Question

If $\sin \left(2^n\theta \right)and\cos \left(2^n\theta \right)$ are the roots of the equation $a{x}^2+bx+c=0,$ then

• A. $\frac{a+b}{2}=\frac{ac}{b-a}$
  
• B. $a^2+b^2=2ac$
• C. $b^2-a^2=ac$
• D. $\frac{a-b}{2}=\frac{a+b}{ac}$
  

Answer 1

The correct option is A $\frac{a+b}{2}=\frac{ac}{b-a}$


Given that $\sin \left(2^n\theta \right)and\cos \left(2^n\theta \right)$ are the roots of the equation $a{x}^2+bx+c=0.$

Then, sum of the roots $=\frac{-b}{a}$ and, the product of the roots $=\frac{c}{a}$

$\therefore \sin \left(2^n\theta \right)+\cos \left(2^n\theta \right)=\frac{-b}{a}$ ...(i)

and $\sin \left(2^n\theta \right)\times \cos \left(2^n\theta \right)=\frac{c}{a}$

$\Rightarrow 2\sin \left(2^n\theta \right)\times \cos \left(2^n\theta \right)=\frac{2c}{a}$ ...(ii)

Squaring (i), on both sides we get,

${\sin }^2\left(2^n\theta \right)+{\cos }^2\left(2^n\theta \right)+2\sin \left(2^n\theta \right)\cos \left(2^n\theta \right)={\left(\frac{-b}{a}\right)}^2$

$\Rightarrow 1+\frac{2c}{a}=\frac{b^2}{a^2}$ $[\because {\sin }^2\theta +{\cos }^2\theta =1\text{and from (ii)}]$

$\Rightarrow 1-\frac{b^2}{a^2}=\frac{-2c}{a}$

$\Rightarrow \frac{a^2-b^2}{a^2}=\frac{-2c}{a}$

$\Rightarrow \frac{b^2-a^2}{a^2}=\frac{2c}{a}$

$\Rightarrow \frac{(b-a)(b+a)}{2}=ac$

$\Rightarrow \frac{(a+b)}{2}=\frac{ac}{(b-a)}$

Hence, the correct answer is option (a).