Submitted by Surfer Rosa on Mon, 02/20/2017 - 20:38
Forums:
For the last part of the STEP Question I tried the following - please excuse the basic formatting
(1 + sin 2θ)/(1+cos 2θ) = (cos²θ + sin²θ + 2sinθ cosθ)/(2cos²θ)
which, if you take good care of the stray ½, you can basically rewrite as
((sinθ + cos θ)/(cos θ))²
Remembering that this lives inside a ln (), from here I used the laws of logs and the first part of the question.
Is this a valid approach?
Yep
That's the solution I used in the "hints and partial solutions" document, though I did not put in all the steps as question 2(v) does some of them i.e showing that $\dfrac{1+\sin 2 \alpha}{1 + \cos 2 \alpha}=\frac 1 2 (1+\tan \alpha)^2$.
Ok
Ok. I hadn't seen 2(v). I thought this was a really nice question. I often think that when I get one right :-)
Extension
If you want, another nice question is evaluating $\displaystyle \int_0^{\frac{\pi}{2}}\log \sin x \, \mathrm{d}x$. It uses the same conceptual idea but has a slight twist to it, give it a go if you want.
Also, as an interesting aside, the conceptual underpinning of the question is the fact that $\displaystyle \int_a^b f(x) \, \mathrm{d}x = \int_a^b f(a+b-x) \, \mathrm{d}x$ using the substitution $x\mapsto a+b-x$. This is particularly effective for periodic functions, hence it's effect on trigonometric functions.