Give the exact value of tan(arccos sqrt3/2 + arcsin -3/5)

Question

I believe that calculators are not allowed, because the final answer is supposed to be exact. Thanks

Answer 1

First of all, recognize the arccos's and arcsin's can be treated as separate angles in which are:

 

θ1=acrccos sqrt3/2 = 30^o = pi/6 rad

 

θ2=arcsin -3/5 = -arcsin 3/5

 

This implies

 

tan(θ1-θ2) = sin(θ1-θ2)/cos(θ1-θ2) = (sin30*cos(θ2) -cos30*sin(θ2))/(cos30*cos(θ2)+sin30*sin(θ2))

 

sin(θ2)=3/5  cos(θ2) = 4/5  Note: Recognize this is a 3-4-5 right triangle containing θ2.

 

sin(30) = 1/2  cos(30)= sqrt(3)/2

 

 

Substitute these values.

 

tan(θ1-θ2) = (1/2*4/5 - sqrt(3)/2*1/2)/(sqrt(3)/2*sqrt(3)/2+1/2*3/5)

                =(2/5 - sqrt(3)/4)/(3/4+3/10)

                =((8-5*sqrt(3)/20)/((15+6)/20)

                =(8-5*sqrt(3))/21

 

You could leave this as your final answer. Your teacher is simply testing you on your understanding the concepts of trigonometric identities and inverses.

 

Hope this helps.