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Mathematics 23 Online
OpenStudy (anonymous):

How do you get from sqrt((4sin^(2)2t)+(16cos^(2)2t)) to sqrt(1+3cos^(2)2t)?

Parth (parthkohli):

\[\sqrt{4 \sin^2 (2t) + 16 \cos^2 (2t)}\]\[= \sqrt{4\sin^2 (2t) + 4 \cos^2 (2t) + 12 \cos^2(2t)}\]\[= \sqrt{4 + 12 \cos^2 (2t)}\]\[= 2\sqrt{1 + 3\cos^2(2t)}\]

OpenStudy (campbell_st):

well I would have said to start split cos \[\sqrt{4\sin^2(2t) + 4\cos^2(2t) + 12\cos^2(2t)}\] which becomes \[\sqrt{4(\sin^2(2t) + \cos^2(2t)) + 12 \cos^2(2t)}\] now sin^2 + cos^2 = 1 so it becomes \[\sqrt{4 + 12 \cos^2(2t)}\] factor it and its \[\sqrt{4(1 + 3 \cos^2(2t)}\] can you simplify from here...?

OpenStudy (anonymous):

Thank you guys!

Parth (parthkohli):

No problem :)

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