Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
Cos x + cos y formula
Ex 3.3, 11 Important
Misc 3
Misc 4 Important
Ex 3.3, 14
Ex 3.3, 15
Misc 5
Misc 7 Important You are here
Example 16 Important
Ex 3.3, 16 Important
Ex 3.3, 17
Ex 3.3, 18 Important
Ex 3.3, 19
Ex 3.3, 20
Ex 3.3, 21 Important
Example 17 Important
Misc 6 Important
cos x + cos y formula
Last updated at May 29, 2023 by Teachoo
Misc 7 Prove that: sin 3x + sin2x β sin x = 4 sin x cos π₯/2 cos 3π₯/2 Solving L.H.S sin 3x + sin 2x β sin x = sin 3x + (sin 2x β sin x ) = sin 3x + 2cos ((2π₯ + π₯)/2) . sin ((2π₯βπ₯)/2) = sin 3x + 2cos (3π₯/2) sin π₯/2 Using sin x β sin y = 2 cos (π₯ + π¦)/2 sin (π₯ β π¦)/2 Putting x = 2x & y = x , Rough As (3π₯ + π₯)/2 = 4π₯/2 = 2x & (2π₯ + π₯)/2 = 3π₯/2 As 3π₯/2 is in R.H.S. , we take x & 2x We know that sin 2x = 2 sin x cos x Divide by x by x/2 sin 2x/2 = 2 sin x/2 cos x/2 sin x = 2 sin x/2 cos x/2 Now Replace x by 3x sin 3x = 2 sin ππ±/π cos ππ±/π = 2 sin 3π₯/2 cos 3π₯/2 + ["2 cos " 3π₯/2 " sin " π₯/2] = 2 cos 3π₯/2 ["sin " 3π₯/2 " + sin " π₯/2] Using sin x + sin y = 2 sin (π₯ + π¦)/2 cos (π₯ β π¦)/2 Putting x = 3π₯/2 & y = π₯/2 , = 2 cos 3π₯/2 ["2 sin " ((3π₯/2 " + " π₯/2))/2 " . cos " ((3π₯/2 " β " π₯/2))/2] = 2 cos 3π₯/2 ["2 sin " (((3π₯ + π₯)/2))/2 " . cos " (((3π₯ β π₯)/2))/2] = 2 cos 3π₯/2 ["2 sin " ((4π₯/2))/2 " . cos " ((2π₯/2))/2] = 2 cos 3π₯/2 ["2 sin " ((2π₯/1))/2 " . cos " ((π₯/1))/2] = 2 cos 3π₯/2 ["2 sin " 2π₯/2 " . cos " π₯/2] = 2 cos 3π₯/2 ["2 sin " π₯" . cos " π₯/2] = 4 cos 3π₯/2 sin π₯ cos π₯/2 = R.H.S Hence L.H.S = R.H.S Hence proved