Solving $\cos^2(a)+\cos(a) = \sin^2(a)$ by using a vector product

Question

$$\cos^2(a)+\cos(a) = \sin^2(a)$$

I'm required to solve it using vectors. Let me show my attempt.

We know that

$$u\cdot v=|u||v| \iff v=ku$$

Recall that

$$ u=(1,1,1)\implies |u|=\sqrt{1^2+1^2+1^2}=\sqrt 3$$

$$v=[\cos^2(a), \cos(a), \sin^2(a)]$$

I couldn't define $v$. That's why I couldn't proceed further.

Regards

Answer 1

Rewrite the equation as $$\cos^2(a) - \sin^2(a) = -\cos(a)$$ which is equivalent to the vector equation $$\left<\cos(a), \sin(a)\right> \cdot \left<\cos(a),-\sin(a) \right> = -\cos(a)$$ Now write $\textbf{u}=\left<\cos(a), \sin(a)\right>$ and $\textbf{v}=\left<\cos(a), -\sin(a)\right>$. We observe that both u and v are unit vectors, and the angle between them is $2a$. Therefore from $$\textbf{u}\cdot \textbf{v} = |\textbf{u}||\textbf{v}|\cos(\theta)$$ (where $\theta$ is the angle between the two vectors) we find that the equation is equivalent to $$\cos(2a) = -\cos(a)$$

(Of course, we could have gotten to this same conclusion much simpler by just using the double-angle identity $\cos(2a) = \cos^2(a) - \sin^2(a)$, but OP wanted something that explicitly uses vectors, so we took the long way around.)

The most straightforward way to interpret this latter equation is as follows: we are seeking a point on the unit circle, located by an angle $a$, with the property that doubling $a$ either reflects the point across the $y$-axis or across the origin. By inspection one can see that $a = \frac{\pi}{3}$ is the only angle in the first quadrant with that property, $a = \frac{5\pi}{3}$ is the only angle in the fourth quadrant with that property, and in addition $a=\pi$ has that property; it's not too hard to reason your way to the conclusion that the angle can't be anywhere else.

I don't see any "purely vectorial" way to solve $\cos(2a) = -\cos(a)$, but then again I'm not completely sure what "using vectors" means in the OP's context.

Answer 2

With vectors, one could proceed as follows:

$$\cos^2 a+\cos a=\sin^2 a \iff ( 1,1,1) \cdot \left(\cos^2 a,\cos a,-\sin^2 a\right ) =0$$

The convenient values of $a$ are in the gimusi's answer and the math community is satisfied :) , but I feel you want to see how to continue with $||v||:$

$$||v||=\sqrt{\cos^4 a+\cos^2 a+\sin^4 a}=\sqrt{2\cos^4 a-\cos^2 a+1}.$$

But I do not see how this can help you to continue.