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Last updated at Feb. 13, 2020 by Teachoo
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Ex 2.1, 12 Find the value of cosโ1 (1/2) + 2 sinโ1 (1/2) Solving cosโ1 (๐/๐) Let y = cosโ1 (1/2) cos y = (1/2) cos y = cos (ฯ/3) Range of principal value of cosโ1 is [0 , ๐] Hence, the principal value is ฯ/3. Rough We know that cos 60ยฐ = 1/2 ฮธ = 60ยฐ = 60ยฐ ร ๐/180 = ๐/3 Since 1/2 is positive Principal value is ฮธ i.e. ๐/3 Solving sinโ1 (๐/๐) Let y = sinโ1 (1/2) sin y = 1/2 sin y = sin (ฯ/6) Range of principal value of sinโ1 is [(โ๐)/2 " , " ๐/2] Hence, the principal value is ฯ/6 Rough We know that sin 30ยฐ = 1/2 ฮธ = 30ยฐ = 30 ร ๐/180 = ๐/6 Since 1/2 is positive Principal value is ฮธ i.e. ๐/6 Now we have cosโ1 1/2 = ๐/3 & sinโ1 1/2 = ๐/6 Solving cosโ1 ๐/๐ + 2sinโ1 ๐/๐ = ๐/3 + 2 ร ๐/6 = ๐/3 + ๐/3 = (๐ + ๐)/3 = ๐๐ /๐
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