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Emanuel Keith 2022-06-24 Answered
Prove that sin 6 θ 2 + cos 6 θ 2 = 1 4 ( 1 + 3 cos 2 θ )
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Answers (2)

grcalia1
Answered 2022-06-25 Author has 23 answers
sin 6 θ 2 + cos 6 θ 2 = = 1 3 ( 1 cos 2 θ 2 ) cos 2 θ 2
is correct. From here, note that
sin 6 θ 2 + cos 6 θ 2 = 1 3 ( 1 cos 2 θ 2 ) cos 2 θ 2 = 1 3 ( 1 cos θ + 1 2 ) cos θ + 1 2 = 1 3 1 cos θ 2 1 + cos θ 2 = 4 3 ( 1 cos 2 θ ) 4 = 1 + 3 cos 2 θ 4

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Zion Wheeler
Answered 2022-06-26 Author has 11 answers
You may begin factoring:
sin 6 θ 2 + cos 6 θ 2 = ( sin 2 θ 2 + cos 2 θ 2 ) ( sin 4 θ 2 sin 2 θ 2 cos 2 θ 2 + cos 4 θ 2 ) = ( cos 2 θ 2 sin 2 θ 2 ) 2 + sin 2 θ 2 cos 2 θ 2 = cos 2 θ + 1 4 sin 2 θ = 1 4 ( 3 cos 2 θ + 1 ) .
You also may linearise:
cos 2 θ + 1 4 sin 2 θ = 1 2 ( 1 + cos 2 θ ) + 1 8 ( 1 cos 2 θ ) = 1 8 ( 3 cos 2 θ + 5 ) .

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