Trigonometry & Inverse Trigonometry
Inverse Trigonometric Functions
Grade 12
Question:
<p>Let \(f(x) = 1 + 2\sin\left(\frac{\pi x}{e^x+1}\right)\), \(x > 0\), then \(f^{-1}(x)\) is equal to (assuming \(f\) is bijective)</p>
<p>(a) \(\log\left(\frac{e^{\sin^{-1}\frac{x-1}{2}}}{1-\sin\frac{x-1}{2}}\right)\)</p>
<p>(b) \(\log\left(\frac{e^{\sin^{-1}\frac{x-1}{2}}}{1-\sin^{-1}\frac{x-1}{2}}\right)\)</p>
<p>(c) \(e^{\frac{\sin^{-1}\frac{x-1}{2}}{1-\sin^{-1}\frac{x-1}{2}}}\)</p>
<p>(d) \(e^{\frac{\cos^{-1}\frac{x-1}{2}}{1-\sin^{-1}\frac{x-1}{2}}}\)</p>
Step-by-Step Solution
Key Concept: Finding the inverse function requires isolating the independent variable and using algebraic and trigonometric manipulation.
<p>To find the inverse function, let $y = f(x)$ and solve for $x$ in terms of $y$. From $y = 1 + 2\sin\left(\frac{\pi x}{e^x+1}\right)$, we get $\sin\left(\frac{\pi x}{e^x+1}\right) = \frac{y-1}{2}$. Taking inverse sine: $\frac{\pi x}{e^x+1} = \sin^{-1}\frac{y-1}{2}$. Rearranging and using logarithms yields the inverse function.</p>
Correct Answer: a