STEP Question Help

Can you check this is what you meant to upload? It feels like there's a lot missing.

Also: why don't you just check the given solution (it's clearly from one of the Siklos booklets)?

In your first step of diffing the RHS of (*) you seem to have done $\dfrac{d}{d\theta} \dfrac{f(\theta)}{g(\theta)} = \dfrac{\frac{df}{d\theta}}{\frac{dg}{d\theta}}$

instead of the correct

$\dfrac{d}{d\theta} \dfrac{f(\theta)}{g(\theta)} = \dfrac{g \frac{df}{d\theta} - f \frac{dg}{d\theta}}{g^2}$

In your first step of diffing the RHS of (*) you seem to have done $\dfrac{d}{d\theta} \dfrac{f(\theta)}{g(\theta)} = \dfrac{\frac{df}{d\theta}}{\frac{dg}{d\theta}}$

instead of the correct

$\dfrac{d}{d\theta} \dfrac{f(\theta)}{g(\theta)} = \dfrac{g \frac{df}{d\theta} - f \frac{dg}{d\theta}}{g^2}$

In your first step of diffing the RHS of (*) you seem to have done $\dfrac{d}{d\theta} \dfrac{f(\theta)}{g(\theta)} = \dfrac{\frac{df}{d\theta}}{\frac{dg}{d\theta}}$

instead of the correct

$\dfrac{d}{d\theta} \dfrac{f(\theta)}{g(\theta)} = \dfrac{g \frac{df}{d\theta} - f \frac{dg}{d\theta}}{g^2}$

Okay I did the following:

I can appreciate the simplification needed now, however, what I see is I get similar to the answer they want but the term in n doesn’t seem to product a factor that simplifies, also at the line marked with the asterisk, what if I distribute the sin^2 across the bracket and then again apply the sin^2 + cos^2 to reduce it, this will leave me a sin^2 on the top and bottom and remove the cot^2 all together... leaving a cosec^2 only

So, I'm not sure I'm following your work properly, but from your end point if you add 1/4 to both sides you can replace (2n+1)/4 on the LHS with (n+1)/2 which will combine the other term to give you (n+1)^2/2 which looks close to what you need once you divide through by 2.

Yes! That seems to do it, but now I’ve found this more simplified method (which basically just uses the second part to simplify the differentiation):

But now I can’t see where this goes, it’s missing an “n” term to factor the LHS
Edit: I think I know why, reassure me if I'm correct or correct me otherwise but does the result I begun with only hold for theta=pi/n but then I'm thinking if this is true, I can only substitute pi/n later on which is what I do..

If a result only holds for "theta = pi/n" then you can't differentiate it and expect to get the right answer even at theta = pi/n.

E.g. sin nx = 0 only holds for theta = pi/n,

Differentiating to deduce cos nx = 0 is false at theta = pi/n.