How do you calculate $sin ({tan}^{- 1} (4 x))$ $sin(tan^-1(4x))$ ?

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How do you calculate $sin ({tan}^{- 1} (4 x))$ $sin(tan^-1(4x))$ ?

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Answer 1 · 2015-09-22T15:07:51

$\frac{4 x}{\sqrt{16 x^{2} + 1}} = sin (arctan (4 x))$ $(4x)/sqrt(16x^2+1) = sin(arctan(4x))$

Explanation:

Assuming that by ${tan}^{- 1} (4 x)$ $tan^(-1)(4x)$ you mean $arctan (4 x)$ $arctan(4x)$ and not $cot (4 x)$ $cot(4x)$ .

We know the pythagorean identity ${sin}^{2} (θ) + {cos}^{2} (θ) = 1$ $sin^2(theta) + cos^2(theta) = 1$ , so if we divide both sides by #sin^2(theta) we'd have:

$\frac{{sin}^{2} (θ)}{{sin}^{2} (θ)} + \frac{{cos}^{2} (θ)}{{sin}^{2} (θ)} = \frac{1}{{sin}^{2} (θ)} \to$ $sin^2(theta)/sin^2(theta) + cos^2(theta)/sin^2(theta) = 1/sin^2(theta) rarr$
$1 + {cot}^{2} (θ) = {csc}^{2} (θ)$ $1 + cot^2(theta) = csc^2(theta)$

So, if we switch in $θ$ $theta$ for $arctan (4 x)$ $arctan(4x)$ , we have
$1 + \frac{1}{{tan}^{2} (arctan (4 x))} = \frac{1}{{sin}^{2} (arctan (4 x))}$ $1 + 1/tan^2(arctan(4x)) = 1/sin^2(arctan(4x))$

Knowing that $tan (arctan (θ)) = θ$ $tan(arctan(theta)) = theta$

$1 + \frac{1}{{(4 x)}^{2}} = \frac{1}{{sin}^{2} (arctan (4 x))}$ $1 + 1/(4x)^2 = 1/sin^2(arctan(4x))$

Taking the least common multiple so we can sum the two fractions on the right side:

$\frac{16 x^{2} + 1}{16 x^{2}} = \frac{1}{{sin}^{2} (arctan (4 x))}$ $(16x^2+ 1)/(16x^2) = 1/sin^2(arctan(4x))$

Invert both sides
$\frac{16 x^{2}}{16 x^{2} + 1} = {sin}^{2} (arctan (4 x))$ $(16x^2)/(16x^2+1) = sin^2(arctan(4x))$

Taking the square root

$\sqrt{\frac{16 x^{2}}{16 x^{2} + 1}} = sin (arctan (4 x))$ $sqrt((16x^2)/(16x^2+1)) = sin(arctan(4x))$

Simplifying

$\frac{4 x}{\sqrt{16 x^{2} + 1}} = sin (arctan (4 x))$ $(4x)/sqrt(16x^2+1) = sin(arctan(4x))$

You can rationalize it if you want
$\frac{4 x \cdot \sqrt{16 x^{2} + 1}}{16 x^{2} + 1} = sin (arctan (4 x))$ $(4x*sqrt(16x^2+1))/(16x^2+1) = sin(arctan(4x))$

And last but not least, gotta list out the values $x$ $x$ can't be because of the liberities we took during our process. We can't have denominators being zero nor negatives in square roots, so we have:
${(4 x)}^{2} \neq 0$ $(4x)^2 != 0$
$16 x^{2} + 1 > 0$ $16x^2 +1 > 0$
$sin (arctan (4 x)) \neq 0 \to \frac{4 x}{\sqrt{16 x^{2} + 1}} \neq 0$ $sin(arctan(4x)) != 0 rarr (4x)/sqrt(16x^2+1) != 0$

From these we see that $x \neq 0$ $x != 0$

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How do you calculate $sin ({tan}^{- 1} (4 x))$ $sin(tan^-1(4x))$ ?

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