How do you simplify the expression #sin(arcsin(3/5)+arctan(-2))#?

1 Answer
Mar 6, 2018

#sin(arcsin(3/5)+arctan(-2))=-1/sqrt5#

Explanation:

Let us note that the range of #arcsinx# and #arctanx# is #[-pi/2,pi/2]#

Now let #arcsin(3/5)=A# and #arctan(-2)=B#,

then #sinA=3/5# and #tanB=-2#

therefore #cosA=sqrt(1-(3/5)^2)=sqrt(16/25)=4/5#

and #secB=sqrt(1+(-2)^2)=sqrt5# i,e. #cosB=1/sqrt5#

and as #sinA=tanBcosB=-2/sqrt5#

we have #sin(arcsin(3/5)+arctan(-2))#

= #sin(A+B)#

= #sinAcosB+cosAsinB#

= #3/5xx1/sqrt5+4/5xx(-2/sqrt5)#

= #(3-8)/(5sqrt5)#

= #-5/(5sqrt5)#

= #-1/sqrt5#