Find the value of γsinγ^(-1) [cos(33π/6)]
This question is similar to Example-9 Chapter-2 Inverse trigonometry
Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
CBSE Class 12 Sample Paper for 2024 Boards
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CBSE Class 12 Sample Paper for 2024 Boards
Last updated at Aug. 9, 2023 by Teachoo
This question is similar to Example-9 Chapter-2 Inverse trigonometry
Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
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