If $\sum_{p=0}^{2020}{\sin(2^p\theta)\sec(3^{p+1}\theta})=a\tan(b\theta)+c\tan(d\theta)$, find $ac,ad,cd,bd$

Question

If $\sum_{p=0}^{2020}{\sin(2^p\theta)\sec(3^{p+1}\theta})=a\tan(b\theta)+c\tan(d\theta)$, find $ac,ad,cd,bd$

I converted secant to cosine, opened the sigma and tried to take LCM pairwise. But couldn't conclude.

I tried multiplying and dividing by $\cos(2^p\theta)$ to make tan or to make $\sin2\theta$ in numerator, but couldn't conclude.

I guess the method of differences could be used here, but not able to manipulate.

Answer 1

So here's my try (assuming $sin(3^p \theta$));

We have S = $\sum_{p=0}^{2020} sin(3^p \theta)sec(3^{p+1} \theta)$

Let $T_r = sin(3^r \theta)sec(3^{r+1} \theta)$

Taking r = 0, we get $T_0 = sin( \theta)sec(3 \theta)$ = $\frac{sin( \theta)}{cos(3 \theta)}$

Clearly this question can be done using method of difference, so to create that we can try using $cos(3 \theta)$ identity but that leads to a messy path so we multiply by 2 $cos(\theta)$ in $T_0$ and use 2$sin(\theta)$$cos(\theta)$ = $sin(2\theta)$, then again it still seems not so neat to create a difference,so we rewrite $sin(2\theta)$ as $sin(3\theta-\theta)$ and then using sin(a+b) identity;

$T_r$=$\frac{1}{2}(tan(3^{r+1} \theta)-tan(3^{r} \theta)$) which on telescoping,

S = $\frac{1}{2}(tan(3^{2021} \theta)-tan( \theta)$)

compare to get a,b,c and d