If $tan (x) = \frac{5}{12}$ $tan (x) =5/12$ , then what is sin x and cos x?

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If $tan (x) = \frac{5}{12}$ $tan (x) =5/12$ , then what is sin x and cos x?

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Answer 1 · 2015-04-15T12:23:36

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

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

Answer 2 · 2015-09-01T19:33:16

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, $h^{2} = p^{2} + b^{2}$ $h^2= p^2 + b^2$
$\Rightarrow$ $rArr$ $h^{2}$ $h^2$ = $5^{2}$ $5^2$ + $12^{2}$ $12^2$
$\Rightarrow$ $rArr$ $h^{2}$ $h^2$ =25+144
$\Rightarrow$ $rArr$ $h^{2}$ $h^2$ =169
$\Rightarrow$ $rArr$ h= $\sqrt{169}$ $sqrt169$
$\Rightarrow$ $rArr$ h=13
..............................................................................................................
Hence , sinx = perpendicular/hypotenuse=p/h = 5/13
cosx = base/hypotenuse=b/h = 12/13 .... (Answer)**

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

Answer 3 · 2016-06-24T22:58:04

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

Explanation:

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

$tan (x) = \frac{5}{12} \Rightarrow x = arctan (\frac{5}{12})$ $tan(x)=5/12" "=>" "x=arctan(5/12)$

So, we see that:

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

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

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

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

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

Rewriting:

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

Common denominator:

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

Inverting:

${sin}^{2} (x) = \frac{{tan}^{2} (x)}{{tan}^{2} (x) + 1} \Rightarrow sin (x) = \frac{tan (x)}{\sqrt{{tan}^{2} (x) + 1}}$ $sin^2(x)=tan^2(x)/(tan^2(x)+1)" "=>" "sin(x)=tan(x)/sqrt(tan^2(x)+1)$

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

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

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

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

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

${sin}^{2} (x) + {cos}^{2} (x) = 1 \Rightarrow {(\frac{5}{13})}^{2} + {cos}^{2} (x) = 1$ $sin^2(x)+cos^2(x)=1" "=>" "(5/13)^2+cos^2(x)=1$

So:

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

Answer 4 · 2016-06-25T00:55:08

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

Explanation:

Use the Pythagorean identity:

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

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

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

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

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

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

Thus:

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

So:

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

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

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

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

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

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.

Answer 5 · 2018-06-01T17:01:56

One other method:

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

By the definition of tangent,

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

Then we can say:

$sin (x) = \frac{5}{12} cos (x) (⋆)$ $sin(x)=5/12cos(x)" "" "" "" "(star)$

We can then substitute $(⋆)$ $(star)$ into the Pythagorean identity:

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

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

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

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

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

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

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

Answer 6 · 2018-06-02T00:01:40

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

Explanation:

$tan x = \frac{5}{12}$ $tan x = 5/12$ .
x could be either in Quadrant 1 or Quadrant 3.
${cos}^{2} x = \frac{1}{1 + {tan}^{2} x} = \frac{1}{1 + \frac{25}{144}} = \frac{144}{169}$ $cos^2 x = 1/(1 + tan^2 x) = 1/(1 + 25/144) = 144/169$
$cos x = \pm \frac{12}{13}$ $cos x = +- 12/13$
${sin}^{2} x = 1 - {cos}^{2} x = 1 - \frac{144}{169} = \frac{25}{169}$ $sin^2 x = 1 - cos^2 x = 1 - 144/169 = 25/169$
$sin x = \pm \frac{5}{13}$ $sin x = +- 5/13$
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.

Answer 7 · 2018-08-02T22:12:18

$sin x = \pm \frac{5}{13}$ $sinx=pm5/13$ , $cos x = \pm \frac{12}{13}$ $cosx=pm12/13$

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

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

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

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

$\Rightarrow 25 + 144 = c^{2}$ $=>25+144=c^2$

$\Rightarrow 169 = c^{2}$ $=>169=c^2$

$c = 13$ $c=13$

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

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

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

Recall that $tan x = \frac{sin x}{cos x}$ $tanx=(sinx)/(cosx)$ . Since the negatives cancel out, we can also say

$tan x = \frac{- sin x}{- cos x}$ $tanx=(-sinx)/(-cosx)$ . This means that $sin x$ $sinx$ and $cos x$ $cosx$ can be either both positive or both negative.

Hope this helps!

Answer 8 · 2018-08-03T02:58:16

Disambiguation answer.

Explanation:

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

$x \in Q_{1} and = k π + arctan (\frac{5}{12}), k = 0, 2, 4, 6, . .$ $x in Q_1 and = kpi + arctan (5/12), k = 0, 2, 4, 6, ..$ or

$in Q_3 and = kpi + arctan (5/12), k = 1. 3, 5, ...$ .

Easily,

$sin x = 5/(+-sqrt( 12^2 + 5^2 )) = +-5/13$ and

$cos x = +-sqrt ( 1 - sin ^2x)= +- 12/13$

If $x in$ (all positive ) $Q_1$ , choose positive values.

Otherwise, both are negative.

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

it is either $arctan (3/12) =22.62^o = 0.3948 rad$ , nearly, or

$180^o + 22.62^o = 202.62^o = 3.5364 rad$ , nearly.

See the combined graph of $y = tan x, y = sin x and y = cos x$ ,

depicting all these aspects. Slide the graph $larr and rarr$ , to see

x-solutions, as the meet of $y = 5/12$ 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.

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