could you help me please. any help will be appreciated...
I have become interested in this problem and have spent a great deal of time trying to get induction (among other things) to work, but have not been successful. Hopefully, someone can show me where I am going wrong. Obviously the statement is true for the base case of n = 1. So I stated the induction hypothesis $\displaystyle P_n$:
$\displaystyle \int_0^{\pi}\frac{\cos(nx)-\cos(na)}{\cos(x)-\cos(a)}\,dx=\pi\frac{\sin(na)}{\sin(a)}$
Assuming this is true, I then added $\displaystyle P_{n+2}$ to get:
$\displaystyle \int_0^{\pi}\frac{\cos(nx)-\cos(na)}{\cos(x)-\cos(a)}\,dx+\int_0^{\pi}\frac{\cos\left((n+2)x \right)-\cos\left((n+2)a \right)}{\cos(x)-\cos(a)}\,dx=$
$\displaystyle \pi\frac{\sin(na)}{\sin(a)}+\pi\frac{\sin\left((n+ 2)a\right)}{\sin(a)}$
$\displaystyle \int_0^{\pi}\frac{\left(\cos\left((n+2)x \right)+\cos(nx)\right)-\left(\cos\left((n+2)a\right)+\cos(na) \right)}{\cos(x)-\cos(a)}\,dx=\frac{\pi}{\sin(a)}\left(\sin\left((n +2)a\right)+\sin(na)\right)$
$\displaystyle \int_0^{\pi}\frac{\left(2\cos\left((n+1)x \right)\cos(x)\right)-\left(2\cos\left((n+1)a\right)\cos(a)\right)}{\cos (x)-\cos(a)}\,dx=\frac{\pi}{\sin(a)}\(2\sin\left((n+1) a\right)\cos(a)\)$
$\displaystyle \int_0^{\pi}\frac{\cos\left((n+1)x\right) \frac{\cos(x)}{\cos(a)}-\cos\left((n+1)a\right)}{\cos(x)-\cos(a)}\,dx=\pi\frac{\sin\left((n+1)a \right)}{\sin(a)}$
I can't figure out how to get rid of the $\displaystyle \frac{\cos(x)}{\cos(a)}$ in the numerator of the integrand.
edit: sorry about my edits...this board has some unexpected idiosyncrasies concerning the
\left and \right operators in $\displaystyle \LaTeX$...
I should add that this problem has also been posted on the forum which I help moderate, and I have posted the same problem I have in using induction.
The OP is an infrequent visitor there, and it may be a while before we hear from them to get any feedback on the context of the problem.
The following fix was given to me:
$\displaystyle \int_0^{\pi}\frac{\cos((n+1)x)}{\cos(a)}\,dx=0$
$\displaystyle \int_0^{\pi}\frac{\cos((n+1)x)\left(\frac{\cos(x)} {\cos(a)}-1\right)+\cos((n+1)a)-\cos((n+1)a)}{\cos(x)-\cos(a)}\,dx=0$
$\displaystyle \int_0^{\pi}\frac{\cos((n+1)x)\frac{\cos(x)}{\cos( a)}-\cos((n+1)a)}{\cos(x)-\cos(a)}\,dx=\int_0^{\pi}\frac{\cos((n+1)x)-\cos((n+1)a)}{\cos(x)-\cos(a)}\,dx$