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

2 Answers
Binayaka C.
May 27, 2018

# 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]

MattyMatty
May 27, 2018

#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#

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