Question

\[\cos(0.5A) . \cos(0.25A) . \cos(0.125A) ........n = ?\]

Answer 1

You know the answer?

Answer 2

Yup...

\[\frac{ 2^{-n}sinA }{ Sin \frac{ A }{ 2^n} }\]

Answer 3

Ohh. Its not till infinity,
My bad,
Then it's simple.
Hint: Write it as:
\[[2\sin(A/2^n)\cos(A/2^n)\cos(A/2^{n-1})......\cos(A/2)]/2\sin(A/2^n)\]

Answer 4

Now use:
\[2sinAcosA=\sin2A \]

Answer 5

And watch the terms combining :)

Answer 6

Thats Helps