Q.77. Inverse trigonometric functions

Question 77.
Find the exact value of sin[ (arcsin 1/3) + (arcsin 2/3) ].

Answer 77.
Let arc sin (1/3) = α and arc sin (2/3) = β,
where α and β belong to [ - π /2, π /2 ]
=> sin α = 1/3 and sin β = 2/3
=> cos α = √(1 - 1/9) = (2√2) / 3 and
cos β = √(1 - 4/9) = (√5) / 3
[ Note: cos α and cos β are positive because α and β belong to [ - π /2, π /2 ] ]

Now, sin[ (arcsin 1/3) + (arcsin 2/3) ]
= sin ( α + β )
= sin α cos β + cos α sin β
= (1/3) * [(√5) / 3] + [(2√2) / 3] * (2/3)
= (√5) / 9 + (4√2) / 9
= (1/9)(√5 + 4√2)

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