Question

How do you evaluate `sin(arcsin(1/2) + arccos (0))`?

Answer 1

Under the multivalued interpretation,

`sin (arcsin(1/2) + arccos(0)) =pm \sqrt{3}/2`

Explanation:

Let's use the multivalued interpretation of `arcsin` and `arccos.`

The real question is what is

`sin arccos (a /b)` and `cos arcsin (a /b)`

When we have the cosine of an inverse sine or the sine of an inverse cosine the denominator `b` is the hypotenuse and the other side is `sqrt{b^2-a^2},` with indeterminate sign. So,

`sin arccos(a/b) = pm \sqrt{b^2-a^2}/b`

`cos arcsin (a/b) = pm sqrt(b^2-a^2)/b`

`sin (arcsin(1/2) + arccos(0))`

`= sin arcsin(1/2) cos arccos 0 + cos arcsin(1/2) sin arccos 0`

`= 1/2 (0) + (pm \sqrt{3}/2)(pm 1)`

`=pm \sqrt{3}/2`