Value of \( \cos\alpha\cos2\alpha\cos4\alpha\cdots\cos2^{\,n-1}\alpha \)
Question
If \(n=1,2,3,\ldots\), then
\[ \cos\alpha\cos2\alpha\cos4\alpha\cdots\cos2^{n-1}\alpha \]
is equal to
(a) \(\dfrac{\sin(2^n\alpha)}{2^n\sin\alpha}\)
(b) \(\dfrac{\sin(2^n\alpha)}{2^n\sin(2^{n-1}\alpha)}\)
(c) \(\dfrac{\sin(4^{\,n-1}\alpha)}{4^{\,n-1}\sin\alpha}\)
(d) \(\dfrac{\sin(2^n\alpha)}{2^n\sin\alpha}\)
Solution
Use the standard identity:
\[ \sin(2^n\alpha) = 2^n\sin\alpha \cos\alpha \cos2\alpha \cos4\alpha \cdots \cos2^{n-1}\alpha \]
Dividing both sides by \(2^n\sin\alpha\),
\[ \cos\alpha \cos2\alpha \cos4\alpha \cdots \cos2^{n-1}\alpha = \frac{\sin(2^n\alpha)} {2^n\sin\alpha} \]
Final Answer
\[ \boxed{ \cos\alpha\cos2\alpha\cos4\alpha\cdots\cos2^{n-1}\alpha = \frac{\sin(2^n\alpha)} {2^n\sin\alpha} } \]
Hence, the correct option is (a) (option (d) is identical if printed the same way).