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

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.

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