2x 3x formula - Proving

Chapter 3 Class 11 Trigonometric Functions
Concept wise

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### Transcript

Example 22 Prove that cos2 π₯+cos2 (π₯+π/3) + cos2 (π₯βπ/3) = 3/2 Lets first calculate all 3 terms separately We know that cos 2x = 2 cos2 x β 1 cos 2x + 1 = 2cos2 x πππ β‘γ2π₯ + 1γ/2 = cos2 x So, cos2 x = ππ¨π¬β‘γππ + πγ/π Replacing x with ("x + " π/3) is about cos2 ("x" +π/3) = cosβ‘γ2(π₯ + π/3)+1γ/2 = cosβ‘γ(2π₯ + 2π/3) + 1γ/2 Similarly, Replacing x with ("x β" π/3) in cos2 x = cosβ‘γ2π₯ + 1γ/2 cos2 ("x" βπ/3) = cosβ‘γ2(π₯ β π/3)+ 1γ/2 = cosβ‘γ(2π₯ β 2π/3)+ 1γ/2 Solving L.H.S cos2 x + cos2 (π₯+ π/3) + cos2 (π₯βπ/3) = (1 + cosβ‘2π₯)/2 + (1 + cosβ‘(2π₯ + 2π/3))/2 + (1 + cosβ‘(2π₯ β 2π/3))/2 = 1/2 [1+cosβ‘γ2π₯+1+πππ (2π₯+2π/3)+1+πππ (2π₯β2π/3)γ ] = 1/2 [3+cosβ‘γ2π₯+πππ (2π₯+2π/3)+πππ (2π₯β2π/3)γ ] = 1/2 [3+cosβ‘γ2π₯+2πππ ((2π₯ + 2π/3 + 2π₯ β 2π/3)/2).πππ ((2π₯ + 2π/3 β(2π₯ β 2π/3))/2)γ ] Using cos x + cos y = 2 cos ((π₯ + π¦)/2). cos ((π₯ β π¦)/2) Replace x by ("2" π₯" + " 2π/3) & y by ("2x β " 2π/3) = 1/2 [3+cosβ‘γ2π₯+2πππ ((4π₯ + 0)/2).πππ ((0 + 4π/3)/2)γ ] = 1/2 [3+cosβ‘γ2π₯+2πππ (4π₯/2).πππ ((4π/3)/2)γ ] = 1/2 [3+cosβ‘γ2π₯+2 cosβ‘2π₯ cosβ‘γ2π/3γ γ ] = 1/2 [3+cosβ‘γ2π₯+2 cosβ‘2π₯ cosβ‘(πβπ/3) γ ] = 1/2 [3+cosβ‘γ2π₯+2 cosβ‘2π₯ γ (γβcosγβ‘(π/3) ) ] = 1/2 [3+cosβ‘γ2π₯+2 cosβ‘2π₯ γ (β1/2) ] (As cos (Ο β π) = βcos π) = 1/2 [3+cosβ‘γ2π₯β2 Γ1/2Γcosβ‘2π₯ γ ] = 1/2 [3+cosβ‘γ2π₯βcosβ‘2π₯ γ ] = 1/2 [3+0] = 3/2 = R.H.S. Hence, L.H.S. = R.H.S. Hence Proved