Ex 2.1, 12 - Chapter 2 Class 12 Inverse Trigonometric Functions
Last updated at Dec. 16, 2024 by Teachoo
Finding principal value
Principal Value
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Chapter 2 Class 12 Inverse Trigonometric Functions
Concept wise
Transcript
Ex 2.1, 12 Find the value of cosβ1 (1/2) + 2 sinβ1 (1/2) Solving cosβ1 (π/π) Let y = cosβ1 (1/2) cos y = (1/2) cos y = cos (π /π) β΄ y = π /π Since Range of cosβ1 is [0 , π] Hence, the principal value is π /π (Since cos π/3 = 1/2) Solving sinβ1 (π/π) Let y = sinβ1 (1/2) sin y = 1/2 sin y = sin (π /π) β΄ y = π /π Since Range of sinβ1 is [(βπ)/2 " , " π/2] Hence, the Principal Value is π /π (Since sin π/6 = 1/2) Now we have cosβ1 1/2 = π/3 & sinβ1 1/2 = π/6 Solving cosβ1 π/π + 2 sinβ1 π/π = π/3 + 2 Γ π/6 = π/3 + π/3 = (π + π)/3 = ππ /π
