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
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.
Hi, it looks like you're using AdBlock :(
Displaying ads are our only source of revenue. To help Teachoo create more content, and view the ad-free version of Teachooo... please purchase Teachoo Black subscription.
Please login to view more pages. It's free :)
Teachoo gives you a better experience when you're logged in. Please login :)
Solve all your doubts with Teachoo Black!
Teachoo answers all your questions if you are a Black user!