How to prove with matematical induction: $\cos1 + \cos 2 + \cdots + \cos n = -\frac{1}{2} + \frac{\sin(n+\frac{1}{2})}{2\sin\frac{1}{2}}$

Question

I'm having truble calculating right side of this equation

$$\cos1 + \cos 2 + \cdots + \cos n = -\frac{1}{2} + \frac{\sin(n+\frac{1}{2})}{2\sin\frac{1}{2}}$$ for $n = 1$.

I've managed to reach that $$cos1 = -\frac{1}{2} + \frac{\cos^2 1 -\cos1 -\sin^2 1}{2(1-\cos1)}$$

Can i get any tips on how to solve this equation? (If it's even correct at this point).

Thanks.

Answer 1

For $n=1$, the right-hand side is $$ -\frac{1}{2}+\frac{\sin(3/2)}{2\sin(1/2)} $$ Let $\alpha=1/2$ for simplicity; you get $$ -\frac{1}{2}+\frac{\sin3\alpha}{2\sin\alpha} = -\frac{1}{2}+\frac{3\sin\alpha-4\sin^3\alpha}{2\sin\alpha}= \frac{-1+3-4\sin^2\alpha}{2}=1-2\sin^2\alpha=\cos2\alpha=\cos1 $$

Answer 2

Something went wrong in your computation for $n=1$, which is going to conclude in $cos1=-1-cos1$.
It shall go like this: $$ \begin{gathered} - \frac{1} {2} + \frac{{\sin \left( {1 + 1/2} \right)}} {{2\sin \left( {1/2} \right)}} = \frac{1} {2}\left( {\frac{{\sin \left( 1 \right)\cos \left( {1/2} \right) + \cos \left( 1 \right)\sin \left( {1/2} \right)}} {{\sin \left( {1/2} \right)}} - 1} \right) = \hfill \\ = \frac{1} {2}\left( {\frac{{\sin \left( 1 \right)\cos \left( {1/2} \right) + \left( {\cos \left( 1 \right) - 1} \right)\sin \left( {1/2} \right)}} {{\sin \left( {1/2} \right)}}} \right) = \hfill \\ = \frac{1} {2}\left( {\frac{{2\sin \left( {1/2} \right)\cos ^{\,2} \left( {1/2} \right) + \left( {2\cos ^{\,2} \left( {1/2} \right) - 2} \right)\sin \left( {1/2} \right)}} {{\sin \left( {1/2} \right)}}} \right) = \hfill \\ = \frac{1} {2}\left( {2\cos ^{\,2} \left( {1/2} \right) + \left( {2\cos ^{\,2} \left( {1/2} \right) - 2} \right)} \right) = \left( {2\cos ^{\,2} \left( {1/2} \right) - 1} \right) = \cos 1 \hfill \\ \end{gathered} $$ After that, we have: $$ \begin{gathered} \cos 1 + \cos 2 + \; \cdots \; + \cos n + \cos (n + 1) = \hfill \\ = - \frac{1} {2} + \frac{{\sin \left( {n + 1/2} \right)}} {{2\sin \left( {1/2} \right)}} + \cos (n + 1) = \hfill \\ = \frac{1} {2}\left( {\frac{{\sin \left( {n + 1/2} \right)}} {{\sin \left( {1/2} \right)}} - 1 + 2\cos (n + 1)} \right) = \hfill \\ = \frac{1} {2}\left( {\frac{{\sin \left( {n + 1/2} \right) - \sin \left( {1/2} \right) + 2\cos (n + 1)\sin \left( {1/2} \right)}} {{\sin \left( {1/2} \right)}}} \right) = \hfill \\ = \frac{1} {2}\left( {\frac{{\sin \left( {n + 1} \right)\cos \left( {1/2} \right) + \cos \left( {n + 1} \right)\sin \left( {1/2} \right) - \sin \left( {1/2} \right)}} {{\sin \left( {1/2} \right)}}} \right) = \hfill \\ = \frac{1} {2}\left( {\frac{{\sin \left( {n + 1 + 1/2} \right) - \sin \left( {1/2} \right)}} {{\sin \left( {1/2} \right)}}} \right) = - \frac{1} {2} + \frac{{\sin \left( {n + 1 + 1/2} \right)}} {{\sin \left( {1/2} \right)}} \hfill \\ \end{gathered} $$ which is the demonstration you are looking for.