Question

Find the equation of whose roots are sin18 and cos36

Answer 1

Please see the attachment

Answer 2

The quadratic equation is $4x^{2}-2\sqrt{5}x+4=0.$

Step-by-step explanation:

We know that,

$\sin 18=\dfrac{\sqrt{5}-1}{4}$ and

$\cos 36=\dfrac{\sqrt{5}+1}{4}$

To find, the equation of whose roots are $\sin 18$ and $\cos 36$=?

We know that,

The quadratic equation has

$x^{2}-(\alpha+\beta)x+\alpha\beta=0$

∴ $\sin 18+\cos 36=\dfrac{\sqrt{5}-1}{4}+\dfrac{\sqrt{5}+1}{4}$

$=\dfrac{\sqrt{5}-1+\sqrt{5}+1}{4}=\dfrac{2\sqrt{5}}{4}$

and,

$\sin 18.\cos 36=\dfrac{\sqrt{5}-1}{4}.\dfrac{\sqrt{5}+1}{4}=\dfrac{\sqrt{5} ^{2}-1^{2}}{4}$

$=\dfrac{5-1}{4}=1$

The quadratic equation is:

$x^{2}-(\dfrac{2\sqrt{5}}{4})x+1=0$

$4x^{2}-2\sqrt{5}x+4=0$

Hence, the quadratic equation is $4x^{2}-2\sqrt{5}x+4=0.$