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

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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

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This can be solved easily once we visualise it.

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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,

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

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

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I hope this helps :)

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Use the Pythagorean identity:

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

Divide through by

#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.

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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#

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