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OpenStudy (anonymous):
what is the slope of r=squareroot of cos^2 theta at theta= pi/4
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OpenStudy (amistre64):
r = sqrt(cos^2(pi/4)) cos(45) = sqrt(2)/2 square that and we get 2/4 r = sqrt(2/4) r = sqrt(2)/2
OpenStudy (amistre64):
the slope? it is the slope of 45 degrees if I recall correctly. slope = 1
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