SOLUTION: Hi, I am not sure how to approach this question, any help is appreciated. Thanks!
Prove the identity;
cos 2A − cos 4A = 2cos^2 A − 2cos^2 2A
By substituting A
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-> SOLUTION: Hi, I am not sure how to approach this question, any help is appreciated. Thanks!
Prove the identity;
cos 2A − cos 4A = 2cos^2 A − 2cos^2 2A
By substituting A
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Question 847218: Hi, I am not sure how to approach this question, any help is appreciated. Thanks!
Prove the identity;
cos 2A − cos 4A = 2cos^2 A − 2cos^2 2A
By substituting A= 36, show without using s calculator that
cos36 - cos72 = 1/2
Hence find the value of cos36 in the form a+b√5 where a and b are found. Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! Prove the identity;
cos 2A − cos 4A = 2cos^2 A − 2cos^2 2A
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Working only with the left side::
2cos^2(A)-1 - cos(2(2A))
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2cos^2(A)-1 -[2cos^2(2A)-1]
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= 2cos^2(A) - 2cos^2(2A)
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By substituting A= 36, show without using s calculator that
cos36 - cos72 = 1/2
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Factoring the right side of the identity you get:
2(cos(A)-(cos(2A))(cos(A)+cos(2A)) = cos(2A)-cos(4A)
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Let A = 36
(cos(36)-cos(72) = [cos(72)-cos(144))/[2(cos(36)+cos(72)]
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Note: cos(144) = -cos(36)
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cos(36)-cos(72) = [cos(72)+cos(36)]/[2(cos(36)+cos(72))
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cos(36)-cos(72) = 1/2
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Hence find the value of cos(36) in the form a+b√5 where a and b are found.
Since cos(36)-cos(72) = 1/2
cos(36)- cos(2(36)) = 1/2
cos(36)- [2cos^(36)-1] = 1/2
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2cos^2(36)-cos(36)-1/2) = 0
4cos^2(36) - 2cos(36) -1 = 0
cos(36) = [2 +- sqrt(4-4*4*-1)]/8
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cos(36) = [2 +- sqrt(20)]/8
cos(36) = [2 +- 2sqrt(5)]/8
cos(36) = (1+-sqrt(5))/4
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Cheers,
Stan H.
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