To find an algebraic expression for [tex]\(\sin(\tan^{-1}(x))\)[/tex] that involves [tex]\(x\)[/tex] only, let's start by understanding the given expression step by step.
### Step 1: Define [tex]\(\theta\)[/tex]
Let [tex]\(\theta = \tan^{-1}(x)\)[/tex]. This implies that:
[tex]\[ \tan(\theta) = x \][/tex]
### Step 2: Trigonometric Identity
We want to express [tex]\(\sin(\theta)\)[/tex] in terms of [tex]\(x\)[/tex]. We can use the following trigonometric identity:
[tex]\[ \sin^2(\theta) + \cos^2(\theta) = 1 \][/tex]
From [tex]\(\tan(\theta) = x\)[/tex], we have:
[tex]\[ x = \frac{\sin(\theta)}{\cos(\theta)} \][/tex]
### Step 3: Solve for [tex]\(\cos(\theta)\)[/tex]
Express [tex]\(\cos(\theta)\)[/tex] in terms of [tex]\(\sin(\theta)\)[/tex]:
[tex]\[ \cos(\theta) = \frac{\sin(\theta)}{x} \][/tex]
Now, substitute this expression into the Pythagorean identity:
[tex]\[ \sin^2(\theta) + \left(\frac{\sin(\theta)}{x}\right)^2 = 1 \][/tex]
Simplify the equation:
[tex]\[ \sin^2(\theta) + \frac{\sin^2(\theta)}{x^2} = 1 \][/tex]
Factor out [tex]\(\sin^2(\theta)\)[/tex]:
[tex]\[ \sin^2(\theta) \left(1 + \frac{1}{x^2}\right) = 1 \][/tex]
Combine the terms inside the parentheses:
[tex]\[ \sin^2(\theta) \left(\frac{x^2 + 1}{x^2}\right) = 1 \][/tex]
Multiply through by [tex]\(x^2\)[/tex]:
[tex]\[ x^2 \sin^2(\theta) + \sin^2(\theta) = x^2 \][/tex]
[tex]\[ (x^2 + 1) \sin^2(\theta) = x^2 \][/tex]
Divide both sides by [tex]\(x^2 + 1\)[/tex]:
[tex]\[ \sin^2(\theta) = \frac{x^2}{x^2 + 1} \][/tex]
### Step 4: Take the Square Root
Since [tex]\(\sin(\theta)\)[/tex] is non-negative when [tex]\(\theta = \tan^{-1}(x)\)[/tex]:
[tex]\[ \sin(\theta) = \sqrt{\frac{x^2}{x^2 + 1}} \][/tex]
This simplifies to:
[tex]\[ \sin(\theta) = \frac{x}{\sqrt{x^2 + 1}} \][/tex]
Hence, the algebraic expression for [tex]\(\sin(\tan^{-1}(x))\)[/tex] is:
[tex]\[ \boxed{\frac{x}{\sqrt{x^2 + 1}}} \][/tex]
