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I need to find the angle $\theta$ so that: $$\max(\cos^2(\theta),1-\cos^2(45-\theta))$$ is minimized.

OK, so I wrote \begin{align*}f(\theta)&=\max(\cos^2(\theta),1-\cos^2(45-\theta))\\ &=\frac{\cos^2(\theta)+1-\cos^2(45-\theta)+|\cos^2(\theta)-1+\cos^2(45-\theta)|}{2}\end{align*} Then I'm stuck.

cmk
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user 6663629
  • 174

3 Answers3

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Is known that

$$2\cos^2a = 1+\cos 2a.$$

So $$f(\theta) = \dfrac12\max\left(1+\cos2\theta,2-\left(1+\cos\left(\dfrac\pi2-2\theta\right)\right)\right),$$

$$f(\theta) = \dfrac12+\dfrac12\max(\cos2\theta,-\sin2\theta).$$

Since $$\cos2\theta+\sin2\theta = \sqrt2\left(\sin\dfrac\pi4\cos2\theta+\cos\dfrac\pi4\sin2\theta\right) = \sqrt2\sin\left(2\theta+\dfrac\pi4\right),$$ then $$f(\theta) = \begin{cases} \dfrac12+\dfrac12\cos2\theta,\quad\text{if}\quad 2\theta\in\left[2k\pi-\dfrac\pi4,2k\pi+\dfrac{3\pi}4\right]\\ \dfrac12-\dfrac12\sin2\theta,\quad\text{if}\quad 2\theta\in\left[2k\pi+\dfrac{3\pi}4,2k\pi+\dfrac{7\pi}4\right], \end{cases}$$ where $$k\in\mathbb Z.$$

If $$2\theta\in\left(2k\pi-\dfrac\pi4,2k\pi+\dfrac{3\pi}4\right),$$ then $$\min f(\theta) = \dfrac12-\dfrac{\sqrt2}4\quad\text{at}\quad 2\theta=2k\pi+\dfrac{3\pi}4.$$ If $$2\theta\in\left[2k\pi+\dfrac{3\pi}4,2k\pi+\dfrac{7\pi}4\right],$$ then $$\min f(\theta) = \dfrac12-\dfrac{\sqrt2}4\quad\text{at}\quad 2\theta=2k\pi+\dfrac{3\pi}4.$$ Therefore, $$\boxed{\min f(\theta) = \dfrac12-\dfrac{\sqrt2}4\quad\text{at}\quad \theta=k\pi+\dfrac{3\pi}8.}$$

Yuri Negometyanov
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Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$

$$D=\cos^2t-(1-\cos^2(45^\circ-t))=\cos^2t-\sin^2(45^\circ-t)=\cos(45^\circ)\cos(2t-45^\circ)$$

Case $\#1:$

$\cos^2t$ will be maximum if $D\ge0$ if $\cos(2t-45^\circ)\ge0$

$\iff360^\circ n-90^\circ\le2t-45^\circ\le360^\circ n+90^\circ$

$\iff360^\circ n-45^\circ\le2t\le360^\circ n+135^\circ$

Now in the above range $\cos^2t=\dfrac{1+\cos2t}2$ will be minimum $\iff$ if $\cos2t$ is minimum

which happens if $2t=360^\circ n+135^\circ$

Case $\#2:$

Check when $$1-\cos^2(45^\circ-t)=\sin^2(45^\circ-t)$$ will be maximum

lab bhattacharjee
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A little nudge in the right direction (hopefully!):

Now, using the required formula (which?), we can simplify: \begin{align} 1 - \cos^2(45^\circ - \theta) & = \sin^2(45^\circ - \theta) \\ & = \frac{1 - \sin 2\theta}{2}\end{align}

We are now required to find: $$\min(\max(\frac{1+\cos 2\theta}{2}, \frac{1-\sin 2\theta}{2}))$$

We can see that in the interval $[0, \pi]$, the solutions to $\cos 2\theta = - \sin 2\theta$ are: $\frac{3\pi}{8}, \frac{7\pi}{8}$. Also, the points $\frac{3\pi}{4}$ and $\pi$ may be of interest (why?)

Can you take it from here?