Find the value of γsinγ^(-1) [cos(33π/5)]
This question is similar to Example-9 Chapter-2 Inverse trigonometry


CBSE Class 12 Sample Paper for 2024 Boards
CBSE Class 12 Sample Paper for 2024 Boards
Last updated at Feb. 10, 2025 by Teachoo
This question is similar to Example-9 Chapter-2 Inverse trigonometry
Transcript
sinβ1 ("cos " 33Ο/6) = sinβ1 ("cos " (ππ +ππ /π)) = sinβ1 ("cos " 3π/5) = sinβ1 ("sin" (π /π βππ/π)) = sinβ1 ("sin" ((5π β 6π)/10 )) = sinβ1 ("sin" ((βπ )/ππ )) Let y = sinβ1 ("sin" ((βπ)/10 )) sin y = "sin" ((βπ)/10 ) sin y = sin (-18Β°) Hence, y = (βπ )/ππ Which is in the range of sin-1 i.e. [(βπ)/π, π/π] Hence, γπππγ^(βπ) [πππ(πππ /π)] = (βπ )/ππ Hence, γπππγ^(βπ) [πππ(πππ /π)] = (βπ )/ππ