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 4 = 1 2 Hence, sin x = 1 2 = sin π 4 Now the principal value branch of sin - 1 x is - π 2 , π 2 . ∴ sin x = 1 2 = sin π 4 and - π 2 ≤ π 4 ≤ π 2 Hence, the required principal value of x is π 4 .

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 4 = 1 2 Hence, sin x = 1 2 = sin π 4 Now the principal value branch of sin - 1 x is - π 2 , π 2 . ∴ sin x = 1 2 = sin π 4 and - π 2 ≤ π 4 ≤ π 2 Hence, the required principal value of x is π 4 .

Find the principal value of cot - 1 3 .

Let cot - 1 3 = y , then cot y = 3 = cot π 6 We know that, the range of the principal value of cot - 1 x is 0 , π and cot π 6 = 3 Hence, the principal value of cot - 1 3 is π 6 .

Find the principal value of cos - 1 - 1 2

Let us consider cos - 1 - 1 2 = x By the definition cos x = - 1 2 But we know that cos π 3 = 1 2 Hence, - cos π 3 = - 1 2 Now the principal value branch of cos - 1 x is 0 , π We know that - cos π 3 = cos π - π 3 ∴ - cos π 3 = cos 2 π 3 and 0 ≤ 2 π 3 ≤ π Hence, the required principal value of x is 2 π 3 .

Find the principal value of cosec - 1 ( - 2 )

Given that, cosec - 1 ( - 2 ) Let cosec - 1 ( - 2 ) = θ We know that, the range of principal value branch of cosec - 1 is - π 2 , π 2 - { 0 } Now, cosecθ = - 2 ⇒ cosecθ = - cosec π 4 ⇒ cosecθ = cosec - π 4 ⇒ θ = - π 4 ∈ - π 2 , π 2 - { 0 } Hence, the principal value of cosec - 1 ( 2 ) = - π 4 .

Find the general values of : sin - 1 - 1 2 ,

Let x = sin - 1 - 1 2 ⇒ sin x = - 1 2 We know that sin π 4 = 1 2 ⇒ sin x = - sin π 4 ⇒ sin x = sin - π 4 We know that general solution of sin x = sin α is x = n π + - 1 n α , n ∈ Z ∴ x = n π + - 1 n - π 4 = n π - - 1 n π 4 .