Question

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

Answer 1

`-(3sqrt7)/7`

Explanation:

Let's think about it this way:

We're trying to find the tangent of the angle whose sine is `-3/4`.

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

`sin^2theta+cos^2theta=1`

`(-3/4)^2+cos^2theta=1`

`9/16 + cos^2theta=1`

`cos^2theta = 7/16`

`costheta = +-sqrt7/4`

Since `sintheta` is negative, `theta` must be in quadrant 4 since `arcsin(x)` is only defined for quadrants 1 (+) and 4 (-). In quadrant 4, `costheta` is always positive, so it must be `sqrt7/4`.

We now know `sintheta=-3/4` and `costheta = sqrt7/4`. This means we can solve for `tantheta`, which is what we're looking for.

`tantheta=sintheta/costheta`

`tantheta = (-3/4)/(sqrt7/4) = -3/sqrt7 = -(3sqrt7)/7`

Therefore, the exact value of `tan(arcsin(-3/4))` is `-(3sqrt7)/7`.

Final Answer

Answer 2

`(-3)/sqrt(7)` or `(-3sqrt(7))/7`

Explanation:

Domain of arcsin is the 1st quadrant is positive, 4th quadrant is negative. So, `-3/4` is in the 4th quadrant.

Arcsin corresponds to the ratio of `(opp)/(hyp)`

`color(white)(0)`

`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)(|)`

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)`