Question

How do you calculate `cos(tan^-1(3/4))`?

Answer 1

`cos ( tan^-1 (3/4))= 0.8`

Explanation:

`cos ( tan^-1 (3/4))= ?` Let `tan^-1 (3/4)= theta`

`:. tan theta = 3/4 =P/B, P and B` are perpendicular and base

of right triangle , then `H^2= P^2+B^2= 3^2+4^2=25`

`:.H=5 ; :. cos theta = B/H=4/5 =0.8`

`cos ( tan^-1 (3/4))= cos theta= 0.8`

`:. cos ( tan^-1 (3/4))= 0.8` [Ans]

Answer 2

`4/5`

Explanation:

`tan(tan^-1(3/4)) = 3/4`
`"Name "y = tan^-1(3/4)`
`"Then we have"`
`tan(y) = 3/4`
`"Now use "sec²(x) = 1 + tan²(x)`
`=> sec²(y) = 1 + tan²(y) = 1 + 9/16 = 25/16`
`=> sec(y) = 1/cos(y) = pm 5/4`
`=> cos(y) = pm 4/5`
`=> cos(tan^-1(3/4)) = pm 4/5`
`"We have to take the solution with + sign as"`
`-pi/2 <= arctan(x) <= pi/2`
`"and"`
`cos(x) > 0, if -pi/2<=x<=pi/2`
`=> cos(tan^-1(3/4)) = 4/5`

`"Note that we could also have used"`
`tan(y)=sin(y)/cos(y)`
`"and"`
`sin^2(y) + cos^2(y) = 1`
`tan(y)=sin(y)/cos(y) = 3/4`
`=> pm sqrt(1-cos^2(y))/cos(y) = 3/4`
`=> 1-cos^2(y) = ((3/4) cos(y))^2`
`=> (1+9/16) cos^2(y) = 1`
`=> cos^2(y) = 16/25`
`=> cos(y) = 4/5`