Question

How do you find the exact value of `sin (arcsin(2/3) + arccos(1/3))`?

Answer 1

`sin(arcsin(2/3)+arccos(1/3)) = (2(1+sqrt10))/9`

Explanation:

Using the formula for the sine of the sum of two angles:

`sin(arcsin(2/3)+arccos(1/3)) = sin(arcsin(2/3))cos(arccos(1/3))+cos(arcsin(2/3))sin(arccos(1/3))`

Now, clearly:

`sin(arcsin(2/3)) = 2/3`

`cos(arccos(1/3)) = 1/3`

while:

`cos(arcsin(2/3)) = sqrt(1-sin^2(arcsin(2/3)))= sqrt(1-4/9) = sqrt5/3`

`sin(arccos(1/3)) = sqrt(1-cos^2(arccos(1/3))) = sqrt(1-1/9) = (2sqrt2)/3`

Where we take the positive value because the angles are in the first quadrant.

Then:

`sin(arcsin(2/3)+arccos(1/3)) = 2/3*1/3+ sqrt5/3*(2sqrt2)/3 = (2(1+sqrt10))/9`