Last updated at Sept. 3, 2021 by
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Example 26 Prove that cos 2x cos π₯/2 β cos 3x cos 9π₯/2 = sin 5x sin 5π₯/2 Solving L.H.S Solving cos 2x cos x/2 and cos 3x cos 9π₯/2 separately We know that 2 cos x cos y = cos (x + y) + cos (x β y) cos x cos y = 1/2 ("cos (x + y) + cos (x β y)" ) cos 2x cos π/π Replacing x with 2x and y with π₯/2 = 1/2 ("cos " ("2x + " x/2)" + cos" ("2x" βx/2)) = 1/2 ("cos " ((4x + x )/2)" + cos " ((4x β x)/2)) = 1/2 ("cos " (5x/2)" + cos " (3x/2)) cos 3x cos ππ/π Replacing x with 3x and y with 9π₯/2 = 1/2 ("cos" ("3x + " 9π₯/2)" + cos " ("3x β " 9π₯/2)) = 1/2 ("cos " ((6π₯ + 9π₯ )/2)" + cos " ((6π₯ β 9π₯)/2)) = 1/2 ("cos " (15π₯/2)" + cos " ((β3π₯)/2)) Now, cos 2x cos π₯/2 β cos 3x cos 9π₯/2 Putting values = 1/2 ("cos " (5x/2)" + cos " (3x/2)) β 1/2 ("cos " (15π₯/2)" + cos " ((β3π₯)/2)) = 1/2 ("cos " (5x/2)" + cos " (3x/2)) β 1/2 ("cos " (15π₯/2)" + cos " (3π₯/2)) = 1/2 ("cos " (5x/2)" + cos " (3x/2)"β cos " (15π₯/2)"β cos " (3π₯/2)) = 1/2 ("cos " (5x/2)"β cos " (15π₯/2)"+ cos " (3x/2)"β cos " (3π₯/2)) = 1/2 ("cos " (5x/2)"β cos " (15π₯/2)"+ " 0) = 1/2 ("cos " (5x/2)"β cos " (15π₯/2)) = 1/2 ("β 2 sin " ((5π₯/2 +15/2 π₯)/2)". sin " ((5π₯/2 β 15/2 π₯)/2)) = β"sin " ((5π₯ + 15π₯)/(2 Γ 2))" . sin " ((5π₯ β 15π₯)/(2 Γ 2)) = "β sin " (20π₯/4)" . sin " (( β10π₯)/4) Using cos x β cos y = β2 sin (π₯ + π¦)/2 sin (π₯ β π¦)/2 Putting x = 5π₯/2 , y = 15π₯/2 = β "sin " (5π₯)" . sin " (( β5π₯)/2) = β "sin " (5π₯) Γ β" sin " (( 5π₯)/2) = sin (5π₯) sin (( 5π₯)/2) = R.H.S Hence L.H.S. = R.H.S. Hence proved (As sin(βx) = β sin x)
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