How do you find the exact value of tan(arcsin(-3/4))?

May 22, 2017

$- \frac{3 \sqrt{7}}{7}$

Explanation:

Let's think about it this way:

We're trying to find the tangent of the angle whose sine is $- \frac{3}{4}$.

Let's call this angle $\theta$. Using the Pythagorean identity, we can solve for $\cos \theta$:

${\sin}^{2} \theta + {\cos}^{2} \theta = 1$

${\left(- \frac{3}{4}\right)}^{2} + {\cos}^{2} \theta = 1$

$\frac{9}{16} + {\cos}^{2} \theta = 1$

${\cos}^{2} \theta = \frac{7}{16}$

$\cos \theta = \pm \frac{\sqrt{7}}{4}$

Since $\sin \theta$ is negative, $\theta$ must be in quadrant 4 since $\arcsin \left(x\right)$ is only defined for quadrants 1 (+) and 4 (-). In quadrant 4, $\cos \theta$ is always positive, so it must be $\frac{\sqrt{7}}{4}$.

We now know $\sin \theta = - \frac{3}{4}$ and $\cos \theta = \frac{\sqrt{7}}{4}$. This means we can solve for $\tan \theta$, which is what we're looking for.

$\tan \theta = \sin \frac{\theta}{\cos} \theta$

$\tan \theta = \frac{- \frac{3}{4}}{\frac{\sqrt{7}}{4}} = - \frac{3}{\sqrt{7}} = - \frac{3 \sqrt{7}}{7}$

Therefore, the exact value of $\tan \left(\arcsin \left(- \frac{3}{4}\right)\right)$ is $- \frac{3 \sqrt{7}}{7}$.

May 22, 2017

$\frac{- 3}{\sqrt{7}}$ or $\frac{- 3 \sqrt{7}}{7}$

Explanation:

Domain of arcsin is the 1st quadrant is positive, 4th quadrant is negative. So, $- \frac{3}{4}$ is in the 4th quadrant.

Arcsin corresponds to the ratio of $\frac{o p p}{h y p}$

$\textcolor{w h i t e}{0}$

$\textcolor{w h i t e}{4} \textcolor{b l a c k}{- - - - - - - - -}$
$\textcolor{w h i t e}{4} \textcolor{b l a c k}{\setminus \setminus} \textcolor{b l a c k}{\theta} \textcolor{w h i t e}{- - - - - - - -} \textcolor{b l a c k}{|}$
color(white)(4)color(white)( - )color(black)(\\)color(white)(theta)color(white)(- - -)color(white)( - - - / - .)color(white)(/)color(black)(|)
color(white)(4)color(white)(- -)color(black)(\\)color(white)(theta)color(white)(- - - - - - /)color(black)(|)
color(white)(4)color(white)(- - -)color(black)(\\)color(white)(theta)color(white)(- - - - -/)color(black)(|)
color(white)(- - -)color(black)(4)color(white)(-)color(black)(\\)color(white)(theta)color(white)(- - - -/)color(black)(|)-3
color(white)(4)color(white)(- - - - -)color(black)(\\)color(white)(theta)color(white)(- - -/)color(black)(|)
color(white)(4)color(white)(- - - - - -)color(black)(\\)color(white)(theta)color(white)(- -/)color(black)(|)
$\textcolor{w h i t e}{4} \textcolor{w h i t e}{- - - - - - -} \textcolor{b l a c k}{\setminus \setminus} \textcolor{w h i t e}{\theta} \textcolor{w h i t e}{-} \textcolor{b l a c k}{|}$
$\textcolor{w h i t e}{4} \textcolor{w h i t e}{- - - - - - - -} \textcolor{b l a c k}{\setminus \setminus} \textcolor{w h i t e}{\theta} \textcolor{b l a c k}{|}$

Let's solve for the remaining side:
${4}^{2} - {3}^{2} = {7}^{2} = \sqrt{7}$

$\textcolor{w h i t e}{000000000} \sqrt{7}$

$\textcolor{w h i t e}{4} \textcolor{b l a c k}{- - - - - - - - -}$
$\textcolor{w h i t e}{4} \textcolor{b l a c k}{\setminus \setminus} \textcolor{b l a c k}{\theta} \textcolor{w h i t e}{- - - - - - - -} \textcolor{b l a c k}{|}$
color(white)(4)color(white)( - )color(black)(\\)color(white)(theta)color(white)(- - -)color(white)( - - - / - .)color(white)(/)color(black)(|)
color(white)(4)color(white)(- -)color(black)(\\)color(white)(theta)color(white)(- - - - - - /)color(black)(|)
color(white)(4)color(white)(- - -)color(black)(\\)color(white)(theta)color(white)(- - - - -/)color(black)(|)
color(white)(- - -)color(black)(4)color(white)(-)color(black)(\\)color(white)(theta)color(white)(- - - -/)color(black)(|)-3
color(white)(4)color(white)(- - - - -)color(black)(\\)color(white)(theta)color(white)(- - -/)color(black)(|)
color(white)(4)color(white)(- - - - - -)color(black)(\\)color(white)(theta)color(white)(- -/)color(black)(|)
$\textcolor{w h i t e}{4} \textcolor{w h i t e}{- - - - - - -} \textcolor{b l a c k}{\setminus \setminus} \textcolor{w h i t e}{\theta} \textcolor{w h i t e}{-} \textcolor{b l a c k}{|}$
$\textcolor{w h i t e}{4} \textcolor{w h i t e}{- - - - - - - -} \textcolor{b l a c k}{\setminus \setminus} \textcolor{w h i t e}{\theta} \textcolor{b l a c k}{|}$

Now we need to find $\tan \left(\theta\right)$, which is a ratio again

$\tan \left(\theta\right) = \frac{- 3}{\sqrt{7}}$