# If tan x = y what is sin x in terms of y.

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

It is given that `tan x = y` . sin x has to be expressed in terms of y.

`tan x = y`

=> `tan^2x = y^2`

=> `(sin^2x)/(cos^2x) = y^2`

=> `(sin^2x)/(1 - sin^2x) = y^2`

=> `1/(1/(sin^2x) - 1) = y^2`

=> `1/(sin^2x) - 1 = 1/y^2`

=> `1/(sin^2x) = 1 + 1/y^2`

=> `1/sin x = +- sqrt(1 + 1/y^2)`

=> `1/sin x = +- sqrt((y^2 + 1)/y^2)`

=> `sin x = +-y/sqrt(1 +y^2)`

In terms of y, `sin x = +-y/(sqrt(y^2 + 1))`

thiny12 | Student, Grade 11 | (Level 1) eNoter

Posted on

think of a right angled triangle when you are given tan x= y you know that tan x = perpendicular divided by the base. so y is the perpendicular and 1 is the base. now using Pythagoras' Theorem find the hypotenuse that will be √(y^2 + 1) and thus sin x = perpendicular over hypotenuse sin x = y / √(y^2 + 1)

najm1947 | Elementary School Teacher | (Level 1) Valedictorian

Posted on

tanx = y

But tanx is also equal to height/base if we make a right triangle with angle x at base and height y.

In this case base=1 to maake tanx = y and

hypotenuse is equal to +sqrt(y^2+1) and -sqrt(y^2+1)

sinx = hieght/hypotenuse = y/sqrt(y^2+1) and -y/sqrt(y^2+1)

Therefore in terms of y:

sinx = y/sqrt(y^2+1) and -y/sqrt(y^2+1)