Question

find the exact value of the expression. (hint: sketch a right triangle. enter your answer in radians.) tanarcsin(-18/19)

Explanation:

Step1: Let $\theta=\arcsin(-\frac{18}{19})$.

By definition of the inverse - sine function, $\sin\theta =-\frac{18}{19}$, and we consider a right - triangle. Let the opposite side to the angle $\theta$ be $y=- 18$ and the hypotenuse $r = 19$ (in a right - triangle in the coordinate plane, for the angle $\theta$ in standard position).

Step2: Use the Pythagorean theorem to find the adjacent side $x$.

We know that $x^{2}+y^{2}=r^{2}$. Substituting $y=-18$ and $r = 19$, we get $x^{2}+(-18)^{2}=19^{2}$. Then $x^{2}=19^{2}-18^{2}=(19 + 18)(19 - 18)=37$. So $x=\pm\sqrt{37}$. Since $\sin\theta=-\frac{18}{19}<0$, $\theta$ is in either the third or fourth quadrant. In the context of the right - triangle for the inverse - sine function, when $\sin\theta<0$, the adjacent side $x=\sqrt{37}$ (in the fourth quadrant if considering the principal value range of $\arcsin$).

Step3: Calculate $\tan\theta$.

We know that $\tan\theta=\frac{y}{x}$. Substituting $y=-18$ and $x = \sqrt{37}$, we get $\tan\theta=-\frac{18}{\sqrt{37}}=-\frac{18\sqrt{37}}{37}$.

Answer:

$-\frac{18\sqrt{37}}{37}$