How do you find the exact value of $tan (arcsin (- \frac{3}{4}))$ $tan(arcsin(-3/4))$ ?

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How do you find the exact value of $tan (arcsin (- \frac{3}{4}))$ $tan(arcsin(-3/4))$ ?

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Answer 1 · 2017-05-22T02:55:03

$- \frac{3 \sqrt{7}}{7}$ $-(3sqrt7)/7$

Explanation:

Let's think about it this way:

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

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

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

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

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

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

$cos θ = \pm \frac{\sqrt{7}}{4}$ $costheta = +-sqrt7/4$

Since $sin θ$ $sintheta$ is negative, $θ$ $theta$ must be in quadrant 4 since $arcsin (x)$ $arcsin(x)$ is only defined for quadrants 1 (+) and 4 (-). In quadrant 4, $cos θ$ $costheta$ is always positive, so it must be $\frac{\sqrt{7}}{4}$ $sqrt7/4$ .

We now know $sin θ = - \frac{3}{4}$ $sintheta=-3/4$ and $cos θ = \frac{\sqrt{7}}{4}$ $costheta = sqrt7/4$ . This means we can solve for $tan θ$ $tantheta$ , which is what we're looking for.

$tan θ = \frac{sin θ}{cos θ}$ $tantheta=sintheta/costheta$

$tan θ = \frac{- \frac{3}{4}}{\frac{\sqrt{7}}{4}} = - \frac{3}{\sqrt{7}} = - \frac{3 \sqrt{7}}{7}$ $tantheta = (-3/4)/(sqrt7/4) = -3/sqrt7 = -(3sqrt7)/7$

Therefore, the exact value of $tan (arcsin (- \frac{3}{4}))$ $tan(arcsin(-3/4))$ is $- \frac{3 \sqrt{7}}{7}$ $-(3sqrt7)/7$ .

Final Answer

Answer 2 · 2017-05-22T03:05:36

$\frac{- 3}{\sqrt{7}}$ $(-3)/sqrt(7)$ or $\frac{- 3 \sqrt{7}}{7}$ $(-3sqrt(7))/7$

Explanation:

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

Arcsin corresponds to the ratio of $\frac{o p p}{h y p}$ $(opp)/(hyp)$

$0$ $color(white)(0)$

$4 - - - - - - - - -$ $color(white)(4)color(black)(---------)$
$4 θ - - - - - - - - ∣$ $color(white)(4)color(black)(\)color(black)(theta)color(white)(- - - - - - - -)color(black)(|)$
$4 - θ - - - - - \frac{-}{- .} / ∣$ $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)(|)$
$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(black)(|)$

Let's solve for the remaining side:
$4^2-3^2=7^2=sqrt7$

$color(white)(000000000)sqrt7$

$color(white)(4)color(black)(---------)$
$color(white)(4)color(black)(\)color(black)(theta)color(white)(- - - - - - - -)color(black)(|)$
$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)(|)$
$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(black)(|)$

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

$tan(theta)=(-3)/sqrt(7)$

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