Question

The value of \[(4{\cos ^2}{9^ \circ } - 1)(4{\cos ^2}{81^ \circ } - 1)(4{\cos ^2}{27^ \circ } - 1)(4{\cos ^2}{243^ \circ } - 1)\] is
A.1
B.-1
C.2
D.None of these

Answer 1

Hint: We have given \[(4{\cos ^2}{9^ \circ } - 1)(4{\cos ^2}{81^ \circ } - 1)(4{\cos ^2}{27^ \circ } - 1)(4{\cos ^2}{243^ \circ } - 1)\]
To find the value of above trigonometric expressions first, we have to use this trigonometric identity $(4{\cos ^2}\theta - 1)$

Complete step-by-step answer:
  The trigonometric identity is $(4{\cos ^2}\theta - 1)$ =$\dfrac{{\sin 3\theta }}{{\sin \theta }}$
Here is the further explanation of this trigonometric identity or we can say the proof of this formula
In the first part, we use this identity where R.H.S. \[\dfrac{{\operatorname{Sin} 3\theta }}{{\sin \theta }} = \dfrac{{3\sin \theta - 4{{\sin }^3}\theta }}{{\sin \theta }}\]
In the next part, we take out the common part and it will cancel with the denominator
$ = \dfrac{{\sin \theta (3 - 4{{\sin }^2}\theta )}}{{\sin \theta }} = 3 - 4{\sin ^2}\theta $
So, In this area, we use another identity of trigonometry
$ = 3 - 4(1 - {\cos ^2}\theta )$ [∴ ${\sin ^2}\theta = 1 - {\cos ^2}\theta $]
So, here we open the brackets by multiply 4 with both digits
$ = 3 - 4 + 4{\cos ^2}\theta $
$
   = - 1 + 4{\cos ^2}\theta \\
   = 4{\cos ^2}\theta - 1 \\
$ = L.H.S.
$(4{\cos ^2}\theta - 1)$ =$\dfrac{{\sin 3\theta }}{{\sin \theta }}$
For $\theta = {9^ \circ }$
$4{\cos ^2}{9^ \circ } - 1 = \dfrac{{\sin 3 \times {9^ \circ }}}{{\sin {9^ \circ }}} = \dfrac{{\sin {{27}^ \circ }}}{{\sin {9^ \circ }}}.......(i)$
For $\theta = {27^ \circ }$
$4{\cos ^2}{27^ \circ } - 1 = \dfrac{{\sin 3 \times {{27}^ \circ }}}{{\sin {{27}^ \circ }}} = \dfrac{{\sin {{81}^ \circ }}}{{\sin {{27}^ \circ }}}.........(ii)$
$\theta = {81^ \circ }$
$4{\cos ^2}{81^ \circ } - 1 = \dfrac{{\sin 3 \times {{81}^ \circ }}}{{\sin {{81}^ \circ }}} = \dfrac{{\sin {{243}^ \circ }}}{{\sin {{81}^ \circ }}}.........(iii)$
$\theta = {243^ \circ }$
$4{\cos ^2}{243^ \circ } - 1 = \dfrac{{\sin 3 \times {{243}^ \circ }}}{{\sin {{243}^ \circ }}} = \dfrac{{\sin {{729}^ \circ }}}{{\sin {{243}^ \circ }}}.........(iv)$
Multiplying (i) (ii) (iii) and (iv)
\[(4{\cos ^2}{9^ \circ } - 1)(4{\cos ^2}{81^ \circ } - 1)(4{\cos ^2}{27^ \circ } - 1)(4{\cos ^2}{243^ \circ } - 1)\]
= \[\]\[
  \dfrac{{\sin {{27}^ \circ }}}{{\sin {9^ \circ }}} \times \dfrac{{\sin {{81}^ \circ }}}{{\sin {{27}^ \circ }}} \times \dfrac{{\sin {{243}^ \circ }}}{{\sin {{81}^ \circ }}} \times \dfrac{{\sin {{729}^ \circ }}}{{\sin {{243}^ \circ }}} \\
   = \dfrac{{\sin {{729}^ \circ }}}{{\sin {9^ \circ }}} = \dfrac{{\sin {{(720 + 9)}^ \circ }}}{{\sin {9^ \circ }}} = \dfrac{{\sin {9^ \circ }}}{{\sin {9^ \circ }}} = 1 \\
\] \[\because \sin (2\pi + \theta ) = \sin \theta \]
Option A is correct.

Note: Identities involving only angles are known as trigonometric identities , related both the sides and angles of a given triangle.
Periodic function: On changing a variable θ to θ +α. α being the least positive constant, the value of a function of θ remains unchanged, the function is said to be periodic and α is called the periodic function.
$\sin (\theta + 2\pi ) = \sin \theta $