tinfoQ

2021-08-22

Please, rewrite the expression as an algebraic expression in x.
$\mathrm{sin}\left({\mathrm{tan}}^{-}1x\right)$

sweererlirumeX

Let $\theta ={\mathrm{tan}}^{-1}\left(x\right)$, then $\mathrm{tan}\theta =x$
Draw a right triangle with $\theta$ as one of it’s acute angles:
1) $AC=\sqrt{A{B}^{2}+D{C}^{2}}$
$AC=\sqrt{{x}^{2}+{1}^{2}}=\sqrt{1+{x}^{2}}$
2) $\mathrm{sin}\left({\mathrm{tan}}^{-1}\left(x\right)\right)=\frac{x}{\sqrt{1+{x}^{2}}}$

karton

Step 1: Recall the identity ${\mathrm{tan}}^{-1}x=\mathrm{arctan}\left(x\right)$.
Step 2: Use the definition of the tangent function: $\mathrm{tan}\left(\theta \right)=\frac{\mathrm{sin}\left(\theta \right)}{\mathrm{cos}\left(\theta \right)}$. In this case, $\theta ={\mathrm{tan}}^{-1}x$. So we can rewrite ${\mathrm{tan}}^{-1}x$ as $\mathrm{arctan}\left(x\right)=\frac{\mathrm{sin}\left(\mathrm{arctan}\left(x\right)\right)}{\mathrm{cos}\left(\mathrm{arctan}\left(x\right)\right)}$.
Step 3: Apply the identity $\mathrm{sin}\left(\mathrm{arctan}\left(x\right)\right)=\frac{x}{\sqrt{1+{x}^{2}}}$. This can be derived using the right triangle definition of the trigonometric functions.
Step 4: Apply the identity $\mathrm{cos}\left(\mathrm{arctan}\left(x\right)\right)=\frac{1}{\sqrt{1+{x}^{2}}}$. Again, this can be derived using the right triangle definition of the trigonometric functions.
Step 5: Substitute the values of $\mathrm{sin}\left(\mathrm{arctan}\left(x\right)\right)$ and $\mathrm{cos}\left(\mathrm{arctan}\left(x\right)\right)$ back into the expression $\frac{\mathrm{sin}\left(\mathrm{arctan}\left(x\right)\right)}{\mathrm{cos}\left(\mathrm{arctan}\left(x\right)\right)}$.
Combining all the steps, we get:

Therefore, the expression $\mathrm{sin}\left({\mathrm{tan}}^{-1}x\right)$ can be rewritten as the algebraic expression $\overline{)x}$.

star233

To rewrite the expression $\mathrm{sin}\left({\mathrm{tan}}^{-1}x\right)$ as an algebraic expression in $x$, we can use trigonometric identities. First, we'll start by using the identity ${\mathrm{tan}}^{-1}x=\mathrm{arctan}\left(x\right)$:
$\mathrm{sin}\left({\mathrm{tan}}^{-1}x\right)=\mathrm{sin}\left(\mathrm{arctan}\left(x\right)\right)$
Next, we can use the identity $\mathrm{sin}\left(\mathrm{arctan}\left(x\right)\right)=\frac{x}{\sqrt{1+{x}^{2}}}$:
$\mathrm{sin}\left({\mathrm{tan}}^{-1}x\right)=\mathrm{sin}\left(\mathrm{arctan}\left(x\right)\right)=\frac{x}{\sqrt{1+{x}^{2}}}$
Therefore, the algebraic expression in $x$ equivalent to $\mathrm{sin}\left({\mathrm{tan}}^{-1}x\right)$ is $\frac{x}{\sqrt{1+{x}^{2}}}$.

user_27qwe

$\frac{x}{\sqrt{1+{x}^{2}}}$
Explanation:
${\mathrm{tan}}^{-1}x=\mathrm{arctan}x$
Using this identity, we have:
$\mathrm{sin}\left({\mathrm{tan}}^{-1}x\right)=\mathrm{sin}\left(\mathrm{arctan}x\right)$
Now, let's use another trigonometric identity to express $\mathrm{sin}\left(\mathrm{arctan}x\right)$ in terms of $x$. The identity is:
$\mathrm{sin}\left(\mathrm{arctan}x\right)=\frac{x}{\sqrt{1+{x}^{2}}}$
Therefore, the algebraic expression in $x$ equivalent to $\mathrm{sin}\left({\mathrm{tan}}^{-1}x\right)$ is:
$\overline{)\frac{x}{\sqrt{1+{x}^{2}}}}$

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