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Let $f(x,y)=\cos^{2}(x)+\cos^{2}(y)$ over the restriction $x-y=\frac{\pi}{4}$. I was wondering if there is some fast way to solve this problem?? The way im trying this is that over my given constrict $y=x-\frac{\pi}{4}$, this way

$$f(x,y)=f(x,x-\frac{\pi}{4})=\cos^{2}x+cos^2{x-\frac{\pi}{4}}.$$

But I dont clearly see how to proceed from here? Basically, Im trying to solve this in the fashion of the following answer as both functions are similar:

Find extreme values for $f(x,y)=\sin^2(x)+\sin^2(y)$ over the constraint: $(x^2-y^2)^3+(x^2-y^2)=0$.

Thanks!

Cos
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  • $f'(x)=0$ only indicates $f(x)$ has an stationary point there. It can be maximum or minimum or not either of them ( a inflexion point). What Rhys Hughes gave $tan(2x) = 1$ , 2x must be either in first quadrant or in third quadrant. Considering it in first quadrant $2x = \pi/4$ or $x = \pi/8$(which gives maximum) , similarly in third quadrant $2x = \pi+ \pi/4$ or $x = \pi/2 + \pi/8$ (which gives maximum). Also CY Aries gives the same without using calculus. – amitava May 06 '19 at 07:03
  • I wanted to mean $x = \pi/2 + \pi/8$ which gives minimum – amitava May 06 '19 at 07:17

3 Answers3

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This can be found without using calculus.

\begin{align*} f(x,y)&=\cos^2x+\cos^2\left(x-\frac{\pi}{4}\right)\\ &=\cos^2x+\left(\cos x\cos\frac{\pi}{4}+\sin x\sin\frac{\pi}{4}\right)^2\\ &=\cos^2x+\frac{1}{2}(\cos x+\sin x)^2\\ &=\frac{1}{2}(2\cos^2x+\cos^2x+2\sin x\cos x+\sin^2x)\\ &=\frac{1}{2}(\cos2x+1+1+\sin 2x)\\ &=\frac{\sqrt{2}}{2}\cos\left(2x-\frac{\pi}{4}\right)+1 \end{align*}

CY Aries
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If $$f(x)=\cos^2(x)-\cos^2(x-\frac\pi4)$$ Then: $$f'(x)=\cos(2x)-\sin(2x)$$ $$f'(x)=0\implies \tan(2x)=1$$

Use the $\cos(x-y)=\ldots$ formula then differentiate with the product rule to get this derivative.

Rhys Hughes
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  • Thanks a lot! But still have a doubt. You got that f'(x)=$tan(2x)=1$, but $tan(x)=1$ whenever $x=\frac{\pi}{4}$. So $2x=2 (\frac{\pi}{4})+ \pi n)$ as $tan(x)$ has period $\pi$, then $x=\frac{\pi}{2}+ 2 \pi n $ but I need other value to talk about a minimum and maximum. How can I finish the proof? @Rhys Hughes – Cos May 06 '19 at 02:20
  • So $f'(x)=0$ when $\tan(2x)=1$. This means $2x=(\frac\pi4 +n\pi)\to x=\frac \pi8+\frac{n\pi}{2}=\frac{(4n+1)\pi}{8}$. – Rhys Hughes May 06 '19 at 11:18
  • There are no other values, these are the only points which are maxima and minima. – Rhys Hughes May 06 '19 at 11:19
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Using Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$,

$\cos^2x+\cos^2y=1+\cos(x-y)\cos(x+y)=1+\cos\dfrac\pi4\cos(2y-\dfrac\pi4)$

Now for real $A,$ $$-1\le\cos A\le1$$

lab bhattacharjee
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