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# If $\cos \dfrac{{12}}{{13}}$, $\sin < 0$, how do you find tan in simplest form?

Last updated date: 21st Jun 2024
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Hint: Sin is comparable to the side inverse a given point in a correct triangle to the hypotenuse. Cos is identical to the proportion of the side nearby an intense point in a right-calculated triangle to the hypotenuse.

Use the definition of cosine to find the known sides of the unit circle right triangle. The quadrant determines the sign on each of the values.
$\cos (x) = \dfrac{{adjacent}}{{hypotenuse}}$
Let's find the opposite side of the unit circle triangle. Since the adjacent side and hypotenuse are known, use the Pythagorean theorem to find the remaining side.
$opposite = \sqrt {hypotenuse{e^2} - adjacent{t^2}}$
Replace the known value of the in the equation.
$opposite = \sqrt {{{13}^2} - {{12}^2}}$
Simplify $\sqrt {{{13}^2} - {{12}^2}}$
Raise 13 to the power of 2.
$Opposite = \sqrt {169 - {{(12)}^2}}$
Raise 12 to the power of 2.
$Opposite = \sqrt {169 - 1.144}$
Multiply $- 1$ by 144
$Opposite = \sqrt {169 - 144}$
Subtract 144 from 169.
$Opposite = \sqrt {25}$
Rewrite 25 as ${5^2}$.
$Opposite = \sqrt {{5^2}}$
Pull terms out from under the radical, assuming positive real numbers.
$Opposite = 5$
Use the definition of $\sin$to find the value of $\sin (x)$.
$\sin (x) = \dfrac{{opp}}{{hyp}}$
Substitute in the known values.
$\sin (a) = \dfrac{5}{{13}}$.
$cos(a) = \dfrac{{12}}{{13}}$. Angle is in either 1st quadrant or in the 4th.
$\sin (a) = \pm \dfrac{5}{{13}}$. As, $\sin (a) < 0$, a is in the 4th quadrant. So, $\sin (a) = - \dfrac{5}{{13}}$.
Use the definition of tangent to find the value of $\tan (x)$
$\tan (x) = \dfrac{{opp}}{{adj}}$
Substitute in the known values.
$\tan (a) = - \dfrac{5}{{12}}$.

Thus, the ratio of $\dfrac{{\sin (a)}}{{\cos (a)}} = \tan (a) = - \dfrac{5}{{12}}$.

Note: We note that the domain of sin inverse function is $( - 1,1)$ and since $\dfrac{{12}}{{13}} \in ( - 1,1)$ the value $\sin = - \dfrac{5}{{13}}$ is well defined.