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

2 Answers
May 22, 2017

#-(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

May 22, 2017

#(-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)#