Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class

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Chapter 3 Class 11 Trigonometric Functions

Serial order wise

Last updated at June 22, 2023 by Teachoo

Example 22 Prove that cos2 ๐ฅ+cos2 (๐ฅ+๐/3) + cos2 (๐ฅโ๐/3) = 3/2 Lets first calculate all 3 terms separately We know that cos 2x = 2 cos2 x โ 1 cos 2x + 1 = 2cos2 x ๐๐๐ โกใ2๐ฅ + 1ใ/2 = cos2 x cos2 x = ๐๐จ๐ฌโกใ๐๐ + ๐ใ/๐ Replacing x with ("x + " ๐ /๐) cos2 ("x" +๐ /๐) = cosโกใ2(๐ฅ + ๐/3)+1ใ/2 = ๐๐๐โกใ(๐๐ + ๐๐ /๐) + ๐ใ/๐ Similarly, Replacing x with ("x โ" ๐ /๐) in cos2 x = ๐๐จ๐ฌโกใ๐๐ + ๐ใ/๐ cos2 ("x" โ๐/3) = cosโกใ2(๐ฅ โ ๐/3)+ 1ใ/2 = cosโกใ(2๐ฅ โ 2๐/3)+ 1ใ/2 Solving L.H.S cos2 x + cos2 (๐ฅ+ ๐/3) + cos2 (๐ฅโ๐/3) = (๐ + ๐๐จ๐ฌโก๐๐)/๐ + (๐ + ๐๐๐โก(๐๐ + ๐๐ /๐))/๐ + (๐ + ๐๐๐โก(๐๐ โ ๐๐ /๐))/๐ = 1/2 [1+cosโกใ2๐ฅ+1+๐๐๐ (2๐ฅ+2๐/3)+1+๐๐๐ (2๐ฅโ2๐/3)ใ ] = 1/2 [3+cosโกใ2๐ฅ+๐๐๐(๐๐+๐๐ /๐)+๐๐๐(๐๐โ๐๐ /๐)ใ ] = 1/2 [3+cosโกใ2๐ฅ+2๐๐๐((๐๐ + ๐๐ /๐ + ๐๐ โ ๐๐ /๐)/๐).๐๐๐((๐๐ + ๐๐ /๐ โ(๐๐ โ ๐๐ /๐))/๐)ใ ] = 1/2 [3+cosโกใ2๐ฅ+2๐๐๐ ((4๐ฅ + 0)/2).๐๐๐ ((0 + 4๐/3)/2)ใ ] = 1/2 [3+cosโกใ2๐ฅ+2๐๐๐ (4๐ฅ/2).๐๐๐ ((4๐/3)/2)ใ ] = 1/2 [3+cosโกใ2๐ฅ+๐ ๐๐จ๐ฌโก๐๐ ๐๐จ๐ฌโกใ๐๐ /๐ใ ใ ] = 1/2 [3+cosโกใ2๐ฅ+2 cosโก2๐ฅ cosโก(๐โ๐/3) ใ ] = 1/2 [3+cosโกใ2๐ฅ+2 cosโก2๐ฅ ใ (ใโ๐๐๐ใโก(๐ /๐) ) ] = 1/2 [3+cosโกใ2๐ฅ+2 cosโก2๐ฅ ใ (โ1/2) ] = 1/2 [3+cosโกใ2๐ฅโ2 ร1/2รcosโก2๐ฅ ใ ] = ๐/๐ [๐+๐๐๐โกใ๐๐โ๐๐๐โก๐๐ ใ ] = 1/2 [3+0] = 3/2 = R.H.S. Hence, L.H.S. = R.H.S. Hence Proved