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EEJ MAIN Mathematics QUESTION #6966
Question 1

If $\cos^{-1}x - \cos^{-1}\!\dfrac{y}{2}=\alpha$ where $-1\leq x\leq1,\ -2\leq y\leq2,\ x\leq\dfrac{y}{2}$, then $4x^2-4xy\cos\alpha+y^2$ equals:

  • $4\sin^2\alpha$✔️
  • $2\sin^2\alpha$
  • $4\sin^2\alpha-2x^2y^2$
  • $4\cos^2\alpha+2x^2y^2$
Correct Answer Explanation

Let $\cos^{-1}x=A$ and $\cos^{-1}(y/2)=B$, so $A-B=\alpha$.

Then $\cos A=x,\ \cos B=y/2,\ \sin A=\sqrt{1-x^2},\ \sin B=\sqrt{1-y^2/4}$.

Using $\cos\alpha=\cos(A-B)=\cos A\cos B+\sin A\sin B=\dfrac{xy}{2}+\sqrt{1-x^2}\sqrt{1-\frac{y^2}{4}}$

$4x^2-4xy\cos\alpha+y^2 = (2x-y\cos\alpha)^2+y^2(1-\cos^2\alpha)$... expanding directly: $= 4x^2+y^2-4xy\cos\alpha$.

After substitution and simplification using the expression for $\cos\alpha$: $= 4(1-\cos^2\alpha) = \mathbf{4\sin^2\alpha}$.