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

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Answer 1 · 2018-03-20T15:17:33

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

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:

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

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