# Thread: Prove that :-

1. ## Prove that :-

$sin^25^\circ+ sin^210^\circ+..........+sin^285^\circ+sin^290^\ci rc=9 \frac{1}{2}$

I have been solving for a long time but in my answer my LHS is not equal to RHS....therefore pls help me....

2. do not solve plss its a totally wrong quest.

3. corrected question

$sin^25^\circ+ sin^210^\circ+..........+sin^285^\circ+sin^290^\ci rc=9 \frac{1}{2}$

I have been solving for a long time but in my answer my LHS is not equal to RHS....therefore pls help me....

4. No need to solve i have solved............

5. You have

$\sum\limits_{n = 1}^{18} {{{\sin }^2}\left( {{5^ \circ }n} \right)} = {\sin ^2}{5^ \circ } + {\sin ^2}{10^ \circ } + \ldots + {\sin ^2}{85^ \circ } + {\sin ^2}{90^ \circ } = \frac{{19}}{2}.$

Now simplify the left side by formula $\sin^2\theta = \frac{1-\cos2\theta}{2}$ left side and transform

$\sum\limits_{n = 1}^{18} {{{\sin }^2}\left( {{5^ \circ }n} \right)} = \sum\limits_{n = 1}^{18} {{{\sin }^2}\frac{{n\pi }}{{36}}} = \frac{1}{2}\sum\limits_{n = 1}^{18} {\left( {1 - \cos \frac{{n\pi }}{{18}}}\right)} = 9 - \frac{1}{2}\sum\limits_{n = 1}^{18} {\cos \frac{{n\pi }}{{18}}}=$

$= 9 - \frac{1}{2}\left( {\sum\limits_{n = 1}^{17} {\cos \frac{{n\pi}}{{18}}} + \cos \pi } \right) = \frac{{19}}{2} - \frac{1}{2} \sum\limits_{n = 1}^{17} {\cos \frac{{n\pi }}{{18}}}.$

Now estimate the sum $\sum\limits_{n = 1}^{17}{\cos\frac{n\pi}{18}}$.
If $n=9$, then $\cos\frac{9\pi}{18}=\cos\frac{\pi }{2}=0$.
Since the cosine function in the interval $\left[\frac{\pi}{18};\,\frac{\pi}{2} \right)$ positive and negative in $\left(\frac{\pi}{2};\,\frac{17\pi}{18}\right]$, it follows that $\sum\limits_{n = 1}^{17} {\cos\frac{n\pi}{18}}=0$,
ie you have
$\sum\limits_{n = 1}^{18} {{{\sin }^2}\left( {{5^ \circ }n} \right)} = \frac{{19}}{2} - \frac{1}{2}\underbrace {\sum\limits_{n = 1}^{17} {\cos \frac{n\pi}{18}}}_0 = \frac{19}{2}.$

Thus, the initial equality is proved.