If cos θ = (4x² – 1)/(1 + 4x²) then find the value of sin q.

We know that cos A = (Base/Hypotenuse)

So, Base of the right-angled triangle is (4x ²  - 1) units.

Hypotenuse of the right-angled triangle is (1 + 4x ² ) units.

Using Pythagoras theorem, we get,

(Perpendicular)  ²  + (Base)  ²  = (Hypotenuse)  ²

Or, (Perpendicular)  ²  + (4x ²  - 1)  ²  = (1 + 4x ² )  ²

Or, (Perpendicular)  ²  = (1 + 4x ² )  ²  - (4x ²  - 1)  ²

We know that, a ²  - b ²  = (a + b) X (a - b) .

Or, (Perpendicular)  ²  = (1 + 4x ²  + 4x ²  - 1) X (1 + 4x ²  - 4x ²  + 1)

Or, (Perpendicular)  ²  = 8x ²  X 2 = (16x ² )

Since, perpendicular of a triangle cannot be negative.

So, Perpendicular = '4x' units

We know that sine A = (Perpendicular/Hypotenuse)

So, sin θ = 4x/(1 + 4x²)