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