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Principal Value

Last updated at May 29, 2023 by

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 )

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/3) Range of principal value of cos−1 is between 0 & π Hence principal value is 𝝅/𝟑 Rough We know that cos 60° = 1/2 θ = 60° = 60 × 𝜋/180 = 𝜋/3 Since 1/2 is positive Principal value is θ i.e. 𝜋/3 Find the principal value of cos–1 ((−𝟏)/𝟐). Let y = cos-1 ((−1)/2) Hence, cos y = (−1)/2 cos y = cos (𝜋−𝜋/3) cos y = cos (2𝜋/3) Range of principal value of cos−1 is between 0 & π Hence principal value is 𝟐𝝅/𝟑 Rough We know that cos 60° = 1/2 θ = 60° = 60 × 𝜋/180 = 𝜋/3 Since (−1)/2 is negative Principal value is π – θ i.e. 𝜋−𝜋/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/6) Range of principal value of sin −1 is between (−𝜋)/2 and ( 𝜋)/2 Hence principal value is 𝝅/𝟔 Rough We know that sin 30° = 1/2 θ = 30° = 30 × 𝜋/180 = 𝜋/6 Since 1/2 is positive Principal value is θ i.e. 𝜋/6 Find the principal value of sin–1 ((−𝟏)/𝟐). Let y = sin-1 ((−1)/2) Hence, sin y = (−1)/2 sin y = sin ((−𝜋)/6) Range of principal value of sin −1 is between (−𝜋)/2 and ( 𝜋)/2 Hence principal value is (−𝝅)/𝟔 Rough We know that sin 30° = 1/2 θ = 30° = 30 × 𝜋/180 = 𝜋/6 Since (−1)/2 is negative Principal value is –θ i.e. (−𝜋)/6 Find the principal value of tan–1 (1). Let y = tan-1 (1) Hence, tan y = 1 tan y = tan (𝜋/4) Range of principal value of tan −1 is between (−𝜋)/2 and ( 𝜋)/2 Hence principal value is 𝝅/𝟒 Rough We know that tan 45° = 1 θ = 45° = 45 × 𝜋/180 = 𝜋/4 Since 1 is positive Principal value is θ i.e. 𝜋/4 Find the principal value of tan–1 (–1). Let y = tan-1 (–1) Hence, tan y = –1 tan y = tan ((−𝜋)/4) Range of principal value of tan −1 is between (−𝜋)/2 and ( 𝜋)/2 Hence principal value is (−𝝅)/𝟒 Rough We know that tan 45° = 1 θ = 45° = 45 × 𝜋/180 = 𝜋/4 Since 1 is positive Principal value is –θ i.e. (−𝜋)/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.