Misc 2
Prove that: (sin 3π₯ + sin π₯) sin π₯ + (cos 3π₯ β cos π₯) cos π₯ = 0
Lets calculate (sin 3x + sin x) and (cos 3x β cos x) separately
We know that
sin x + sin y = sin ((π₯ + π¦)/2) cos ((π₯ β π¦)/2)
Replacing x with 3x and y with x
sin 3x + sin x = 2sin ((3π₯ + π₯)/2) cos ((3π₯ β π₯)/2)
sin 3x + sin x = 2 sin 2x cos x
Similarly ,
We know that
cos x β cos y = β2 sin ((π₯ + π¦)/2) sin ((π₯ β π¦)/2)
Replacing x with 3x and y with x
cos 3x β cos x = β2 sin ((3π₯ + π₯)/2) sin ((3π₯ β π₯)/2)
cos 3x β cos x = β2 sin 2x sin x
Now solving L.H.S
(sin 3x + sin x) sin x + (cos 3x β cos x) cos x
Putting values from (1) & (2)
= (2 sin 2x cos x) (sin x) + (β2sin 2x) (sin x) (cos x)
= 2 sin 2x cos x sin x β 2 sin 2x sin x cos x
= 0
= R.H.S
Hence L.H.S = R.H.S
Hence proved

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.