%%

$$\alpha = \sin 36^\circ $$ is a root of which of the following equation?

JEE · Math · previous-year question

  1. A.$$16{x^4} - 10{x^2} - 5 = 0$$
  2. B.$$16{x^4} + 20{x^2} - 5 = 0$$
  3. C.$$16{x^4} - 20{x^2} + 5 = 0$$correct
  4. D.$$4{x^4} - 10{x^2} + 5 = 0$$

Answer

C. $$16{x^4} - 20{x^2} + 5 = 0$$

Explanation

Given that $\alpha = \sin 36^\circ$, we need to determine which equation it is a root of. We start with the known relationship for $\cos 72^\circ$: $ \cos 72^\circ = \frac{\sqrt{5}-1}{4} $ Using the double-angle formula for cosine: $ \cos 72^\circ = 1 - 2 \sin^2 36^\circ $ Substitute $\alpha$ for $\sin 36^\circ$: $ 1 - 2\alpha^2 = \frac{\sqrt{5}-1}{4} $ Multiply both sides by 4: $ 4 - 8\alpha^2 = \sqrt{5} - 1 $ Add 1 to both sides: $ 5 - 8\alpha^2 = \sqrt{5} $ Square both sides to eliminate the radical: $ (5 - 8\alpha^2)^2 = 5 $ Expand the left side: $ 25 + 64\alpha^4 - 80\alpha^2 = 5 $ Simplify by subtracting 5 from both sides: $ 64\alpha^4 - 80\alpha^2 + 20 = 0 $ Divide the entire equation by 4: $ 16\alpha^4 - 20\alpha^2 + 5 = 0 $ Thus, the equation $16\alpha^4 - 20\alpha^2 + 5 = 0$ is the one for which $\alpha = \sin 36^\circ$ is a root.

Practice more JEE questions

Answer thousands more real JEE previous-year questions free, get graded instantly, and climb the global ranked leaderboard.

Practice JEE free →

More JEE Math questions