In the interval [0, π], what are the inflection points for y=cos²x-sin²x?

[lovetolearn]

In the interval [0, π], what are the inflection points for y=cos²x-sin²x?

In the interval [0, π], what are the inflection points for
y=cos²x-sin²x?


At pi/4 & 3pi/4
x = the zeroes of the function
y' = 2cosx*(-sinx) - 2sinxcosx = -4sinxcosx =-2sin(2x)
y'' = -2*2cos(2x) = 0 cos(2x) = 0
2x = pi/2, 3pi/2
x = pi/4, 3pi/4
x = pi/4, 3pi/4 in that interval

Am I on the correct path?

 

[tkhunny]

Let's rethink (-2)*(2). Are you sure that is zero?

 

[lovetolearn]

Let's rethink (-2)*(2). Are you sure that is zero?

Thank you so much! I corrected it. Is it correct now?

At pi/4 & 3pi/4
x = the zeroes of the function
y' = 2cosx*(-sinx) - 2sinxcosx = -4sinxcosx =-2sin(2x)
y'' = -2*2cos(2x) = -4 cos(2x) = -4
2x = pi/2, 3pi/2
x = pi/4, 3pi/4
x = pi/4, 3pi/4 in that interval

 

[tkhunny]

Just two things:

1) I would have gone straight for cos(2x), rather than the wonderful exercise in the chain rule with sines and cosines. The second derivatve certainly gives this away. Don't forget your trigonometry identities.

2) I would have preferred a general solution and THEN select the ones in the desired interval.

It looks fine, but you could increase in style points.