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I have F(theta)=[cos(pi.cos(theta)+n*pi/4)+(-1)^n.cos(3pi.cos(theta)+n*pi/4)]² How do i find theta as function of n pi/4 to maximize F(theta)?
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n can assume values 0,1,2,3
If n is even, we can try simplifying F using the identity: cos(A) + cos(B) = 2cos((A+B)/2)cos((A-B)/2) For odd n it will be the sine function with a -2 in the front.
i was trying to derivate it
Yes, \(\Large \frac{dF}{d\theta} = 0\). But I was wondering if this has to be solved separately for even n and odd n and perhaps the solution, if possible, combined in the end.
Can you solve for n odd??
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is this the function? \[F(\theta)=\left[\cos\left(\pi\cos\theta+\frac{n\pi}{4}\right)+(-1)^n\cos\left(3\pi\cos\theta+\frac{n\pi}{4}\right)\right]^2\]
@Orion1213 , yes
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