How do you prove this trig. identity?

Question

cos x + cos2x + cos3x = cos2x(1 + 2cos)

Answer 1

cosx + cos2x + cos3x

rearrange

cosx + cos3x + cos2x

[cos(2x - x) + cos(2x + x)] + cos(2x)

2cos(2x)cos(x) + cos(2x) (since cos(a-b) + cos(a+b) = 2cosa cosb)

=>cos(2x)[2cos(x) + 1]

Answer 2

cos x + cos2x + cos3x = cos2x(1 + 2cosx)
take the left hand side

cos x + cos2x + cos3x
cos x + cos3x + cos2x

using cosA+cosB=2cos{1/2(A+B)}cos{1/2(A-B)}
add cos x& cos 3x
so

cos x + cos3x + cos2x
[2cos{1/2(x+3x)}cos{1/2(3x-x)}]+ cos 2x
[2cos{1/2(4x)}cos{1/2(2x)}]+ cos 2x
[2cos{2x}cos{x}]+ cos 2x
2cos2xcosx+ cos 2x
cos2x(2cosx+ 1)

this is the left hand side, hence proved.

Answer 3

cos2x + cos(3x+x/2)cos(3x-x/2)
=cos 2x + cos2xcosx
=cos2x(1+2cosx)

Answer 4

try using these three trig. identity formulas
sin2x + cos2x=1
1+tan2x=sec2x
cot2x+1=csc2x

maybe that could help you...

Answer 5

cos(A+B) = cosAcosB - sinAsinB
cos(A-B) = cosAcosB + sinAsinB
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cos(A+B) + cos(A-B) = 2cosAcosB

LHS
= cos(x) + cos(2x) + cos(3x)
= cos(2x) + cos(x) + cos(3x)
= cos(2x) + cos(2x-x) + cos(2x+x)
= cos(2x) + 2cos(2x)cos(x)
= cos(2x)(1 + 2cos(2x))
= RHS