Question

The quadratic equation whose roots are \[ \cos72^\circ \quad \text{and} \quad \sin54^\circ \] is

• A. $4x^2+\sqrt{5}x-1=0$
• B. $4x^2+2\sqrt{5}x+1=0$
• C. $4x^2-2\sqrt{5}x+1=0$
• D. $x^2-2\sqrt{5}x+4=0$

Hint:

Whenever trigonometric values are roots of an equation, first convert them into exact algebraic forms before forming the polynomial.

Answer 1

The Correct Option is C

Concept: For roots $\alpha,\beta$, the quadratic equation is: \[ x^2-(\alpha+\beta)x+\alpha\beta=0 \] Useful identities: \[ \sin54^\circ=\cos36^\circ \] \[ \cos72^\circ=\frac{\sqrt5-1}{4} \] \[ \cos36^\circ=\frac{\sqrt5+1}{4} \]

Step 1: Find the roots explicitly.
\[ \alpha=\cos72^\circ=\frac{\sqrt5-1}{4} \] \[ \beta=\sin54^\circ=\cos36^\circ=\frac{\sqrt5+1}{4} \]

Step 2: Find sum and product.
Sum: \[ \alpha+\beta = \frac{\sqrt5-1+\sqrt5+1}{4} = \frac{2\sqrt5}{4} = \frac{\sqrt5}{2} \] Product: \[ \alpha\beta = \frac{(\sqrt5-1)(\sqrt5+1)}{16} \] \[ = \frac{5-1}{16} = \frac14 \]

Step 3: Form the quadratic equation.
\[ x^2-\frac{\sqrt5}{2}x+\frac14=0 \] Multiplying by $4$: \[ 4x^2-2\sqrt5x+1=0 \] Hence, \[ \boxed{ 4x^2-2\sqrt5x+1=0 } \]