So I've been asked to find $\sin(\theta)$ and $\cos(\theta)$ when $\tan(\theta)=\cfrac{12}{5}$; my question is if $\tan (\theta)=\cfrac{\sin (\theta) }{\cos (\theta)}$ does this mean that because $\tan (\theta)=\cfrac{12}{5}$ then $\sin (\theta) =12$ and $\cos(\theta)=5$?
It doesn't seem to be the case in this example, because $\sin (\theta)\ne 12 $ and $\cos (\theta)\ne 12 $.
Can someone tell me where the error is in my thinking?
On
This is a good start; you're right up to a scaling factor. In other words, $\sin\theta$ and $\cos\theta$ are in a ratio of 12 to 5, but aren't necessarily equal to those values. (And in fact cannot equal those values.)
Set $\sin\theta = 12k$ and $\cos\theta = 5k$, then use the fact that $\sin^2\theta + \cos^2\theta=1$.
On
The definition of the sine, cosine, and tangent of an angle $\theta$ involves finding a point $(x,y)$ on the ray that makes an angle $\theta$ with the positive $x$-axis. In your case, the point may be taken to be $(5,12)$. Then $\tan(\theta)=y/x=12/5$, correctly. But the definition of sine and cosine is that $\sin(\theta)=y/r$, $\cos(\theta)=x/r$, where $r$ is the distance of your point from the origin, $r=\sqrt{x^2+y^2}$, always taken positive. So here, you have to observe that the distance of your point from the origin is $13$, and you’re ready to go.
On
These are trigonometric ratios. tan of any angle gives the ratio of the opposite and adjacent sides to that angle.
Now that you have tan theta = 12/5, this simply means that tan theta = y/x thus y = 12, x = 5. Now if you use Pythagoras, you will get:
$$ \begin{align} & r^2 = y^2 + x^2 \\ & r^2 = 12^2 + 5^2 \\ \text{which is} \ & r = \sqrt{(144+25)} = \sqrt{(169)} = 13 \end{align} $$
Thus, $ r = 13. $
So $$ \sin \theta = \cfrac{\text{opposite}}{\text{hypotenuse}} $$
Which is $\cfrac{y}{r} = \cfrac{12}{13} $
and $$\cos \theta = \cfrac{\text{adjacent}}{\text{hypotenuse}}$$
which is $\cfrac{x}{r} = \cfrac{5}{13}.$
On
The error is that $\tan\theta$ merely gives the ratio of $\sin\theta$ and $\cos\theta$. If it is given as a fraction, that does not mean that the numerator must be $\sin\theta$ and the denominator must be $\cos\theta$. It is necessary to use trigonometric identities to solve for $\sin\theta$ and $\cos\theta$.
We have
$$1+\tan^2\theta=\sec^2\theta$$
so
$$\cos\theta=\sqrt[]{\frac{1}{1+\tan^2\theta}}=\sqrt[]{\frac{1}{1+(\frac{12}{5})^2}}=\sqrt[]{\frac{25}{169}}=\pm\frac{5}{13}$$
Then by the trigonometric identity $\sin^2\theta+\cos^2\theta=1$ we have
$$\sin\theta=\sqrt[]{1-\cos^2\theta}=\sqrt[]{1-\frac{25}{169}}=\sqrt[]{\frac{144}{169}}=\pm\frac{12}{13}$$
Now since $\tan\theta>0$, $\theta$ is in either the first or third quadrant, so $\sin\theta$ and $\cos\theta$ have the same sign. So
$$(\sin\theta,\cos\theta)=(\frac{12}{13},\frac{5}{13}) \text{ or } (\sin\theta,\cos\theta)=(-\frac{12}{13},-\frac{5}{13})$$
Verifying our answer,
$$\tan\theta=\frac{\sin\theta}{\cos\theta}=\frac{\pm\frac{12}{13}}{\pm\frac{5}{13}}=\frac{12}{5}$$
as the problem stated.
On
Most of the above solutions assume, wrongly, that $\theta$ is acute, i.e. $0^{\circ} < \theta < 90^{\circ}.$ Here I give all the cases for $0^{\circ}<\theta <360^{\circ}.$ I like to use simple identities to answer these questions. We know that $1+\tan^2\theta \equiv \sec^2\theta$. If $\tan\theta = \frac{12}{5}$ then we see that:
$$\begin{eqnarray*} 1+\tan^2\theta &\equiv& \sec^2\theta \\ \\ 1+\left(\frac{12}{5}\right)^{\! 2} &=& \sec^2\theta \\ \\ \frac{169}{25} &=& \sec^2\theta \end{eqnarray*}$$ It follows that $\sec\theta = \pm \frac{13}{5}$, and hence $\cos\theta = \pm \frac{5}{13}$. The question is: which is it, plus or minus? Sadly, there is not enough information in the question.
Looking at the graph $y=\tan\theta$ we see that $\tan \theta > 0$ if $0^{\circ} < \theta < 90^{\circ}$ or $180^{\circ} < \theta < 270^{\circ}$. That means $\tan \theta = \frac{12}{5} > 0$ has solutions in the range $0^{\circ} < \theta < 90^{\circ}$ or $180^{\circ} < \theta < 270^{\circ}$. However, $\cos \theta > 0$ when $0^{\circ} < \theta < 90^{\circ}$ and $\cos \theta < 0$ when $180^{\circ} <\theta < 270^{\circ}$.
To conclude: $\cos\theta = \frac{5}{13}$ if $0^{\circ} < \theta < 90^{\circ}$ and $\cos\theta = -\frac{5}{13}$ if $180^{\circ} < \theta < 270^{\circ}$.
There are many ways to find $\sin \theta$. You could use the identity $1+\cot^2\theta \equiv \operatorname{cosec}^2\theta$, noting that $\cot$ is the reciprocal of $\tan$ and $\tan\theta = \frac{12}{5} \iff \cot\theta = \frac{5}{12}$. Alternatively, you could use your value of $\cos\theta$ along with the identity $\sin^2\theta + \cos^2\theta \equiv 1$. It's up to you; I'd use the second one.
On
Mathematically, if
$\frac{a}{b}$=$\frac{c}{d}$ does not imply that $a=c$ and $b=d$.
In Trigonometry, there is a terminology called Pythagorean triples. These are the triples that satisfy the Pythagorean formula $a^2+b^2=c^2$. Some of the famous triples are 3, 4, 5;1, $\sqrt{3}$, 2; 5, 12, 13.
Now, with regard to your question, the missing number in the triplet is 13. And secondly, the ratio $\frac{opposite}{adjacent}$=$\frac{12}{5}$. Hence, hypotenuse must be 13units. Hence $sin(\theta)=\frac{12}{13}$ and $cos(\theta)=\frac{5}{13}$
On
One general approach to problems where you know
$$ \frac{x}{y} = \frac{a}{b} $$
is to introduce a new variable $z$, and convert the above equation into the following three:
$$ x = az \qquad y = b z \qquad z \neq 0 $$
Exercise: prove that every solution for $(x,y)$ to the original equation is also a solution for the system of equations.
Exercise: prove that every solution for $(x,y,z)$ to the system of equations is also a solution to the original equation.
The error in your thinking is that $x=a$ and $y=b$ is just one among the many possible solutions to the original equation for $(x,y)$. If all you know is that the $(x,y)$ you're looking for is a solution, you have no reason to believe it has to be the particular solution $x=a$ and $y=b$.
The point of doing this conversion is that you've reduced two variables to one variable: you can substitute $x \to az$ and $y \to bz$ in all other equations and there is now one less variable overall to solve for.
Of course, you could instead just solve for $x$:
$$ x = \frac{ay}{b} $$
but sometimes the variation of the substitution I describe above is simpler to work with.
Either way, going back to the original problem, you need to find another equation that involves $\sin(\theta)$ and $\cos(\theta)$ before you can proceed to solve for them.
Just because $\frac{a}{b}=\frac{c}{d},$ it does not necessarily follow that $a=c$ or that $b=d.$
e.g. $\frac{2}{3}=\frac{4}{6}$, but $2 \neq 4$ and $3 \neq 6$.
In your case, since $\tan(\theta)=\frac{12}{5},$ we have that $$\frac{\sin(\theta)}{\cos(\theta)}=\frac{12}{5} \iff \boxed{5 \sin(\theta)=12\cos(\theta)},$$ which is not the same thing as saying that $\color{red}{\sin(\theta)=12 \ \rm{and} \ \cos(\theta)=5 \ (\rm{wrong})}.$