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Show that the function $f$ defined by $f(x)=\sin(x-\pi/3)\sin(x+\pi/3)$ for all $x \in R $ has a minima at $x=\pi/6$ and maxima at $x=\pi/2$ and $x=-\pi/6$

My attempt: $$f'(x)=\cos(x-\pi/3)\sin(x+\pi/3)+\sin(x-\pi/3)\cos(x+\pi/3)$$ $$=\sin(2x)$$ $$f'(x)=0\Rightarrow \sin2x=0$$ $$2x=n\pi,n\in Z$$

I can't figure out how $\pi/6$ comes as a critical point too

John Hughes
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PiGamma
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2 Answers2

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$$ \sin a \sin b = \frac{1}{2} \left( \cos (a-b) - \cos (a+b)\right) $$ So your function is $$ f(x) = \frac{1}{2} ( \cos (-2\pi/3 ) - \cos(2x)) $$ which is just a cosine, slightly scaled and translated vertically.

As such, its maxima and minima come at multiples of $\pi$. So those $\pi/6$ values are just plain wrong. Your work is fine.

José Carlos Santos
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John Hughes
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Using Prove $ \sin(A+B)\sin(A-B)=\sin^2A-\sin^2B $,

$$\sin\left(x-A\right)\sin\left(x+A\right)=\sin^2x-\sin^2A$$

If $A$ is constant, so will be $\sin^2A$ and we know $$0\le\sin^2x\le1$$

For $\sin^2x=1\iff\cos x=0,x=(2n+1)\dfrac\pi2$ where $n$ is any integer

For $\sin x=0,x=m\pi$ where $m$ is any integer

lab bhattacharjee
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  • Nice -- I think I like your formula even more than the one I used, although it's not as obvious that the result is a period-$2\pi$ sinusoid unless you know the double angle formulas. Then again, it's pretty clear where the maxima and minima occur, so it's about a wash. – John Hughes Nov 19 '17 at 19:35