Show that ${{\sin }^{4}}A={{\sin }^{2}}A-\frac{1}{4}{{\sin }^{2}}2A$. It is given that ${{S}_{n}}=\s

Question

Show that ${{\sin }^{4}}A={{\sin }^{2}}A-\frac{1}{4}{{\sin }^{2}}2A$. It is given that ${{S}_{n}}=\sum\limits_{r=0}^{n}{\frac{1}{{{4}^{r}}}{{\sin }^{4}}\left[ {{2}^{r}}\left( \frac{\pi }{3} \right) \right]}$. By using the result in (i), prove that ${{S}_{n}}=\frac{3}{4}-\frac{1}{{{4}^{n+1}}}{{\sin }^{2}}\left[ {{2}^{n+1}}\left( \frac{\pi }{3} \right) \right]$. Hence give a reason why ${{S}_{n}}$ converges and state the sum to infinity.

Accepted Answer

$S_\infty = \dfrac{3}{4}$

2013 RI P1 Q8 [Modified]

2013 RI P1 Q8 [Modified]

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