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Suppose we are given

cos -1 (1/2)

Β 

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

the angle x where cos xΒ  = 1/2

Β 

We use this table for our help

Β 

Range

Positive

Negative

sin -1

[-Ο€/2, Ο€/2]

ΞΈ

 – ΞΈ

cos -1

[0,Ο€]

ΞΈ

Β Ο€ – ΞΈ

tan -1

(-Ο€/2, Ο€/2)

ΞΈ

 – ΞΈ

Β 

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 ofΒ ΞΈ
  4. Using the table, Find principal value according to the table

Β 

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 )

Principal Value - Part 2

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

Principal Value - Part 3

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

Principal Value - Part 4

Find the principal value of tan –1 (1)

Principal Value - Part 5

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

Principal Value - Part 6

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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

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.