Find the maximum of $(\cos\theta +1)(\cos\phi +1)-(\sin\theta + \sqrt{3})(\sin\phi + \sqrt{3})$

Question

If I could find the MAXIMUM of following, I'll finish some problem concerned with vector inner product.

$$(\cos\theta +1)(\cos\phi +1)-(\sin\theta + \sqrt{3})(\sin\phi + \sqrt{3})$$

If the CONSTANTS are ZERO, then

$$\cos(\theta+\phi)$$

SO, the maximum will be 1.

but in this case... HOW DO I modify ...

Thanks in advance.

Answer 1

The function is

$$ f(\theta,\phi) = \cos(\theta+\phi) - 2\sin\left(\theta-\frac{\pi}{6}\right) - 2\sin\left(\phi-\frac{\pi}{6}\right) - 2 $$

Taking the partial derivatives

$$ \frac{\partial f}{\partial\theta} = -\sin(\theta+\phi) - 2\cos\left(\theta-\frac{\pi}{6}\right) $$ $$ \frac{\partial f}{\partial\phi} = -\sin(\theta+\phi) - 2\cos\left(\phi-\frac{\pi}{6}\right) $$

Both have to be $0$, so it follows that $$ \cos\left(\theta-\frac{\pi}{6}\right) = \cos\left(\phi-\frac{\pi}{6}\right) $$

Here we have two scenarios

• $\phi-\dfrac{\pi}{6} = \theta - \dfrac{\pi}{6} + 2n\pi \implies \phi = \theta + 2n\pi $

• $\phi-\dfrac{\pi}{6} = \dfrac{\pi}{6} - \theta + 2n\pi \implies \phi = \dfrac{\pi}{3} -\theta + 2n\pi$

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If $\phi = \theta + 2n\pi$ then $$ \sin(2\theta) + 2\cos\left(\theta-\frac{\pi}{6}\right) = 0 $$

There are two solution sets

• $\theta = -0.574 +2m\pi$, where $f = 1.971$

• $\theta = 1.837 + 2m\pi$, where $f = -6.730$

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If $\phi = \dfrac{\pi}{3} -\theta + 2n\pi$ then $$ \sin \frac{\pi}{3} + 2\cos\left(\theta-\frac{\pi}{6}\right) = 0 $$

You don't need to solve this, as $$f\left(\theta,\frac{\pi}{3}-\theta\right) = \cos \frac{\pi}{3} - 2 = -\frac{3}{2}$$