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

##### 8 Answers

Viewed as a right angled triangle

can be thought of as the ratio of opposite to adjacent sides in a triangle with sides

So

and

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,

..............................................................................................................

Hence , **sinx = perpendicular/hypotenuse=p/h** = **5/13**

**cosx = base/hypotenuse=b/h** = **12/13** .... **(Answer)

.............................................................................................................

I hope this helps :)

#### Explanation:

The best way is to visualize the

#tan(x)=5/12" "=>" "x=arctan(5/12)#

So, we see that:

#sin(x)=sin(arctan(5/12))#

We can relate

#sin^2(x)+cos^2(x)=1#

Dividing through by

#1+cot^2(x)=csc^2(x)#

Rewriting:

#1+1/tan^2(x)=1/sin^2(x)#

Common denominator:

#(tan^2(x)+1)/tan^2(x)=1/sin^2(x)#

Inverting:

#sin^2(x)=tan^2(x)/(tan^2(x)+1)" "=>" "sin(x)=tan(x)/sqrt(tan^2(x)+1)#

So, plugging this into

#sin(arctan(5/12))=tan(arctan(5/12))/sqrt(tan^2(arctan(5/12))+1)#

Since

#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

#sin^2(x)+cos^2(x)=1" "=>" "(5/13)^2+cos^2(x)=1#

So:

#cos^2(x)=1-25/169" "=>" "cos(x)=sqrt(144/169)=12/13#

#### Explanation:

Use the Pythagorean identity:

#sin^2(x)+cos^2(x)=1#

Divide through by

#sin^2(x)/cos^2(x)+cos^2(x)/cos^2(x)=1/cos^2(x)#

#tan^2(x)+1=sec^2(x)#

Since

#25/144+1=sec^2(x)=169/144#

Thus:

#sec(x)=13/12#

So:

#cos(x)=12/13#

Using the Pythagorean identity again with

#sin^2(x)+144/169=1#

#sin^2(x)=25/169#

#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.

One other method:

#tan(x)=5/12#

By the definition of tangent,

#sin(x)/cos(x)=5/12#

Then we can say:

#sin(x)=5/12cos(x)" "" "" "" "(star)#

We can then substitute

#sin^2(x)+cos^2(x)=1#

#(5/12cos(x))^2+cos^2(x)=1#

#25/144cos^2(x)+cos^2(x)=1#

#169/144cos^2(x)=1#

#cos^2(x)=144/169#

#cos(x)=sqrt(144/169)=12/13#

Then, use

#### Explanation:

x could be either in Quadrant 1 or Quadrant 3.

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.

#### 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

Plugging our

From this, we know the opposite side is

From SOH-CAH-TOA, we know sine is equal to the opposite side over the hypotenuse. This means that

We also know that cosine is equal to the adjacent side over the hypotenuse. Thus,

Recall that

Hope this helps!

Disambiguation answer.

#### Explanation:

Here,

Easily,

If

Otherwise, both are negative.

If x is chosen in

it is either

See the combined graph of

depicting all these aspects. Slide the graph

x-solutions, as the meet of

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.