Question

If \[ \alpha=\tan\left(2\sin^{-1}\left(\frac23\right)\right) \quad\text{and}\quad \beta=\sin\left(2\tan^{-1}\left(\frac13\right)\right), \] then the maximum value of \[ \alpha\sin\theta+\beta\cos\theta \] is

• A. \(1\)
• B. \(\dfrac{\sqrt{2009}}{5}\)
• C. \(\dfrac{\sqrt{2024}}{3}\)
• D. \(4\sqrt5+\dfrac35\)

Hint:

The maximum value of \[ a\sin\theta+b\cos\theta \] is always \[ \boxed{\sqrt{a^2+b^2}}. \] First evaluate the constants \(a\) and \(b\), then apply this standard result.

Answer 1

The Correct Option is B

Step 1: Find the value of \(\alpha\). Let \[ A=\sin^{-1}\left(\frac23\right). \] Then \[ \sin A=\frac23, \qquad \cos A=\frac{\sqrt5}{3}. \] Using \[ \tan2A=\frac{2\tan A}{1-\tan^2A}, \] where \[ \tan A=\frac{2}{\sqrt5}, \] we get \[ \alpha = \tan2A = 4\sqrt5. \]

Step 2: Find the value of \(\beta\). Let \[ B=\tan^{-1}\left(\frac13\right). \] Using \[ \sin2B = \frac{2\tan B}{1+\tan^2B}, \] we obtain \[ \beta = \frac{2\left(\frac13\right)} {1+\frac19} = \frac35. \]

Step 3: Find the maximum value. The maximum value of \[ \alpha\sin\theta+\beta\cos\theta \] is \[ \sqrt{\alpha^2+\beta^2}. \] Hence, \[ \sqrt{(4\sqrt5)^2+\left(\frac35\right)^2} = \sqrt{80+\frac9{25}} = \sqrt{\frac{2009}{25}} = \frac{\sqrt{2009}}5. \] Therefore, \[ \boxed{\frac{\sqrt{2009}}5}. \] Thus, \[ \boxed{(B)} \] is the correct answer.