Finding principal value

Chapter 2 Class 12 Inverse Trigonometric Functions
Concept wise

### Transcript

Ex 2.1, 12 Find the value of cosโ1 (1/2) + 2 sinโ1 (1/2) Solving cosโ1 (๐/๐) Let y = cosโ1 (1/2) cos y = (1/2) cos y = cos (๐/๐) โด y = ๐/๐ Since Range of cosโ1 is [0 , ๐] Hence, the principal value is ๐/๐ (Since cos ๐/3 = 1/2) Solving sinโ1 (๐/๐) Let y = sinโ1 (1/2) sin y = 1/2 sin y = sin (๐/๐) โด y = ๐/๐ Since Range of sinโ1 is [(โ๐)/2 " , " ๐/2] Hence, the Principal Value is ๐/๐ (Since sin ๐/6 = 1/2) Now we have cosโ1 1/2 = ๐/3 & sinโ1 1/2 = ๐/6 Solving cosโ1 ๐/๐ + 2 sinโ1 ๐/๐ = ๐/3 + 2 ร ๐/6 = ๐/3 + ๐/3 = (๐ + ๐)/3 = ๐๐/๐

#### Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.