Practice: Finding the sum of several cosines

Simplify the equation as we may obtain:

$\displaystyle \sin N^\circ - \sin 5^\circ = 2 \cos 10^\circ \sin 5^\circ + 2 \cos 20^\circ \sin 5^\circ + 2 \cos 30^\circ \sin 5^\circ + \ldots + 2 \cos 80^\circ \sin 5^\circ$

From the identity $\displaystyle 2 \cos A \sin B = \sin (A+B) - \sin (A-B)$ we can get:

$\displaystyle \sin N^\circ - \sin 5^\circ = ( \sin 15^\circ - \sin 5^\circ) + (\sin 25^\circ - \sin 15^\circ) + ( \sin 35^\circ - \sin 25^\circ) + \ldots + (\sin 85^\circ - \sin 75^\circ)$

$\displaystyle \sin N^\circ = \sin 85^\circ$

Hence, the answer is $\displaystyle \boxed{N =85}.$

First change all the expressions in terms of sine instead of cosine using:

$\sin(\theta) = \cos(90-{\theta})$

This gives us: $\cos10 + \cos20 + \cos30... +\cos 80 = \sin80 + \sin70 + \sin60 ... + \sin10 = \sin10 + \sin20 + \sin30 ... + \sin80$

multiply both expressions by $2 \sin5$ . This gives:

$\sin N^{\circ} - \sin5^{\circ} = 2\sin10^{\circ}\sin5^{\circ} + 2\sin20^{\circ}\sin5^{\circ}...2\sin80^{\circ} \sin5^{\circ}$

Using: $\cos(a-b) - \cos(a+b) = 2 \sin a \sin b$

We can simplify the expression to:

$(\cos5 - \cos15) + (\cos15 -\cos25) + (\cos25 - \cos30) ... (\cos75 - \cos85) = \cos5 - \cos85$

Rewriting this in terms of sine gives us:

$\sin N^{\circ} - \sin 5^{\circ} = \sin 85^{\circ} - \sin 5^{\circ}$

The only solution for N on the interval [0, 90] is $N = 85$

We shall attempt to derive Lagrange's trigonometric identity, which states that:

$\sum_{n=1}^N \cos(n \theta) = -\frac{1}{2} + \frac{\sin(N+ \frac{1}{2}) \theta}{2 \sin(\frac{1}{2} \theta)}$

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Let us first recall that:

$1 + z + z^2 +...+ z^n = \frac{1 - z^{n+1}}{1 - z} (\star)$ where $z \neq 1$ .

Firstly, remark that by setting $z = e^{i \theta}$ (motivated by DeMoivre's Formula ), consider the real part of $z^k$ . Thus, $\sum_{n=1}^N \cos(n \theta)$ is the real part of $1+ z + z^2 + \ldots + z^N$ . Thus by $\star$ , we get that the LHS of Lagrange's identity must be the real part of

$\frac{1 - [\cos((n + 1)x) + i \sin((n + 1)x]}{1 - [ \cos(x) + i \sin(x)]}$

In order to make the denominator of the above "nicer", we rationalise it. In essence, we multiply it by its complex conjugate:

$[1 - \cos(x)] + i \sin(x)$ .

You can verify easily that the denominator is: $[1 - \cos( \theta )]^2 + \sin^2( \theta) = 4 \sin^2( \frac{1}{2} \theta)$ and that the real part of the numerator is:

$[1 - \cos((n + 1) \theta)][1 - \cos( \theta)] + \sin[(n + 1) \theta] \sin( \theta) = [1 - \cos((n + 1) \theta)][2 \sin^2(\frac{1}{2} \theta)] + \sin[(n + 1) \theta][2\sin(\frac{1}{2} \theta) \cos(\frac{1}{2} \theta)]$

$= 2 \sin^2(\frac{1}{2} \theta) - 2 \sin(\frac{1}{2} \theta) \{ \cos[(n + 1) \theta] \sin(\frac{1}{2} \theta) - \sin [(n + 1) \theta ]\cos(\frac{1}{2} \theta) \}$

Finally by considering the difference formula, the interested reader may easily verify that when put over the numerator, we indeed get the desired.

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By directly applying the identity mentioned at the beginning, we can conclude that the RHS $= \frac{\sin(85°)}{2 \sin(5°)} - \frac{1}{2}$ , thus our answer is $N = \fbox{85}$ .

$\sin N^{\circ}-\sin 5^{\circ}=\sin 15^{\circ}-\sin 5^{\circ}+\sin 25^{\circ}-\sin 15^{\circ}+....+\sin 85^{\circ}-\sin 75^{\circ}$

$\sin N^{\circ}=\sin 85^{\circ}$

$N=\fbox{85}$

N=COS(10+20+30+40+50+60+70+80)+0.50 X 5SIN5 N=85

Using Lagrange's trigonometric identities (http://en.wikipedia.org/wiki/Lagrange%27s trigonometric identities#Lagrange.27s trigonometric identities), we get the R.H.S. = - (1/2) + [sin(85°)/{2*sin(5°)}]. Therefore, N = 85.