What is #cos[sin^(-1)(-1/2 ) + cos^(-1)(5/13) ]#?

#rarrcos[cos^(-1)(5/13)+sin^(-1)(-1/2)]#

#=cos[cos^(-1)(5/13)-sin^(-1)(1/2)]#

#=cos[cos^(-1)(5/13)-cos^(-1)(sqrt3/2)]#

Now, using #cos^(-1)x-cos^(-1)y=xy+sqrt((1-x^2)*(1-y^2))#, we get,

#rarrcos[cos^(-1)(5/13)-sin^(-1)(1/2)]#

#=cos(cos^(-1)(5/13*sqrt3/2+sqrt((1-(5/13)^2)*(1-(sqrt(3)/2)^2))))#

#=(5sqrt3)/26+12/26#

#=(12+5sqrt3)/26#

#x = cos(arcsin(−1/2)+arccos(5/13))#

These questions are confusing enough with the funky inverse function notation. The real problem with questions like this is it's generally best to treat the inverse functions as multivalued, which may mean the expression has multiple values as well.

We can also look at the value of #x# for the principal value of the inverse functions, but I'll leave that to others.

Anyway, this is the cosine of the sum of two angles, and that means we employ the sum angle formula:

#cos(a+b) = cos a cos b - sin a sin b#

# x = cos( arcsin(-1/2)) cos (arccos(5/3)) - sin( arcsin(-1/2)) sin(arccos(5/13)) #

Cosine of inverse cosine and sine of inverse sine are easy. The cosine of inverse sine and sine of inverse cosine are also straightforward, but there's where the multivalued issue comes in.

There will be generally be two non-coterminal angles that share a given cosine, negations of each other, whose sines will be negations of each other. There will generally be two non-coterminal angles that share a given sine, supplementary angles, which will have cosines that are negations of each other. So both ways we up with a #pm#. Our equation will have two #\pm# and it's important to note they're independent, unlinked.

Let's take #arcsin(-1/2)# first. This is of course one of trig's cliches, #-30^circ# or #-150^circ#. The cosines will be #+ sqrt{3}/2# and #- sqrt{3}/2# respectively.

We don't really need to consider the angle. We can think about the right triangle with opposite 1 and hypotenuse 2 and come up with adjacent #\sqrt{3}# and cosine #\pm \sqrt{3}/2#. Or if that's too much thinking, since #cos^2theta + sin ^2 theta = 1# then #cos(theta) = \pm \sqrt{1 - sin^2 theta}# which mechanically lets us say:

# cos(arcsin(-1/2)) = \pm \sqrt{1 - (-1/2)^2} = \pm sqrt{3}/2#

Similarly, #5,12,13# is Pythagorean Triple employed here so

#sin(arccos(5/3)) = \pm \sqrt{1 - (5/13)^2} = \pm 12/13 #

# x =(\pm sqrt{3}/2)(5/3) - (-1/2) (\pm 12/13) #

#x = \pm {5\sqrt{3}}/6 \pm 6/13#