Just out of interest, I was experimenting different angles of $A$, $B$ such that $0\leq A\leq B\leq\frac{\pi}{2}$ and it seems to me that everytime this following inequality holds:
$$\sin(A-B)\geq \sin(A)-\sin(B).$$
I was wondering why is this the case? I feel like I am just missing something really trivial. I tried to use double angle formula but I was stuck when bounding $\sin(A) \cos(B)-\sin(B)\cos(A)$.
Could someone please point it out what is missing for me? Thank you so much!!
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$\begingroup$You cannot prove it, since it is not true. If $B=\frac\pi2$ and $0<A<\frac\pi2$, then\begin{align}\sin(A-B)-\sin(A)+\sin(B)&=\sin\left(A-\frac\pi2\right)-\sin(A)+1\\&=1-\cos(A)-\sin(A)\\&=1-\sqrt2\cos\left(A-\frac\pi4\right)\\&<0.\end{align}
$\endgroup$ 0 $\begingroup$As noted in other answers, OP has the inequality (or else the angles) reversed. Here's trigonographic proof:
$$0\leq A\leq B\leq 90^\circ \qquad\to\qquad\sin(B-A)\;\geq\;\sin B-\sin A$$
$\endgroup$ 1 $\begingroup$It is exactly the other way around: For $0\leq A\leq B\leq\frac{\pi}{2}$ is$$ \sin(B) - \sin(A) = 2 \cos\left( \frac{B+A}{2}\right)\sin\left( \frac{B-A}{2}\right) \\ \le 2 \cos\left( \frac{B-A}{2}\right)\sin\left( \frac{B-A}{2}\right) = \sin(B-A) \, , $$using a sum-to-product identity and a double-angle formula. The inequality holds because $\cos$ is decreasing on $[0, \pi/2]$.
Multiplication with $(-1)$ gives$$ \sin(A) - \sin(B) \ge \sin(A-B) \, . $$
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