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Hoofbeat
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#1
Report Thread starter 14 years ago
#1
Can anyone help me with this trig identity proof! Attempted it several times from scratch and just can't get anywhere! Thanks

1 + 2cos2x + cos4x = 4cos^2xcos2x
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Jonny W
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#2
Report 14 years ago
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Three steps: (1) rewrite the LHS in terms of cos(x); (2) pull out [cos(x)]^2 as a factor; (3) show that the other factor equals 4 cos(2x).

cos(2x) = 2 [cos(x)]^2 - 1.

cos(4x)
= 2 [cos(2x)]^2 - 1
= 2 (2 [cos(x)]^2 - 1)^2 - 1
= 8 [cos(x)]^4 - 8 [cos(x)]^2 + 1.

So
1 + 2 cos(2x) + cos(4x)
= 8 [cos(x)]^4 - 4 [cos(x)]^2
= 4 [cos(x)]^2 * (2 [cos(x)]^2 - 1)
= 4 [cos(x)]^2 * cos(2x).
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