What is sin(inverse tangent(12/5)?

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What is sin(inverse tangent(12/5)?

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Answer 1 · 2015-09-30T10:57:07

$sin (arctan (\frac{12}{5})) = \pm \frac{12}{13}$ $sin(arctan(12/5))=+-12/13$

Explanation:

$arctan (\frac{12}{5})$ $arctan(12/5)$ implies one of the reference triangles pictured below
(with the hypotenuse caculated using the Pythagorean Theorem).

enter image source here

Since $s = \frac{opposite}{hypotenuse}$ $s=("opposite")/("hypotenuse")$

$sin (arctan (\frac{12}{5})) = \frac{12}{13} or - \frac{12}{13}$ $sin(arctan(12/5)) = color(red)(12/13) " or " color(blue)(-12/13)$

Answer 2 · 2015-09-30T10:59:22

$sin (arctan (\frac{12}{5})) = \frac{12}{13}$ $sin(arctan(12/5)) = 12/13$

Explanation:

From the trig identity ${sin}^{2} (θ) + {cos}^{2} (θ) = 1$ $sin^2(theta) + cos^2(theta) = 1$ , we divide both sides by ${sin}^{2} (θ)$ $sin^2(theta)$

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

Since $\frac{sin (θ)}{cos (θ)} = tan (θ)$ $sin(theta)/cos(theta) = tan(theta)$ we can rewrite the second term

$1 + \frac{1}{{tan}^{2} (θ)} = \frac{1}{{sin}^{2} (θ)}$ $1 + 1/tan^2(theta) = 1/sin^2(theta)$

Taking the least common multiple,

$\frac{{tan}^{2} (θ) + 1}{{tan}^{2} (θ)} = \frac{1}{{sin}^{2} (θ)}$ $(tan^2(theta) + 1)/(tan^2(theta)) = 1/sin^2(theta)$

Inverting both sides

$\frac{{tan}^{2} (θ)}{{tan}^{2} (θ) + 1} = {sin}^{2} (θ)$ $tan^2(theta)/(tan^2(theta) + 1) = sin^2(theta)$

Subsituting $θ = arctan (\frac{12}{5})$ $theta = arctan(12/5)$

$\frac{{(\frac{12}{5})}^{2}}{{(\frac{12}{5})}^{2} + 1} = {sin}^{2} (arctan (\frac{12}{5}))$ $(12/5)^2/((12/5)^2+1) = sin^2(arctan(12/5))$

${sin}^{2} (arctan (\frac{12}{5})) = \frac{144}{25} \cdot \frac{25}{169} = \frac{144}{169}$ $sin^2(arctan(12/5)) = 144/25 * 25/169 = 144/169$

Taking the root

$sin (arctan (\frac{12}{5})) = \pm \frac{12}{13}$ $sin(arctan(12/5)) = +-12/13$

To pick the sign we look at the range of the arctangent, it only takes arguments on the first and fourth quadrants, during which the cosine is always positive. If, for $\frac{12}{5}$ $12/5$ the cosine is positive and the tangent is positive, then the sine must be positive too.

$sin (arctan (\frac{12}{5})) = \frac{12}{13}$ $sin(arctan(12/5)) = 12/13$

Also, protip, you can use either the function with a "^-1" or put an arc before it to notate the inverse trig functions, but usually there's a lot less headache for everybody involved if you use the arc notation. (There's no grounds for mistaking it for other functions).

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What is sin(inverse tangent(12/5)?

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