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

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

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 :)

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:

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

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

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

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

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.

Jun 1, 2018

One other method:

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

By the definition of tangent,

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

Then we can say:

$\sin \left(x\right) = \frac{5}{12} \cos \left(x\right) \text{ "" "" "" } \left(\star\right)$

We can then substitute $\left(\star\right)$ into the Pythagorean identity:

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

${\left(\frac{5}{12} \cos \left(x\right)\right)}^{2} + {\cos}^{2} \left(x\right) = 1$

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

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

${\cos}^{2} \left(x\right) = \frac{144}{169}$

$\cos \left(x\right) = \sqrt{\frac{144}{169}} = \frac{12}{13}$

Then, use ${\sin}^{2} \left(x\right) + {\cos}^{2} \left(x\right) = 1$ again to find $\sin \left(x\right) = \frac{5}{13}$.

Jun 2, 2018

$\sin x = \pm \frac{5}{13}$
$\cos x = \pm \frac{12}{13}$

Explanation:

$\tan x = \frac{5}{12}$.
${\cos}^{2} x = \frac{1}{1 + {\tan}^{2} x} = \frac{1}{1 + \frac{25}{144}} = \frac{144}{169}$
$\cos x = \pm \frac{12}{13}$
${\sin}^{2} x = 1 - {\cos}^{2} x = 1 - \frac{144}{169} = \frac{25}{169}$
$\sin x = \pm \frac{5}{13}$
If x lies in Quadrant 1, cos x and sin x are both positive.
If x lies in Quadrant 3, sin x and cos x are both negative.

Aug 2, 2018

$\sin x = \pm \frac{5}{13}$, $\cos x = \pm \frac{12}{13}$

Explanation:

Recall that in a right triangle, the tangent is equal to the opposite side over the adjacent. We know this thru SOH-CAH-TOA.

We know two sides of a triangle, so we can find the third with the Pythagorean Theorem

${a}^{2} + {b}^{2} = {c}^{2}$

Plugging our $a$ and $b$ values in, we get

${5}^{2} + {12}^{2} = {c}^{2}$

$\implies 25 + 144 = {c}^{2}$

$\implies 169 = {c}^{2}$

$c = 13$

From this, we know the opposite side is $5$, the adjacent side is $12$, and the hypotenuse is $13$.

From SOH-CAH-TOA, we know sine is equal to the opposite side over the hypotenuse. This means that $\sin x = \frac{5}{13}$.

We also know that cosine is equal to the adjacent side over the hypotenuse. Thus, $\cos x = \frac{12}{13}$.

Recall that $\tan x = \frac{\sin x}{\cos x}$. Since the negatives cancel out, we can also say

$\tan x = \frac{- \sin x}{- \cos x}$. This means that $\sin x$ and $\cos x$ can be either both positive or both negative.

Hope this helps!

Aug 3, 2018

Explanation:

Here, $0 < \tan x = \frac{5}{12.}$ So,

$x \in {Q}_{1} \mathmr{and} = k \pi + \arctan \left(\frac{5}{12}\right) , k = 0 , 2 , 4 , 6 , . .$or

$\in {Q}_{3} \mathmr{and} = k \pi + \arctan \left(\frac{5}{12}\right) , k = 1. 3 , 5 , \ldots$.

Easily,

$\sin x = \frac{5}{\pm \sqrt{{12}^{2} + {5}^{2}}} = \pm \frac{5}{13}$ and

$\cos x = \pm \sqrt{1 - {\sin}^{2} x} = \pm \frac{12}{13}$

If $x \in$ (all positive )${Q}_{1}$, choose positive values.

Otherwise, both are negative.

If x is chosen in $\left(0 , 2 \pi\right)$,

it is either $\arctan \left(\frac{3}{12}\right) = {22.62}^{o} = 0.3948 r a d$, nearly, or

${180}^{o} + {22.62}^{o} = {202.62}^{o} = 3.5364 r a d$, nearly.

See the combined graph of $y = \tan x , y = \sin x \mathmr{and} y = \cos x$,

depicting all these aspects. Slide the graph $\leftarrow \mathmr{and} \rightarrow$, to see

x-solutions, as the meet of $y = \frac{5}{12}$ with the periodic graph

$y = \tan x$, in infinitude. The circular dots give the answer as y-

values, respectively.

graph{(y-tan x) ( y- sin x ) ( y- cos x )(y- 5/12+0x)(x-0.395+0.001 y)(x-3.536 + 0.0001 y) ((x-0.395)^2+(y-12/13)^2-0.001)((x-3.536)^2+(y+12/13)^2 - 0.001)((x-0.395)^2+(y-5/13)^2-0.001)((x-3.536)^2+(y+5/13)^2-0.001)=0[0 5 -1.2 1-2]}

Thanks to the Socratic graphis potential., for precision graphs.