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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?

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    $\begingroup$ TeX suggestion - I recommend you to use \cfrac instead of \frac here. As \cfrac or \dfrac will make the frac. look much better & understandable. $\cfrac{12}{5} $ and $\frac{12}{5}$ . $$\text{Notice the difference}$$ $\endgroup$ – Kushashwa Ravi Shrimali Jun 30 '14 at 15:21
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    $\begingroup$ \large{\tan{\theta}} = $\large{\tan{\theta}}$ . May be, this looks better? $\endgroup$ – Kushashwa Ravi Shrimali Jun 30 '14 at 15:24
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    $\begingroup$ As a suggestion for making trigonometric functions more readable, try omitting parentheses when the operation is unambiguous: $\tan\theta$ as opposed to $\tan(\theta)$. $\endgroup$ – Théophile Jun 30 '14 at 15:24
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    $\begingroup$ As Théophile and others said, they are proportional to $12$ and $5$, not equal! $\endgroup$ – Yves Daoust Jun 30 '14 at 15:28
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    $\begingroup$ If $6/8=3/4$, does it follow that $6=3$ and $8=4$? $\endgroup$ – WillO Jun 30 '14 at 15:48
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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 $\color{red}{\sin(\theta)=12 \ \rm{and} \ \cos(\theta)=5 \ (\rm{wrong})}.$

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  • $\begingroup$ Does this mean it's possible for me to scale the the fraction $\cfrac{12}{5}$, and i'll eventually have $\sin(\theta)=a$ and $\cos (\theta)=b$ $\endgroup$ – seeker Jun 30 '14 at 15:31
  • $\begingroup$ Well, $12/5=2.4=\tan(\theta)$. Now use the identities $$1+\tan^2(\theta) \equiv \frac{1}{\cos^2(\theta)}$$ and $$\sin^2(\theta)+\cos^2(\theta) \equiv 1$$ to find $\cos(\theta)$ and $\sin(\theta)$. $\endgroup$ – beep-boop Jun 30 '14 at 15:34
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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 $\sin^2\theta + \cos^2\theta=1$.

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    $\begingroup$ Note also that this will give you two values for $k$, leading to two solutions $\ldots$ $\endgroup$ – Théophile Jun 30 '14 at 15:22
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These are trigonometric ratios. tan of any angle gives the ratio of the opposite and adjacent sides to that angle.

tan theta = y/x

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 $\cfrac{x}{r} = \cfrac{5}{13}.$

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    $\begingroup$ Note that this is only a partial solution $\ldots$ $\endgroup$ – Théophile Jun 30 '14 at 15:25
  • $\begingroup$ This solution assumes that $\theta$ is acute, which need not be true. $\endgroup$ – Fly by Night Jun 30 '14 at 15:37
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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.

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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.

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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 $cos(\theta)=\frac{5}{13}$

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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.

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    $\begingroup$ Why do you assume $\theta$ to be in the first quadrant? $\endgroup$ – Théophile Jun 30 '14 at 15:32
  • $\begingroup$ This "solution" assumes that $\theta$ is acute, which need not be true. $\endgroup$ – Fly by Night Jun 30 '14 at 15:38
  • $\begingroup$ Thanks for the comment, I made the assumption for the sake of leaving one answer. I have edited the response appropriately. Cheers. $\endgroup$ – user155385 Jun 30 '14 at 16:18
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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.

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Okay, $$tan(\theta)=\frac{12}{5}$$. This means in a right angled triangle, you have the ratio of perpendicular to base is $$\frac{P}{B}=\frac{12}5$$ so the Hypotenuse becomes $$\sqrt{(12)^2 + (5)^2}=13$$

Now calculating the sine and cos will be easy fo you I am sure! :)

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