= 2(cos(theta) - sin^2(theta) + cos^2(theta))
= 2[cos(theta) - (sin^2(theta) - cos^2(theta))]
not= 2(cos(theta) - 1) since sin^2(theta) - cos^2(theta) not= 1
i need help finding the horizontal tangents on this curve. i got the verticals down correctly, but i cant figure the horizontals out...
C: r=2+2cos(theta)
y=rsin(theta)
y=(2+2cos(theta))sin(theta)
=2sin(theta)+2cos(theta)sin(theta)
dy/d(theta) = 2cos(theta) - 2 sin^2(theta) + 2cos^2(theta)
= 2(cos(theta) - sin^2(theta) + cos^2(theta))
=2(cos(theta) - 1)
2cos(theta) - 2 = 0
2cos(theta) = 2
cos(theta) = 1
(theta) = 0 , 2pi
i was supposed to somehow get pi/3 , 5pi/3 , and pi. where did i go wrong?
so i would have to put that in parenthesis first before i try the trig identity? how would i solve it to find theta?
2[cos(theta) - (sin^2(theta) - cos^2(theta))]
-2[-cos(theta) + (sin^2(theta) + cos^2(theta))]
-2[-cos(theta) + 1]
2cos(theta) - 2 = 0
cos(theta) = 1
i still end up with the same thing ><
no, you won't. i only did the contraction to show you that you would not get 1, but to actually do the problem, we would expand differently.
2[cos(theta) - (sin^2(theta) - cos^2(theta))] = 0
=> 2cos(theta) - 2sin^2(theta) + 2cos^2(theta) = 0
=> 2cos(theta) - 2(1 - cos^2(theta)) + 2cos^2(theta) = 0
=> 4cos^2(theta) + 2cos(theta) - 2 = 0
=> 2cos^2(theta) + cos(theta) - 1 = 0
=> (2cos(theta) - 1)(cos(theta) + 1) = 0
Note: if you simplified correctly using the method you did before, you'd end up with 2(cos(theta) + cos(2*theta))
yes, that would be correct
you expanded the brackets wrong in the first case, and factored out the -1 wrong in the second case.where did i go wrong in simplifying it?
the red plus sign above should have remained a minus sign. if you expand everything you would realize this is the case2[cos(theta) - (sin^2(theta) - cos^2(theta))]
-2[-cos(theta) + (sin^2(theta) + cos^2(theta))]