How do I solve this question using vectors?

Question

Given trigonometric equation

$$5\cos \theta -12\sin \theta = 13$$

I'll be trying to determine a general solution using vectors. I've recently asked this question two times. However, the answers weren't genuinely understandable in particular. This way would also collide with other ways being used to solve this problem.

Regards!

Answer 1

The equation $5\cos\theta - 12\sin\theta = 13$ is equivalent to the vector equation $$\left< 5, -12 \right> \cdot \left<\cos\theta, \sin\theta\right> = 13$$

Now in general, for any two vectors $\textbf{u}, \textbf{v}$ the dot product obeys $$\textbf{u} \cdot \textbf{v} = |\textbf{u}| |\textbf{v}| \cos\phi$$ where $\phi$ is the angle between the two vectors $\textbf{u}$ and $\textbf{v}$. In this case, we have $\textbf{u}= \left< 5, -12 \right>$ with $|\textbf{u}| = 13$, and $\textbf{v} = \left<\cos\theta, \sin\theta \right>$ with $| \textbf{v} | = 1$, so $$(13)(1)\cos\phi = 13$$ which means that $\phi = 0$. In other words, $\left<\cos\theta, \sin\theta\right>$ is a unit vector that points in the same direction as $\left< 5, -12 \right>$, so $\theta = \arctan \left( - \frac{12}{5} \right) $.

Answer 2

$$\langle 5,-12 \rangle \cdot \langle \cos\theta,\sin\theta \rangle = 13$$ $$\left(\sqrt{5^2+12^2}\sqrt{\cos^2\theta+\sin^2\theta}\right)\cos(\theta-\tan^{-1}(-12/5)) = 13$$ $$\cos(\theta+\tan^{-1}(12/5)) = 1$$ $$\theta = 2\pi k - \tan^{-1}(12/5), k\in\mathbb{Z}$$

Answer 3

We know that given two vectors $u$ and $v$

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

therefore let

• $u=(5,-12)\implies |u|=13$

• $v=(\cos \theta, \sin \theta)\implies |v|=1$

therefore since $u\cdot v=|u||v|$ we have that

$$v=(\cos \theta, \sin \theta)=k(5,-12)=\left(\frac5{13},-\frac{12}{13}\right)$$

that is

• $\cos \theta=\frac5{13}$
• $\sin \theta=-\frac{12}{13}$

and

• $\tan \theta =\frac{\sin \theta}{\cos \theta}=-\frac{12}5$

from which we obtain

$$\theta = \arctan(-12/5) \color{red}{+2k\pi}$$

Note that we take $+2k\pi$ since $\cos \theta>0$ and $\sin \theta<0$.

Answer 4

Here is a geometric solution to the similar equation

$$5\cos\theta+12\sin\theta=13$$

From the Pythagorean triple $5$-$12$-$13$ construct the right triangle $\triangle ABD$ with sidelenghts $AB=5$, $AD=12$, and $BD=13$. Pick a point $C$ lying on $\overline{BD}$ such that $\overline{AC}$ is perpendicular to $\overline{BD}$. By the Inscribed Similar Triangles Theorem, $\triangle ABD$ is similar to $\triangle CAD$, therefore $\angle ABD=\angle CAD$.

Now for some trigonometry: Let $\theta=\angle ABD=\angle CAD$. Then $BC=5\cos\theta$ and $CD=12\sin\theta$. By construction, $BD=13$. Therefore, $$BC+CD=BD\iff 5\cos\theta+12\sin\theta=13$$ By the Inscribed Similar Triangles Theorem, $\triangle CBA$ is similar to $\triangle ABD$, yielding the proportion $$\frac{BC}{AB}=\frac{AB}{BD} \iff \frac{5\cos\theta}{5}=\frac{5}{13}$$ $$\implies \theta=\cos^{-1}(\frac{5}{13})$$

Maybe you can rearrange this setup to match your equation $5\cos\theta-12\sin\theta=13$.