# If tan (x) =5/12, then what is sin x and cos x?

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53
Alan P. Share
Apr 15, 2015

Viewed as a right angled triangle
$\tan \left(x\right) = \frac{5}{12}$
can be thought of as the ratio of opposite to adjacent sides in a triangle with sides $5 , 12$ and $13$ (where $13$ is derived from the Pythagorean Theorem)

So
$\sin \left(x\right) = \frac{5}{13}$
and
$\cos \left(x\right) = \frac{12}{13}$

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25
Sep 1, 2015

This can be solved easily once we visualise it.

#### Explanation:

A)wkt, tanx = perpendicular(p)/base(b) .....(1)
B)We are given tanx=5/12 .....(2)
C)On comparing the quantities we get that p=5 and b=12
D)We require the value of hypotenuse(h),
for this we apply the Pythagoras theorum i.e, ${h}^{2} = {p}^{2} + {b}^{2}$
$\Rightarrow$ ${h}^{2}$ =${5}^{2}$+${12}^{2}$
$\Rightarrow$${h}^{2}$=25+144
$\Rightarrow$${h}^{2}$=169
$\Rightarrow$ h=$\sqrt{169}$
$\Rightarrow$ h=13
..............................................................................................................
Hence , sinx = perpendicular/hypotenuse=p/h = 5/13
cosx = base/hypotenuse=b/h = 12/13 .... (Answer)**

.............................................................................................................
I hope this helps :)

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mason m Share
Jun 25, 2016

$\sin \left(x\right) = \frac{5}{13}$
$\cos \left(x\right) = \frac{12}{13}$

#### Explanation:

Use the Pythagorean identity:

${\sin}^{2} \left(x\right) + {\cos}^{2} \left(x\right) = 1$

Divide through by ${\cos}^{2} \left(x\right)$:

${\tan}^{2} \left(x\right) + 1 = {\sec}^{2} \left(x\right)$

Since $\tan \left(x\right) = \frac{5}{12}$:

$\frac{25}{144} + 1 = {\sec}^{2} \left(x\right) = \frac{169}{144}$

Thus:

$\sec \left(x\right) = \frac{13}{12}$

So:

$\cos \left(x\right) = \frac{12}{13}$

Using the Pythagorean identity again with $\cos \left(x\right) = \frac{12}{13}$:

${\sin}^{2} \left(x\right) + \frac{144}{169} = 1$

${\sin}^{2} \left(x\right) = \frac{25}{169}$

$\sin \left(x\right) = \frac{5}{13}$

Note that the signs of these values (positive/negative) depend on the quadrant of the angle. Since tangent is positive, this could be in either the first or third quadrants, so sine and cosine could be positive or negative, so the positive answers are given as defaults.

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mason m Share
Jun 24, 2016

$\sin \left(x\right) = \frac{5}{13}$
$\cos \left(x\right) = \frac{12}{13}$

#### Explanation:

The best way is to visualize the $5 - 12 - 13$ right triangle, but this is another valid method:

$\tan \left(x\right) = \frac{5}{12} \text{ "=>" } x = \arctan \left(\frac{5}{12}\right)$

So, we see that:

$\sin \left(x\right) = \sin \left(\arctan \left(\frac{5}{12}\right)\right)$

We can relate $\sin$ and $\tan$:

${\sin}^{2} \left(x\right) + {\cos}^{2} \left(x\right) = 1$

Dividing through by ${\sin}^{2} \left(x\right)$:

$1 + {\cot}^{2} \left(x\right) = {\csc}^{2} \left(x\right)$

Rewriting:

$1 + \frac{1}{\tan} ^ 2 \left(x\right) = \frac{1}{\sin} ^ 2 \left(x\right)$

Common denominator:

$\frac{{\tan}^{2} \left(x\right) + 1}{\tan} ^ 2 \left(x\right) = \frac{1}{\sin} ^ 2 \left(x\right)$

Inverting:

${\sin}^{2} \left(x\right) = {\tan}^{2} \frac{x}{{\tan}^{2} \left(x\right) + 1} \text{ "=>" } \sin \left(x\right) = \tan \frac{x}{\sqrt{{\tan}^{2} \left(x\right) + 1}}$

So, plugging this into $\sin \left(x\right)$, we see that it equals:

$\sin \left(\arctan \left(\frac{5}{12}\right)\right) = \tan \frac{\arctan \left(\frac{5}{12}\right)}{\sqrt{{\tan}^{2} \left(\arctan \left(\frac{5}{12}\right)\right) + 1}}$

Since $\tan \left(\arctan \left(\frac{5}{12}\right)\right) = \frac{5}{12}$:

$\sin \left(x\right) = \frac{\frac{5}{12}}{\sqrt{\frac{25}{144} + 1}} = \frac{\frac{5}{12}}{\sqrt{\frac{169}{144}}} = \frac{5}{12} \left(\frac{12}{13}\right) = \frac{5}{13}$

We can now use this to find $\cos \left(x\right)$:

${\sin}^{2} \left(x\right) + {\cos}^{2} \left(x\right) = 1 \text{ "=>" } {\left(\frac{5}{13}\right)}^{2} + {\cos}^{2} \left(x\right) = 1$

So:

${\cos}^{2} \left(x\right) = 1 - \frac{25}{169} \text{ "=>" } \cos \left(x\right) = \sqrt{\frac{144}{169}} = \frac{12}{13}$

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