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


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Ex 3.3, 13 Prove that cos2 2π‘₯ – cos2 6π‘₯ = sin⁑4π‘₯ sin⁑8π‘₯ Solving L.H.S. cos2 2x – cos2 6x = (cos 2x + cos 6x) (cos 2x – cos 6x) Lets calculate (cos 2x + cos 6x) and (cos 2x – cos 6x) separately cos 2x + cos 6x = 2 cos ((2x+6x)/2) cos ((2xβˆ’6x)/2) = 2 cos (8π‘₯/2) cos ((βˆ’4π‘₯)/2) = 2 cos 4x cos (-2x) cos 2x – cos 6x = – 2 sin ((2x+6x)/2) sin((2xβˆ’6x)/2) = – 2 sin (8π‘₯/2) sin ((βˆ’4π‘₯)/2) = – 2 sin 4x sin (–2x) Hence π’„π’π’”πŸ πŸπ’™ – π’„π’π’”πŸ πŸ”π’™ = (cos⁑2π‘₯ + cos⁑6π‘₯) (cos⁑2π‘₯ – 6π‘₯) = (2 cos⁑〖4π‘₯ 𝒄𝒐𝒔⁑〖(βˆ’πŸπ’™)γ€— γ€— ) (βˆ’2 sin⁑4π‘₯ (π’”π’Šπ’β‘γ€–(βˆ’πŸπ’™)γ€— )) = (2 cos⁑〖4π‘₯ 𝒄𝒐𝒔⁑〖(πŸπ’™)γ€— γ€— ) (βˆ’2 sin⁑4π‘₯ (γ€–βˆ’π’”π’Šπ’γ€—β‘γ€–(πŸπ’™)γ€— )) = (2 cos⁑〖4π‘₯ cos⁑〖(2π‘₯)γ€— γ€— ) (2 sin⁑4π‘₯ sin⁑〖(2π‘₯)γ€— ) = (𝟐 π’”π’Šπ’β‘πŸ’π’™ π’„π’π’”β‘πŸ’π’™) (𝟐 π’”π’Šπ’β‘πŸπ’™ π’„π’π’”β‘πŸπ’™) We know that sin 2x = 2 sin x cos x Putting 4x instead of x And putting 2x instead of x = sin⁑8π‘₯ sin⁑4π‘₯ = R.H.S. Hence, L.H.S. = R.H.S. Hence proved

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.