Why doesn't$\frac{\sin a+\sin3a+\sin5a}{\cos a+\cos3a+\cos5a}=\tan3a$ imply $\sin a+\sin5a=0$ and $\cos a+\cos5a=0$? $$\frac{\sin(\alpha)+\sin(3\alpha)+\sin(5\alpha)}{\cos(\alpha)+\cos(3\alpha)+\cos(5\alpha)} = \tan(3\alpha) \tag1$$
I've proven this trigonometric identity by subtracting the RHS from both sides and then applying the rule $$\sin(\alpha+\beta) = \sin(\alpha)\cos(\beta)+\cos(\alpha)\sin(\beta)$$
Now,
$$\frac{\sin(3\alpha)}{\cos(3\alpha)} = \tan(3\alpha) \tag2$$ is trivially true, and since $(1)$ is true as well, it seems like $$\sin(\alpha)+\sin(5\alpha)=0 \qquad\text{and}\qquad \cos(\alpha)+\cos(5\alpha)=0 \tag3$$ should be true, but they obviously aren't.
So my question is:

Why does $(1)$ does not imply $(3)$?

I feel it's something trivial, but I just can't figure it out.
 A: Let's look at a simple example to see why it isn't true.
Let $\alpha=\dfrac{\pi}{2}$, then
$$
\begin{align*}
\sin \alpha + \sin 5\alpha &=\sin \dfrac{\pi}{2} + \sin \dfrac{5\pi}{2}\\
&=1+1\\
&=2 \not= 0.
\end{align*}
$$
A: Consider the following true equality:
$$\frac{5 + 6 + 2}{3 + 6 + 4} = 1$$
But $\frac{6}{6} = 1$ as well. Does this imply that the remaining terms $5,2,3,4$ must be zero? Obviously not. You cannot simply eliminate $6$ from the numerator and the denominator $6$, nor can you conclude that the remaining terms are zero. This doesn't follow any of the rules of arithmetic.
A: Remember that $\dfrac ab=\dfrac cd=k$ then we also have $\dfrac{a+c}{b+d}=k$
Note that the "reverse" is also true (provided $d\neq 0$) since $\dfrac ab=\dfrac{a+c}{b+d}=k$ implies that $\dfrac cd=\dfrac {-c}{-d}=\dfrac{a+(-a-c)}{b+(-b-d)}=k\ $ too.
You can check it on the example you were given $\frac{5+2}{3+4}=1$ as well as $\frac 66$.

Here you have $\quad\dfrac{\sin(x)+\sin(3x)+\sin(5x)}{\cos(x)+\cos(3x)+\cos(5x)}=\dfrac{\sin(3x)}{\cos(3x)}=\tan(3x)$

So the quantity $\quad\dfrac{\sin(x)+\sin(5x)}{\cos(x)+\cos(5x)}\ $ should also be equal to $\tan(3x)$ which you can verify.
