Principal Value

Last updated at July 12, 2018 by Teachoo

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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

Put y = cos ^-1 (1/2)
So, cos y = 1/2
Now ignore the signs and find value of θ
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 (-1 /2 )

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

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

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

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

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

Finding principal value

Principal Value You are here

Formulae based →

Chapter 2 Class 12 Inverse Trigonometric Functions

Concept wise

About the Author

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.