Find the principal value of the following : tan - 1 ( 1 ) + cos - 1 - 1 2 + sin - 1 - 1 2

tan - 1 ( 1 ) + π - cos - 1 1 2 + - sin - 1 1 2 [Formula: cos - 1 ( - x ) = π – cos - 1 ( x ) and sin - 1 ( - x ) = - sin - 1 ( x ) ] Putting the values for each of the inverse trigonometric terms and we know that, cos - 1 1 2 = π 3 and sin - 1 1 2 = π 6 ⇒ π 4 + π - π 3 - π 6 ⇒ 3 π 4

Find the domain of f x = sin - 1 x 2 + 1 2 x .

Given f x = sin - 1 x 2 + 1 2 x We know that domain of sin - 1 x is - 1 , 1 Now, - 1 ≤ x 2 + 1 2 x ≤ 1 Consider, - 1 ≤ x 2 + 1 2 x ⇒ - 1 + 1 ≤ x 2 + 1 2 x + 1 ⇒ x 2 + 1 + 2 x 2 x ≥ 0 ⇒ x + 1 2 2 x ≥ 0 x + 1 2 is non-negative for all values of x . In order to get x + 1 2 2 x ≥ 0 we must have x > 0 ∴ x + 1 2 2 x ≥ 0 ⇒ x = - 1 and x > 0 Consider, x 2 + 1 2 x ≤ 1 ⇒ x 2 + 1 2 x - 1 ≤ 1 - 1 ⇒ x 2 + 1 - 2 x 2 x ≤ 0 ⇒ x - 1 2 2 x ≤ 0 x - 1 2 is non-negative for all values of x . In order to get x - 1 2 2 x ≤ 0 we must have x < 0 ∴ x - 1 2 2 x ≤ 0 ⇒ x = 1 and x < 0 . The only values of x satisfying inequalities are - 1 and 1 ∴ Domain of f x is - 1 , 1 .

Find the Principal value of sin - 1 1 2

Let us consider sin - 1 1 2 = x By the definition sin x = 1 2 But we know that sin π 6 = 1 2 Hence, sin x = 1 2 = sin π 6 Now the principal value branch of sin - 1 x is - π 2 , π 2 . ∴ sin x = 1 2 = sin π 6 and - π 2 ≤ π 6 ≤ π 2 Hence, the required principal value of x is π 6 .