Submitted by maths123 on Sat, 06/13/2015 - 21:45
Forums:
I was able to find the general solution for $y_n $ in terms of $ \theta $:
$y_{n}(\theta)=A\cos n\theta+B\sin n\theta $
using the condition y(0) = 1 I found that A=1
However, I am not able to find the value of B by using the second condition - I have looked at the TSR solutions but I do not understand them.
The official STEP hints and solutions state $ cos (n(\theta + \pi)) = (-1)^{-n} cos (n\theta) $ - how does this help when you are actually substituting in $\theta = cos^{-1} x $ ?
Finding B
First off, be careful not to get confused by your notation. When you write $y_n(\theta)=A \cos(n\theta) + B \sin(n\theta)$, it makes it look like $\theta$ is a dummy variable, so $y_n (x)=A\cos(nx)+B\sin(nx)$ which isn't true. Being very clear about what your variables are makes it clearer how to use the conditions. You've written $y(0)=1$ when you mean is $y=0$ when $x=1$, hence when $\theta=0$. This approach shows how we use the other condition (this is the missing detail from the TSR solution): We know $y(-x)=(-1)^n y(x)$. Fix a $\theta_0$, set $\theta_1=\pi-\theta_0,$ and set $x_0=\cos(\theta_0)$ and $x_1=\cos(\theta_1)$.
Then $x_1=-x_0$ because of the relationship between the thetas.
So $y(x_1)=(-1)^n y(x_0)$, i.e. $$A \cos(n\theta_1)+B \sin(n(\theta_1)) = -[A\cos(n\theta_0) + B \sin(n \theta_0)],$$
or dropping the subscripts (since this worked for any $\theta_0$),
$$A\cos(n(\pi-\theta))+B\sin(n(\pi-\theta))=-[A\cos(n\theta) + B \sin(n \theta)],$$
which we can use to work out $B$.