Principal Value

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