On proving an identity given a system of trig equations

Question

We are given the following:

$$a^2 + b^2 + 2ab\cos\theta = 1 \tag1$$

$$d^2 + c^2 + 2cd\cos\theta = 1 \tag2$$

$$ac + bd + (ad + bc)\cos\theta = 0\tag3$$

It is required to prove that:

$$a^2 + c^2 = \csc^2\theta$$

$$b^2 + d^2 = \csc^2\theta$$

I right away noticed that this problem didn't change if we replaced $a, b$ with $c, d$ respectively. This suggested a that a reduction of the problem might be possible, but I couldn't quite put my finger on it.

I then noticed that the first two equations represented the magnitude of the vector sum of the vectors $a, b$ and $c, d$. But again, this was not very helpful, as I couldn't relate the third equation to the problem.

I lastly tried to specialize the problem to a special case, hoping it would yield some useful insight. For instance, if $\cos\theta = 0$, the equations reduces to:

$$a^2 + b^2 = 1 \tag4$$

$$d^2 + c^2 = 1\tag5$$

$$ac + bd = 0\tag6$$

and we are required to prove::

$$a^2 + c^2 = b^2 + d^2 = 1$$

this can be immediately seen if we solve $(6)$ for $a$ and then substitute in $(4)$ getting, $b = c$.

This however, did not give me any insight into solving the problem. In general, I have a lot of trouble figuring out how to relate the hypothesis given in such (algebraic) problems to the conclusion. Most of the time, I just keep transforming the equations aimlessly until I stumble upon the conclusion.

I would prefer a solution with the motivation behind it, and also, if possible, some general tips on approaching such problems.

Answer 1

Consider the map: $f:(a,b)\to (a + b\cos\theta, b\sin\theta)$, the space being $\Bbb R^2$ with euclidian structure.

The hypothesis writes $$ |f(a,b)|^2 = |f(c,d)|^2 = 1 \\ \langle f(a,b), f(c,d) \rangle = 0 $$

Now let us compute $$ f^{-1}(x,y) = (x - {\cot\theta}y, y \csc\theta)$$

Consider a orthogonal basis $((x_1,y_1), (x_2,y_2))$ (under the hypothesis, $ (f(a,b), f(c,d))$ is one). You can find $\phi$ such as:

$$ (x_1,y_1) = (\cos\phi ,\sin\phi) \\ (x_2,y_2) = (-\sin\phi ,\cos\phi) \\ [f^{-1}(x_1,y_1)\cdot e_x]^2 + [f^{-1}(x_2,y_2)\cdot e_y]^2 = (\csc\phi - \cot\theta\sin\phi)^2 + (-\sin\phi- \cot\theta\cos\phi)^2 = \csc^2\phi $$

Answer 2

So you are given the following:

$$a^2 + b^2 + 2ab\cos\theta = 1 \tag1$$

$$d^2 + c^2 + 2cd\cos\theta = 1 \tag2$$

$$ac + bd + (ad + bc)\cos\theta = 0\tag3$$

And you want to prove:

$$a^2 + c^2 = \csc^2\theta$$

$$b^2 + d^2 = \csc^2\theta$$

To Begin

As you stated above, you started by says that $$\cos(\theta) = 0$$

Then $$a^2 + b^2 = 1 \tag1$$

$$d^2 + c^2 = 1 \tag2$$

$$ac + bd = 0\tag3$$

Realize that if $\cos(\theta) = 0$, then $\csc^2(\theta) = 1$

Therefore

$$a^2 + c^2 = 1$$

$$b^2 + d^2 = 1$$

If you add these 2 equation together

$$a^2 + b^2 + c^2 + d^2 = 2$$

Adding (1) and (2) gives us

$$a^2 + b^2 + c^2 + d^2 = 2$$

And thus it is proven