Question

How do you find the exact value of `Sin(arcsin(3/5)+(arctan-2))`?

Answer 1

`sin(arcsin(3/5)+(arctan (-2)))=color(red)(-sqrt5/5)` and

`sin(arcsin(3/5)+(arctan (-2)))=color(red)(sqrt5/5)`

Explanation:

There are 2 possible answers:

First solution

`sin (arcsin (3/5)+arctan (-2))`

`sin (A+B)`

Let `A=arcsin (3/5)` and `B=arctan (color(blue)(-2)/1)`

then `sin A=3/5` and

computed using Pythagorean relation `c^2=a^2+b^2`

`cos A=4/5`

also

`sin B=(-2)/sqrt5`

`cos B=1/sqrt5`

compute `sin (A+B)`
`sin (A+B)=sin A cos B + cos A sin B`

`sin (A+B)=3/5* 1/sqrt5 + 4/5 *(-2)/sqrt5=-5/(5sqrt5)`

`sin (A+B)=-1/sqrt5=-sqrt5/5`
`color(green) ("The 4th quadrant angle")=A+B=-63.4349^@`

second solution

`sin (arcsin (3/5)+arctan (-2))`

`sin (A+B)`

Let `A=arcsin (3/5)` and `B=arctan (2/color (blue)(-1))`

then `sin A=3/5` and

computed using Pythagorean relation `c^2=a^2+b^2`

`cos A=4/5`

also

`sin B=2/sqrt5`

`cos B=(-1)/sqrt5`

compute `sin (A+B)`
`sin (A+B)=sin A cos B + cos A sin B`

`sin (A+B)=3/5* (-1)/sqrt5 + 4/5 *2/sqrt5=5/(5sqrt5)`

`sin (A+B)=1/sqrt5=sqrt5/5`

`color(green) ("The 2nd quadrant angle")=A+B=116.565^@`
Have a nice day !!! from the Philippines