How to prove cos 3x=4cos³ x-3cos x?

Question

How to prove cos 3x=4cos³ x-3cos x USING De Moivre's formula? thanks

Answer 1

de Moivre's theorem is directly resultant on the polar exponential.
That is e^ix = cos x + i sin x
where i is V(-1)
Note that,
(e^ix)^n = e^inx = e^ i(nx) = cos nx + i sin nx = (cos x + i sin x)^n

Now for n=3
(cos x + i sin x)^3 = cos 3x + i sin 3x
Now as you can see the left hand side of the equation can be expanded using the binomial theorem(or just expand algebraically)
When you expand it group the terms with real coefficients and the terms with imaginary(i) coefficients seperately.
Now equating to the right hand side you should note that the real parts on either side can be equated independantly(as can the imaginary parts but you need the real parts because the required cos 3x has no i coefficient), assuming the identity given is true you will get,
Re {(cos x + i sin x)^3} = Re {cos 3x + isin 3x}
=> 4(cos^3 x) - 3cos x = cos 3x

Hope this helps!

PS- expansion of (x+y)^3 = (x+y) . (x^2 + 2xy+ y^2), etc.

Answer 2

De Moivre's formula states:

(cos(x) + isin(x))^n = cos(nx)+isin(nx)
In particular, when n=3 the above formula gives
(cos(x) + isin(x))^3 = cos(3x)+isin(3x)
Now expand the left side of the equation and equate the real and imaginary parts.
When you equate the real parts on both sides, you get the desired equality.

Answer 3

cos 3x= ( e^(i3x)+e^(-3x) )/2
2 cos 3x = (cos x + i sin x)^3 +(cos x - i sin x)^3
= 2cos³ x - 6 cos x (sin x)^2
= 2cos³ x - 6 cos x (1-cos^2x)
= 8cos³ x - 6 cos x
cos 3x = 4cos³ x-3cos x

Answer 4

= cos 5x = cos (3x + 2x) = (cos 3x).(cos 2x) - (sen 3x).(sen 2x) = (4cos³ x - 3cos x).(cos² x - sen² x) - (3sen x - 4sen³ x).[2.(sen x).(cos x)] = 4.(cos x)^5 - 4.(cos³ x).(sen² x) - 3cos³ x + 3.(cos x).(sen² x) - [6.(sen² x).(cos x) - 8.(cos x).(sen x)^4] = 4.(cos x)^5 - 4cos³ x + 4.(cos x)^5 - 3cos³ x + 3cos x - 3cos³ x - [6cos x - 6cos³ x - 8.(cos x).(1 - cos² x)²] = 8.(cos x)^5 - 10cos³ x + 3cos x - {6cos x - 6cos³ x - 8.(cos x).[1 - 2cos² x + (cos x)^4]} = 8.(cos x)^5 - 10cos³ x + 3cos x - [6cos x - 6cos³ x - 8cos x + 16cos³ x - 8.(cos x)^5] = 8.(cos x)^5 - 10cos³ x + 3cos x - [10cos³ x - 2cos x - 8.(cos x)^5] = 8.(cos x)^5 - 10cos³ x + 3cos x - 10cos³ x + 2cos x + 8.(cos x)^5 = 16.(cos x)^5 - 20cos³ x + 5cos x

Answer 5

write cos3x as cos(2x+x) and then expand using ralation cos(A+B)=cosAcosB-sinAsinB u will get
cos2xcosx-sin2xsinx
again expand
(cosxcosx-sinxsinx)cosx-2sinxcosxsinx
(cos^3)x-3(sin^2)xcosx
(cos^3)x-3cosx(1-(cos^2)x)
4(cos^3)x-3cosx

Answer 6

expand cos(3x)=cos(2x+x) and substitute to sin(2x) and cos(2x)
accordingly.
you can get the answer.
sorry wrong answer.

Answer 7

Look up your math text book

Answer 8

i am no Einstein . why u want to waste time in this complex
situation.

Answer 9

Don't have time for that. Sorry