Question

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

Answer 1

`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`