{\left(\dfrac{1}{4}sin\,18^\circ{\times}\dfrac{1}{sin\,54^\circ}\right)\left(\dfrac{1}{4}sin\,54^\circ{\times}\dfrac{1}{sin\,18^\circ}\right)=k}$

$\implies\mathsf{\dfrac{1}{4}{\times}\dfrac{1}{4}=k}$

$\implies\mathsf{k=\dfrac{1}{16}}$

$\mathsf{Now,}$

$\mathsf{\dfrac{1}{2k}}$

$\mathsf{=\dfrac{16}{2}}$

$\implies\boxed{\mathsf{\dfrac{1}{2k}=8}}$

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