Can someone please help this problem as its REALLY puzzling me Watch

futureaussiecto
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#1
Report Thread starter 12 years ago
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In the attachment (paper + ms included)

Q3b

how do they get that it equals 2root2costheta.2root2costhetadth eta

i understand that dx/dtheta = 2root2costheta


i am lost as what happened to the 8 minus bit???

I thought root(A-B) isnt equal to rootA - rootB
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JeremyC
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They substituted \large{ x = 2 \sqrt2 \sin\theta} into \large{\sqrt{8-x^2}} which gives \large{\sqrt{8-8\sin^2\theta} = \sqrt{8(1-\sin^2\theta)}}.

Since \large{\cos^2\theta = 1 - \sin^2\theta}, we get:

\large{\sqrt{8-x^2} = \sqrt{8(1-\sin^2\theta)} = \sqrt{8\cos^2\theta} = 2\sqrt2 \cos\theta} (taking the positive square root).

I hope that helps.
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