I was doing a question in a STEP maths paper given by
By considering the graphs of $y=kx$ and $y=\sin x$, show that the equation $kx=\sin x$, where $k>0$, may of $0,1,2$ or $3$ roots in the interval $(4n+1)\dfrac\pi2<x<(4n+5)\dfrac\pi2$, where $n$ is a positive integer.
For a certain given value of $n$, the equation has exactly one root in this interval. Show that $k$ lies in an interval which may be written as $\sin\delta<k<\dfrac2{(4n+1)\pi}$, where $0<\delta<\dfrac\pi2$ and $$\cos\delta=\left((4n+5)\frac\pi2-\delta\right)\sin\delta.$$
Show that, if $n$ is large, then $\delta\approx\dfrac2{(4n+5)\pi}$ and obtain a second, improved, approximation.
but I got stuck on the $cos(\delta)$ bit. I originally thought that it would have something to do with the gradient as in most cases the gardient is $\frac{dy}{dx}=tan(\theta)=\frac{sin(\theta)}{cos(\theta)}$ which could be rearanged to get $cos{\theta}$ as the subject but I did this with no success. If anybody could suggest an intuitive way of approaching this I would be really thankful.