show that sin A = Sin18˚ is a root of the equation

Question

 

4 sin^2 A + 2 sin A - 1 = 0

Answer 1

4 sin^2 A + 2 sin A - 1 = 0

Using quadratic equation formula

x=(-b+√(b^2-4ac))/2a

Here a=4,b=2 and c=-1

sinA=(-2+√(2^2-4*4*-1))/2*4

         =(-2+√20)/8

         =(-1+√5)/4

sinA=sin18

A=18

Comments

Using quadratic equation formula

x=(-b±√(b^2-4ac))/2a

Here a=4,b=2 and c=-1

sinA=(-2±√(2^2-4*4*-1))/2*4

         =(-2±√20)/8

         =(-1±√5)/4.

The range of sine function is [- 1, 1].

So, sinA = (-1+√5)/4

sinA=sin18⁰

A=18⁰ .

Therefore, sin A  = sin 18⁰ is a root of the given equation.