<s>[INST] Simple Trig Equations - Why is it Wrong to Cancel Trig Terms? In the following problem, I first did it using a cancellation of $sin^2\theta$, working shown below, which gave the wrong answer. Having looked at the question again, I saw it could be solved by factoring, working again below.
My question then, is why is it wrong to cancel the $\sin^2\theta$ term - the algebra seems correct to me?
The Problem
Solve for $\theta$ in the interval $0 \le \theta \le 360$
$$4\sin\theta = \tan\theta$$
My Solution
$$4\sin\theta = \frac{\sin\theta}{\cos\theta}$$
$$4\sin\theta \cos\theta = \sin\theta$$
Squaring, and substituting, using the identity $\cos^2\theta = 1 - \sin^2\theta$
$$16\sin^2\theta(1-\sin^2\theta) = \sin^2\theta$$
$$1-\sin^2\theta = \frac{\sin^2\theta}{16\sin^2\theta}$$
$$1-\sin^2\theta = \frac{1}{16}$$
Rest of working to final answer omitted.
Correct Solution
$$4\sin\theta = \frac{\sin\theta}{\cos\theta}$$
$$4\sin\theta \cos\theta - \sin\theta = 0$$
$$\sin\theta(4\cos\theta - 1) = 0$$
Rest of working to final answer omitted.
In sum, why is it wrong to cancel as I did first time around - why must these problems be solved by factoring as in the correct solution?
 [/INST] Here's a simpler way to get your answer.  Remember what $sin(\theta)$ and $tan(\theta)$ actually represent on a right triangle.  For a given right triangle, let $a$ be the length of the adjacent side, $b$ be the length of the opposite side, and $c$ be the length of the hypotenuse.  This means:
$sin(\theta) = \frac{a}{c}$ and $tan(\theta) = \frac{a}{b}$
Substituting, your equation becomes:
$4\frac{a}{c} = \frac{a}{b}$
Cancelling the $a$ and rearranging gives:
$\frac{b}{c} = \frac{1}{4}$
Using the definition of cosine gives:
$cos(\theta) = \frac{1}{4}$
And
$acos(\frac{1}{4}) = \theta$
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