Prove that the sum of squares of sin and cos equals 1

Question

I need to prove $\sum_{i=1}^n{{(r_i)}^2}=1$ when:

$r_{i}(\theta)= \ \ \ \cos(\theta_{1}) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $if $i=1$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \cos(\theta_{i})\prod_{k=1}^{i-1} \sin(\theta_{k}) \ \ \ \ \ \ \ \ \ \ \ $if $2 \leq i \leq n-1$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \sin(\theta_{n-1})\prod_{k=1}^{n-2} \sin(\theta_{k}) \ \ \ \ \ \ \ $if $i=n$

So my question is, how do I prove that:

$(\cos(\theta_1))^2+(\cos(\theta_2)\sin(\theta_1))^2+(\cos(\theta_3)\sin(\theta_1)\sin(\theta_2))^2+...+(\cos(\theta_{n-1})\sin(\theta_{1})...\sin(\theta_{n-2}))^2+(\sin(\theta_{1})...\sin(\theta_{n-1}))^2 =1$

It seems obvious to do something with $\sin^2 x + \cos^2 x = 1$, but I cannot understand what exactly I should do with it.

Answer 1

Firsly, note that by considering $2$ last terms ( that is for $i=n-1$ and $i=n$):

$(cos(\theta_{n-1})*\prod_{k=1}^{n-2}sin(\theta_k))^2 + (sin(\theta_{n-1})*\prod_{k=1}^{n-2}sin(\theta_k))^2$ =

$(cos^2(\theta_{n-1})+sin^2(\theta_{n-1}))*(\prod_{k=1}^{n-2}sin(\theta_k))^2 $ =

$(\prod_{k=1}^{n-2}sin(\theta_k))^2$

Now we're left with $n-1$ terms ( the one we get after grouping last $2$ terms and first $n-2$ terms).

Continue analogously, combine the new term with $n-2$'th one:

$(\prod_{k=1}^{n-2}sin(\theta_k))^2$ + $(cos(\theta_{n-2})\prod_{k=1}^{n-3}sin(\theta_k))^2$ =

$(sin(\theta_{n-2})(\prod_{k=1}^{n-3}sin(\theta_k))^2$ + $(cos(\theta_{n-2})\prod_{k=1}^{n-3}sin(\theta_k))^2$ = $(sin^2(\theta_{n-2})+cos^2(\theta_{n-2}))(\prod_{k=1}^{n-3}sin(\theta_k))^2$ =

=$(\prod_{k=1}^{n-3}sin(\theta_k))^2$

By continuing so (until the second term, cause it is of the same nature as every $\{2,3,...,n-1\}$, we get: first term + created term ( which is $sin^2(\theta_{n-1-(n-2)})$ ), so:

$cos^2(\theta_1) + sin^2(\theta_1) = 1 $

$1 = 1 $

Q.E.D

Answer 2

I think this is not too hard to prove inductively. The $n=2$ case is precisely $\sin^2 + \cos^2 = 1$, so we will use this as base case. Let's assume the statement holds for $n$ and prove it for $n+1$. We denote with $r^(n+1)_k$ and $r^(n)_k$ the $r_k$ from the cases $n$ and $n+1$ respectively. $$\begin{align} (r^{(n+1)}_{n+1})^2 + (r^{(n+1)}_{n})^2 &= \sin(\theta_{n})^2 \left( \prod_{k=1}^{n+1-2} \sin(\theta_k) \right)^2 + \cos(\theta_{n})^2 \left( \prod_{k=1}^{n+1-2} \sin(\theta_k) \right)^2 \\ &= \left( \prod_{k=1}^{n-1} \sin(\theta_k) \right)^2 = (r^{(n)}_n)^2 \end{align}$$

Now note that for $k \leq n$: $r^{(n+1)}_k = r^{(n)}_k$ and therefore $$(r^{(n+1)}_1)^2 + \dots (r^{(n+1)}_n)^2 + (r^{(n+1)}_{n+1})^2 = (r^{(n)}_1)^2 + \dots + (r^{(n)}_n)^2 = 1$$ by induction hypothesis.