Asked by Gabriel Gad
Prove That Cos3A = 4Cos A - 3Cos A
Answers
Answered by
Bosnian
That is not correct.
cos 3 A = cos ( 2 A + A )
= cos ( 2 A ) ∙ cos A - sin ( 2 A ) ∙ sin A =
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Remark:
cos ( 2 A ) = cos² A - sin² A =
cos² A - ( 1 - cos² A ) =
cos² A - 1 + cos² A =
2 cos² A - 1
sin ( 2 A ) = 2 ∙ sin A ∙ cos A
_______________________
( 2 cos² A - 1 ) ∙ cos A - 2 sin A ∙ cos A ∙ sin A =
2 cos² A ∙ cos A - 1 ∙ cos A - 2 sin A ∙ sin A ∙ cos A =
2 cos³ A - cos A - 2 ∙ sin² A ∙ cos A =
2 cos³ A - cos A - 2 ∙ cos A ∙ sin² A =
2 cos³ A - cos A - 2 ∙ cos A ∙ ( 1 - cos² A ) =
2 cos³ A - cos A - 2 ∙ cos A ∙ 1 - 2 ∙ cos A ∙ ( - cos² A ) =
2 cos³A - cos A - 2 ∙ cos A + 2 ∙ cos³A =
2 cos³A + 2 ∙ cos³A - cos A - 2 ∙ cos A =
4 cos³ A - 3 cos A
So:
cos 3 A = 4 cos³ A - 3 cos A
cos 3 A = cos ( 2 A + A )
= cos ( 2 A ) ∙ cos A - sin ( 2 A ) ∙ sin A =
_______________________
Remark:
cos ( 2 A ) = cos² A - sin² A =
cos² A - ( 1 - cos² A ) =
cos² A - 1 + cos² A =
2 cos² A - 1
sin ( 2 A ) = 2 ∙ sin A ∙ cos A
_______________________
( 2 cos² A - 1 ) ∙ cos A - 2 sin A ∙ cos A ∙ sin A =
2 cos² A ∙ cos A - 1 ∙ cos A - 2 sin A ∙ sin A ∙ cos A =
2 cos³ A - cos A - 2 ∙ sin² A ∙ cos A =
2 cos³ A - cos A - 2 ∙ cos A ∙ sin² A =
2 cos³ A - cos A - 2 ∙ cos A ∙ ( 1 - cos² A ) =
2 cos³ A - cos A - 2 ∙ cos A ∙ 1 - 2 ∙ cos A ∙ ( - cos² A ) =
2 cos³A - cos A - 2 ∙ cos A + 2 ∙ cos³A =
2 cos³A + 2 ∙ cos³A - cos A - 2 ∙ cos A =
4 cos³ A - 3 cos A
So:
cos 3 A = 4 cos³ A - 3 cos A
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