Let A, B, where A,B (-( )/(2),( )/(2)), be the roots of the quadratic equation x^2-2x-5=0.

Question

Let $	an A,	an B$, where $A,B\in (-\frac{\pi }{2},\frac{\pi }{2})$, be the roots of the quadratic equation ${x}^{2}-2x-5=0$. Then $20{\sin }^{2}(\frac{A+B}{2})$ is equal to:

Accepted Answer

Let the roots of the equation

${x}^{2}-2x-5=0$

be $	an A$ and $	an B$.

So, by using relations between roots and coefficients,

$	an A+	an B=2$

and

$	an A	an B=-5.$

We have to find

$20{\sin }^{2}(\frac{A+B}{2}).$

First, find $	an (A+B)$ using the formula

$	an (A+B)=\frac{	an A+	an B}{1-	an A	an B}.$

Substituting the values:

$	an (A+B)=\frac{2}{1-(-5)}=\frac{2}{6}=\frac{1}{3}.$

Now let

$	heta =A+B.$

Then

$	an 	heta =\frac{1}{3}.$

We need ${\sin }^{2}\frac{	heta }{2}$.

Using the identity

${\sin }^{2}\frac{	heta }{2}=\frac{1-\cos 	heta }{2},$

we first find $\cos 	heta $.

Since

$	an 	heta =\frac{1}{3},$

take a right triangle with opposite side $1$, adjacent side $3$. Then hypotenuse is

$\sqrt{{1}^{2}+{3}^{2}}=\sqrt{10}.$

So,

$\cos 	heta =\frac{3}{\sqrt{10}}.$

Hence,

${\sin }^{2}\frac{	heta }{2}=\frac{1-\frac{3}{\sqrt{10}}}{2}.$

Therefore,

$20{\sin }^{2}\frac{	heta }{2}=20\cdot \frac{1-\frac{3}{\sqrt{10}}}{2}=10(1-\frac{3}{\sqrt{10}})=10-\frac{30}{\sqrt{10}}.$

Now simplify:

$\frac{30}{\sqrt{10}}=3\sqrt{10}.$

So,

$20{\sin }^{2}(\frac{A+B}{2})=10-3\sqrt{10}.$

Therefore, the correct answer is

$10-3\sqrt{10}$

So, Option C is correct.

Solutions · 1