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Question: The value of \[(4{\cos ^2}{9^ \circ } - 1)(4{\cos ^2}{81^ \circ } - 1)(4{\cos ^2}{27^ \circ } - 1)(4...

The value of (4cos⁡29∘−1)(4cos⁡281∘−1)(4cos⁡227∘−1)(4cos⁡2243∘−1)(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

Explanation

Solution

We have given (4cos⁡29∘−1)(4cos⁡281∘−1)(4cos⁡227∘−1)(4cos⁡2243∘−1)(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 (4cos⁡2θ−1)(4{\cos ^2}\theta - 1)

Complete step-by-step answer:
The trigonometric identity is (4cos⁡2θ−1)(4{\cos ^2}\theta - 1) =sin⁡3θsin⁡θ\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. Sin⁡3θsin⁡θ=3sin⁡θ−4sin⁡3θsin⁡θ\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
=sin⁡θ(3−4sin⁡2θ)sin⁡θ=3−4sin⁡2θ= \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⁡2θ)= 3 - 4(1 - {\cos ^2}\theta ) [∴ sin⁡2θ=1−cos⁡2θ{\sin ^2}\theta = 1 - {\cos ^2}\theta ]
So, here we open the brackets by multiply 4 with both digits
=3−4+4cos⁡2θ= 3 - 4 + 4{\cos ^2}\theta
=−1+4cos⁡2θ =4cos⁡2θ−1  = - 1 + 4{\cos ^2}\theta \\\ = 4{\cos ^2}\theta - 1 \\\ = L.H.S.
(4cos⁡2θ−1)(4{\cos ^2}\theta - 1) =sin⁡3θsin⁡θ\dfrac{{\sin 3\theta }}{{\sin \theta }}
For θ=9∘\theta = {9^ \circ }
4cos⁡29∘−1=sin⁡3×9∘sin⁡9∘=sin⁡27∘sin⁡9∘.......(i)4{\cos ^2}{9^ \circ } - 1 = \dfrac{{\sin 3 \times {9^ \circ }}}{{\sin {9^ \circ }}} = \dfrac{{\sin {{27}^ \circ }}}{{\sin {9^ \circ }}}.......(i)
For θ=27∘\theta = {27^ \circ }
4cos⁡227∘−1=sin⁡3×27∘sin⁡27∘=sin⁡81∘sin⁡27∘.........(ii)4{\cos ^2}{27^ \circ } - 1 = \dfrac{{\sin 3 \times {{27}^ \circ }}}{{\sin {{27}^ \circ }}} = \dfrac{{\sin {{81}^ \circ }}}{{\sin {{27}^ \circ }}}.........(ii)
θ=81∘\theta = {81^ \circ }
4cos⁡281∘−1=sin⁡3×81∘sin⁡81∘=sin⁡243∘sin⁡81∘.........(iii)4{\cos ^2}{81^ \circ } - 1 = \dfrac{{\sin 3 \times {{81}^ \circ }}}{{\sin {{81}^ \circ }}} = \dfrac{{\sin {{243}^ \circ }}}{{\sin {{81}^ \circ }}}.........(iii)
θ=243∘\theta = {243^ \circ }
4cos⁡2243∘−1=sin⁡3×243∘sin⁡243∘=sin⁡729∘sin⁡243∘.........(iv)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)
(4cos⁡29∘−1)(4cos⁡281∘−1)(4cos⁡227∘−1)(4cos⁡2243∘−1)(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 $