Posted by Lucy on Tuesday, February 28, 2012 at 12:39am.
You need to find dz/dt=0, so set
4sin2tcos2t-2sintcost =0
Expand using double-angle formulae:
8cos(t)sin(t)(cos(t)^2-sin(t)^2)-2cos(t)sin(t) = 0
Factorize:
-2cos(t)sin(t)(4sin(t)^2-4cos(t)^2+1)
Which gives
t=0, t=π/2, t=π and t=3π/2
when sin(t)=0 or cos(t)=0.
This leaves us with:
4sin(t)^2-4cos(t)^2+1=0
which can be reduced by the substitution
cos(t)^2=1-sin(t)^2
to a quadratic equation equivalent to
sin(t)=sqrt(3/8)
Now that leaves you to check each point as a maximum/minimum or inflection point using second derivatives.
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