Question #c4d83

First, convert the inverse functions into the “regular” function.
#cos(tan^(-)1 (1/4) + cos^-1 (1/2)) #

#tan^(-1) (¼) = tan 4# and #cos^(-1) (½) = cos 2#

#cos(tan(4) + cos(2)) # ; # tan phi = ((sin phi)/(cos phi))#

#cos((sin(4)/(cos(4)) + cos(2)) # ; #(sin(4)) + cos^2(2)) #

From here it depends on whether those are degrees or radians. Without an even multiplier of pi, I'm not sure what "without a calculator" means. SOMETHING has to calculate it.

If any of the previous conversions are incorrect, my apologies, and a reviewer may delete this answer (or let me know, and I will).

From the #color(red)(tan^-1(1/4) # we have the side #color(red)(opposite)# to #color(red)(angle theta=1# and the #color(red)( adjacent# to #color(red)(angle theta= 4 #. Now lets find the #color(red)(hypote n use# using pythagorean theorem. That is,

#color(red)(c=sqrt(1^2 +4^2 )=sqrt17#. Hence, #color(blue)(cos(tan^-1(1/4))=cos theta=(adjacent)/(hypote n use)=4/sqrt17=(4sqrt17)/17#

#color(blue)( sin( tan^-1(1/4))=sin theta=(opposite)/(hypote n use)=1/sqrt17=(sqrt17)/17#

From the #color(orange)(cos^-1 (1/2)# we have the side #color(orange)(adjacent)# to#color(orange)( angle theta = 1# and the #color(orange)(hypote n use = 2#. So this is a #color(orange)(30^@-60^@-90^@# triangle and therefore the #color(orange)(opposite)# to #color(orange)(angle theta=sqrt3#. You can check by using the pythagorean theorem. Thus,

#color(blue)(cos(cos^-1(1/2))=cos theta = (adjacent)/(hypote n use) =1/2#

#color(blue)(sin(cos^-1(1/2))=sin theta = (opposite)/(hypote n use) = sqrt3/2#

Then using the property #color(blue) (cos(A+B)=cosAcosB-sinAsinB)#

we have

#color(blue)(A=tan^- 1 (1/4) and B=cos^(-1)(1/2)#
Then

#cos(tan^-1 (1/4)+cos^-1 (1/2))=color(magenta)(cos(tan^-1(1/4))cos(cos^-1(1/2))-sin(tan^-1(1/4))sin(cos^-1(1/2))#

#color(magenta)(=(4sqrt17)/17 *(1/2)-sqrt17/17*sqrt3/2)#

#color(magenta)(=(4sqrt17)/34-sqrt51/34#

#color(blue)( :.=(4sqrt17-sqrt51)/34#