Question #c4d83

2 Answers
Mar 20, 2017

It may depend on the angular units. IF my conversions are correct.

Explanation:

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).

Mar 20, 2017

see below

Explanation:

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) tocolor(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