Trigonometric functionsHow to determine a trigonometric function if i know tanx=m/n, is possible to find sinx ?

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justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

If you know that tan x = (m/n), you can find sin x.

Go about it this way: tan x = (m/n)

=> sin x/ cos x = m/n

now change cos x to sin x, use (cos x)^2 = 1 - (sin x)^2

=> sin x/ sqrt( 1- (sin x)^2) = m/n

=> (sin x)^2 / (1 - (sin x)^2) = m^2/n^2

=> (1 - (sin x)^2) / (sin x)^2 = n^2/m^2

=> 1/ (sin x)^2 - 1 = n^2/m^2

=> 1/ (sin x)^2 = n^2/m^2 +1 = (n^2 +m^2)/m^2

=> (sin x)^2 = m^2/ (n^2 + m^2)

=> sin x = sqrt [ m^2 / (n^2 + m^2)]

So sin x = m / sqrt (n^2 + m^2)

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

Yes, it is possible. Do not be impressed by the parametric values of the numerator and denominator of tangent function.

We know that the tangent, in a right angle triangle, is the ratio between the opposite cathetus and the joined cathetus.

In this case, is given the tan x=m/n, so we'll conclude that one cathetus is m and the other one is n.

Also, in a right angle triangle,

sin x = opposite cathetus/hypotenuse=m/hypotenuse

But in a right angle triangle, by applying Pythagorean theorem:

(hypotenuse)^2=(cathetus)^2 + (cathetus)^2

(hypotenuse)^2 = m^2 + n^2

We'll substitute the formula above

sin x =p/hypotenuse

sin x = p/sqrt (m^2 + n^2)

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