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

 

And we follow these steps

  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 cos –1 ( 1/2 )

Find the principal value of cos-1 (12).JPG

Find the principal value of cos –1 (-1 /2 )

Find the principal value of cos-1 (-12).JPG

Find the principal value of sin –1 ( 1/2 )

Find the principal value of sin-1 (12).JPG

Find the principal value of sin –1 (-1/2)

Find the principal value of sin-1 (-12).JPG

Find the principal value of tan –1 (1)

Find the principal value of tan-1 (1).JPG

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

Find the principal value of tan-1 (-1).JPG

  1. Chapter 2 Class 12 Inverse Trigonometric Functions
  2. Concept wise

Transcript

Find the principal value of cos–1 (𝟏/𝟐). Let y = cos-1 (1/2) Hence, cos y = 1/2 cos y = cos (πœ‹/3) Range of principal value of cosβˆ’1 is between 0 & Ο€ Hence principal value is 𝝅/πŸ‘ Rough We know that cos 60Β° = 1/2 ΞΈ = 60Β° = 60 Γ— πœ‹/180 = πœ‹/3 Since 1/2 is positive Principal value is ΞΈ i.e. πœ‹/3 Find the principal value of cos–1 ((βˆ’πŸ)/𝟐). Let y = cos-1 ((βˆ’1)/2) Hence, cos y = (βˆ’1)/2 cos y = cos (πœ‹βˆ’πœ‹/3) cos y = cos (2πœ‹/3) Range of principal value of cosβˆ’1 is between 0 & Ο€ Hence principal value is πŸπ…/πŸ‘ Rough We know that cos 60Β° = 1/2 ΞΈ = 60Β° = 60 Γ— πœ‹/180 = πœ‹/3 Since (βˆ’1)/2 is negative Principal value is Ο€ – ΞΈ i.e. πœ‹βˆ’πœ‹/3 Find the principal value of sin–1 (𝟏/𝟐). Let y = sin-1 (1/2) Hence, sin y = 1/2 sin y = sin (πœ‹/6) Range of principal value of sin βˆ’1 is between (βˆ’πœ‹)/2 and ( πœ‹)/2 Hence principal value is 𝝅/πŸ” Rough We know that sin 30Β° = 1/2 ΞΈ = 30Β° = 30 Γ— πœ‹/180 = πœ‹/6 Since 1/2 is positive Principal value is ΞΈ i.e. πœ‹/6 Find the principal value of sin–1 ((βˆ’πŸ)/𝟐). Let y = sin-1 ((βˆ’1)/2) Hence, sin y = (βˆ’1)/2 sin y = sin ((βˆ’πœ‹)/6) Range of principal value of sin βˆ’1 is between (βˆ’πœ‹)/2 and ( πœ‹)/2 Hence principal value is (βˆ’π…)/πŸ” Rough We know that sin 30Β° = 1/2 ΞΈ = 30Β° = 30 Γ— πœ‹/180 = πœ‹/6 Since (βˆ’1)/2 is negative Principal value is –θ i.e. (βˆ’πœ‹)/6 Find the principal value of tan–1 (1). Let y = tan-1 (1) Hence, tan y = 1 tan y = tan (πœ‹/4) Range of principal value of tan βˆ’1 is between (βˆ’πœ‹)/2 and ( πœ‹)/2 Hence principal value is 𝝅/πŸ’ Rough We know that tan 45Β° = 1 ΞΈ = 45Β° = 45 Γ— πœ‹/180 = πœ‹/4 Since 1 is positive Principal value is ΞΈ i.e. πœ‹/4 Find the principal value of tan–1 (–1). Let y = tan-1 (–1) Hence, tan y = –1 tan y = tan ((βˆ’πœ‹)/4) Range of principal value of tan βˆ’1 is between (βˆ’πœ‹)/2 and ( πœ‹)/2 Hence principal value is (βˆ’π…)/πŸ’ Rough We know that tan 45Β° = 1 ΞΈ = 45Β° = 45 Γ— πœ‹/180 = πœ‹/4 Since 1 is positive Principal value is –θ i.e. (βˆ’πœ‹)/4

About the Author

Davneet Singh's photo - Teacher, Computer Engineer, Marketer
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.