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This question already has an answer here:

Prove that $| xy-\sqrt{(1-x^2)(1-y^2)}|\leq1$ where $|x|\leq1$ and $|y|\leq1$

I tried:

$x=\sin\alpha$ and $y=\cos\beta$

$\sqrt{(1-x^2)(1-y^2)}=\sqrt{\cos^2\alpha\sin^2\beta}$ but if I write $\sqrt{\cos^2\alpha\sin^2\beta}=\cos\alpha \sin\beta$, it's not true because $\cos\alpha \sin\beta$ can be negative.

Can someone give me an idea?

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marked as duplicate by Carl Mummert, lab bhattacharjee trigonometry Aug 21 '18 at 14:06

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ I'd write $x=\cos\alpha$ and $y=\cos\beta$ myself, and assume $\alpha$ and $\beta$ are between $0$ and $\pi$. $\endgroup$ – Lord Shark the Unknown Aug 17 '18 at 19:26
  • $\begingroup$ Consider the 2 cases $\leq \pi/2$ and $\geq \pi/2$ and use $\sin(a+b)=\sin(a)\cos(b)+\cos(a)\sin(b)$. $\endgroup$ – Diger Aug 17 '18 at 19:29
  • $\begingroup$ Incidentally, $xy\pm\sqrt{(1-x^2)(1-y^2)}$ are the bounds on $\operatorname{Corr}(A,\,C)$ given $\operatorname{Corr}(A,\,B)=x,\,\operatorname{Corr}(B,\,C)=y$. $\endgroup$ – J.G. Aug 17 '18 at 19:38
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Let $x=\cos\alpha$ and $y=\cos\beta$, where $\{\alpha,\beta\}\subset[0,\pi].$

Thus, $\sin\alpha\geq0$, $\sin\beta\geq0$ and $$\left|xy-\sqrt{(1-x^2)(1-y^2)} \right| = \left| \cos(\alpha + \beta) \right| \leq 1.$$

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\begin{align} & \left| xy-\sqrt{(1-x^2)(1-y^2)}\,\right| \\[10pt] = {} & \left| \sin\alpha\cos\beta - \left| \cos\alpha\sin\beta \right| \, \right| \\[10pt] = {} & \begin{cases} \left| \sin\alpha\cos\beta - \cos\alpha \sin\beta\right| = \left|\sin(\alpha-\beta){} \right| & \text{if } \cos\alpha>0\ \&\ \sin\beta>0, \\[5pt] \left| \sin\alpha\cos\beta + \cos\alpha\sin\beta \right| = \left| \sin(\alpha+\beta) \right| & \text{if } \cos \alpha<0\ \&\ \sin\beta>0, \\[5pt] \text{and similarly in the other two cases.} \end{cases} \end{align}

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$$\begin{align*}xy-\sqrt{(1-x^2)(1-y^2)}&\le 1\\ \Leftrightarrow (xy)^2&\le 1+(1-x^2-y^2+(xy)^2 )+2\sqrt{1-x^2)(1-y^2)} \end{align*}$$

which is true since $|x|\le 1$ then $1-x^2 > 0$ and similarly for $y$.

And do the same for the other side.

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