How do you simplify Cos [tan^-1(-1) + cos^-1(-4/5)]?

2 Answers
Mar 20, 2017

see below

Explanation:

Use the property color(blue)(cos(A+B)=cosAcosB-sinAsinB

For the color(red)(tan^-1 (-1) we know that our triangle is in quadrant 4. Since color(red)(tan^-1 x is restricted to quadrants 1 and 4 and since the argument is negative it means our triangle has to be in quadrant 4.

Hence, the side color(red)(opposite to color(red)(angle theta = -1 and color(red)(adjacent to color(red)(angle theta = 1 and so the color(red)(hypote n use=sqrt2. In this case color(red)(theta=45^@ but we don't need to know the value of color(red)(theta. Therefore,

color(blue)(cos(tan^-1(-1))=cos theta=(adjacent)/(hypote n use)=1/sqrt2 = sqrt2/2

color(blue)(sin(tan^-1(-1))=sin theta=(opposite)/(hypote n use)=-1/sqrt2 = -sqrt2/2

For the color(magenta)(cos^-1(-4/5)the triangle is in quadrant 2 since color(magenta)(cos^-1 x)is restricted to quadrants 1 and 2 and since the argument is negative the triangle has to be in quadrant 2.

Hence, color(magenta)(adjacent to color(magenta)(angle theta=-4 and color(magenta)(hypote n use = 5. Therefore by using pythagorean theorem the color(magenta)(opposite to color(magenta)(angle theta=3

color(blue)((cos(cos^-1(-4/5))=cos theta = (adjacent)/(hypote n use)=-4/5

color(blue)(sin(cos^-1(-4/5)=sin theta=(opposite)/(hypote n use) = 3/5

Then using the property color(orange)(cos(A+B)=cosAcosB-sinAsinB where color(orange)(A=tan^-1 (-1) and B=cos^-1(-4/5) we have

cos(tan^-1(-1)+cos^-1(-4/5))=color(blue)(cos(tan^-1(-1))cos(cos^-1(-4/5))-sin(tan^-1(-1)) sin(cos^-1(-4/5))

color(blue)(=sqrt2/2 * (-4)/5-(-sqrt2)/2*3/5

color(blue)(=(-4sqrt2)/10)+color(blue)((3sqrt2)/10

color(blue)( :. = -sqrt2/10

Mar 21, 2017

LHS=cos(tan^-1(-1)+cos^-1(-4/5))

=cos(tan^-1tan(-pi/4)+cos^-1(-4/5))

=cos((-pi/4)+cos^-1(-4/5))

=cos(-pi/4)cos(cos^-1(-4/5))-sin(-pi/4)sin(cos^-1(-4/5))

=cos(pi/4)(-4/5))+sin(pi/4)sin(sin^-1sqrt(1-(-4/5)^2))

=1/sqrt2xx(-4/5)+1/sqrt2xx3/5

=(-4+3)/(5sqrt2)

=-1/(5sqrt2)=-sqrt2/10