Finding principal value

Chapter 2 Class 12 Inverse Trigonometric Functions
Concept wise

Suppose we are given

cos -1 (1/2)

Principal value of cos -1 (1/2) means

the angle x where cos x  = 1/2

We use this table for our help

 Range Positive Negative sin -1 [-π/2, π/2] θ – θ cos -1 [0,π] θ π – θ tan -1 (-π/2, π/2) θ – θ

1. Put y = cos -1 (1/2)
2. So, cos y = 1/2
3. Now ignore the signs and find value of θ
4. Using the table, Find principal value according to the table

Let's look at some examples

## Find the principal value of tan –1 (-1 )

### Transcript

Find the principal value of cosβ1 (π/π). Let y = cos-1 (1/2) Hence, cos y = 1/2 cos y = cos (π/3) Range of principal value of cosβ1 is between 0 & Ο Hence principal value is π/π Rough We know that cos 60Β° = 1/2 ΞΈ = 60Β° = 60 Γ π/180 = π/3 Since 1/2 is positive Principal value is ΞΈ i.e. π/3 Find the principal value of cosβ1 ((βπ)/π). Let y = cos-1 ((β1)/2) Hence, cos y = (β1)/2 cos y = cos (πβπ/3) cos y = cos (2π/3) Range of principal value of cosβ1 is between 0 & Ο Hence principal value is ππ/π Rough We know that cos 60Β° = 1/2 ΞΈ = 60Β° = 60 Γ π/180 = π/3 Since (β1)/2 is negative Principal value is Ο β ΞΈ i.e. πβπ/3 Find the principal value of sinβ1 (π/π). Let y = sin-1 (1/2) Hence, sin y = 1/2 sin y = sin (π/6) Range of principal value of sin β1 is between (βπ)/2 and ( π)/2 Hence principal value is π/π Rough We know that sin 30Β° = 1/2 ΞΈ = 30Β° = 30 Γ π/180 = π/6 Since 1/2 is positive Principal value is ΞΈ i.e. π/6 Find the principal value of sinβ1 ((βπ)/π). Let y = sin-1 ((β1)/2) Hence, sin y = (β1)/2 sin y = sin ((βπ)/6) Range of principal value of sin β1 is between (βπ)/2 and ( π)/2 Hence principal value is (βπ)/π Rough We know that sin 30Β° = 1/2 ΞΈ = 30Β° = 30 Γ π/180 = π/6 Since (β1)/2 is negative Principal value is βΞΈ i.e. (βπ)/6 Find the principal value of tanβ1 (1). Let y = tan-1 (1) Hence, tan y = 1 tan y = tan (π/4) Range of principal value of tan β1 is between (βπ)/2 and ( π)/2 Hence principal value is π/π Rough We know that tan 45Β° = 1 ΞΈ = 45Β° = 45 Γ π/180 = π/4 Since 1 is positive Principal value is ΞΈ i.e. π/4 Find the principal value of tanβ1 (β1). Let y = tan-1 (β1) Hence, tan y = β1 tan y = tan ((βπ)/4) Range of principal value of tan β1 is between (βπ)/2 and ( π)/2 Hence principal value is (βπ)/π Rough We know that tan 45Β° = 1 ΞΈ = 45Β° = 45 Γ π/180 = π/4 Since 1 is positive Principal value is βΞΈ i.e. (βπ)/4