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#