For [tex]n=1[/tex], on the left we have [tex]\cos\theta[/tex], and on the right,
[tex]\dfrac{\sin2\theta}{2\sin\theta}=\dfrac{2\sin\theta\cos\theta}{2\sin\theta}=\cos\theta[/tex]
(where we use the double angle identity: [tex]\sin2\theta=2\sin\theta\cos\theta[/tex])
Suppose the relation holds for [tex]n=k[/tex]:
[tex]\displaystyle\sum_{n=1}^k\cos(2n-1)\theta=\dfrac{\sin2k\theta}{2\sin\theta}[/tex]
Then for [tex]n=k+1[/tex], the left side is
[tex]\displaystyle\sum_{n=1}^{k+1}\cos(2n-1)\theta=\sum_{n=1}^k\cos(2n-1)\theta+\cos(2k+1)\theta=\dfrac{\sin2k\theta}{2\sin\theta}+\cos(2k+1)\theta[/tex]
So we want to show that
[tex]\dfrac{\sin2k\theta}{2\sin\theta}+\cos(2k+1)\theta=\dfrac{\sin(2k+2)\theta}{2\sin\theta}[/tex]
On the left side, we can combine the fractions:
[tex]\dfrac{\sin2k\theta+2\sin\theta\cos(2k+1)\theta}{2\sin\theta}[/tex]
Recall that
[tex]\cos(x+y)=\cos x\cos y-\sin x\sin y[/tex]
so that we can write
[tex]\dfrac{\sin2k\theta+2\sin\theta(\cos2k\theta\cos\theta-\sin2k\theta\sin\theta)}{2\sin\theta}[/tex]
[tex]=\dfrac{\sin2k\theta+\sin2\theta\cos2k\theta-2\sin2k\theta\sin^2\theta}{2\sin\theta}[/tex]
[tex]=\dfrac{\sin2k\theta(1-2\sin^2\theta)+\sin2\theta\cos2k\theta}{2\sin\theta}[/tex]
[tex]=\dfrac{\sin2k\theta\cos2\theta+\sin2\theta\cos2k\theta}{2\sin\theta}[/tex]
(another double angle identity: [tex]\cos2\theta=\cos^2\theta-\sin^2\theta=1-2\sin^2\theta[/tex])
Then recall that
[tex]\sin(x+y)=\sin x\cos y+\sin y\cos x[/tex]
which lets us consolidate the numerator to get what we wanted:
[tex]=\dfrac{\sin(2k+2)\theta}{2\sin\theta}[/tex]
and the identity is established.
