Trigonometry
Trigonometry
Allen Star Batch
Grade 11

Question:

If $x = \sin(2\tan^{-1}2)$ and $y = \sin\left(\frac{1}{2}\tan^{-1}\frac{4}{3}\right)$, then which of the following options is(are) correct ?
$x = y^2$
$x^2 = 1 - x$
$x^2 = \frac{y}{2}$
$x > y$

Step-by-Step Solution

Key Concept: Use double angle and half-angle formulas to convert inverse trigonometric expressions into exact values.
Given $x = \sin(2\tan^{-1}2)$, let $\tan^{-1}2 = \theta$. Then $x = \sin(2\theta) = \frac{2\tan\theta}{1+\tan^2\theta} = \frac{2(2)}{1+4} = \frac{4}{5}$. For the second part, let $\tan^{-1}\frac{4}{3} = \alpha$, so $\tan\alpha = \frac{4}{3}$, giving $\sin\alpha = \frac{4}{5}$ and $\cos\alpha = \frac{3}{5}$. Thus $y = \sin(\frac{1}{2}\tan^{-1}\frac{4}{3}) = \sin(\frac{\alpha}{2}) = \frac{1}{\sqrt{5}}$ using the half-angle formula.
Correct Answer: 2,4

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