Proof of $\left |\frac{\sin(n+1/2)t}{\sin{t/2}}-\frac{\sin{nt}}{\tan{t/2}}\right| \leq 1$

Question

I need help to proof $$\left |\frac{\sin(n+1/2)t}{\sin{t/2}}-\frac{\sin{nt}}{\tan{t/2}}\right| \leq 1$$

In the direction of chenbai, you should get that $|cos(nt)| \leq 1$ as your final step. — Christopher K, Apr 03 '14 at 02:00

score 1 · Answer 1 · answered Apr 03 '14 at 00:50

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hint:$\left |\dfrac{\sin(n+1/2)t}{\sin{t/2}}-\dfrac{\sin{nt}}{\tan{t/2}}\right|=\left |\dfrac{\sin(n+1/2)t-\sin{nt}\cos{t/2}}{\sin{t/2}}\right|$

answered Apr 03 '14 at 00:50

chenbai

Thank you, I tried the following: $$\left |\dfrac{\sin(n+1/2)t}{\sin{t/2}}-\dfrac{\sin{nt}}{\tan{t/2}}\right|=\left |\dfrac{\sin(n+1/2)t-\sin{nt}\cos{t/2}}{\sin{t/2}}\right|=\left|\dfrac{1/2 \sin((N+ 1/2)t)- 1/2 \sin((N- 1/2 )t)}{\sin (t/2)}\right|$$ with $\sin x \cos y = 1/2 \sin (x-y) +1/2 \sin (x+y)$, but now I am stuck... – Niklas Hebe Apr 03 '14 at 08:36
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$sin(x+y)=sin(x)cos(y)+cos(y)sin(x)$,can you go on this way? – chenbai Apr 03 '14 at 12:11
Hi, now I got it. For other readers: you mean $\sin(x+y)=\sin x \cos y +\sin y \cos x$ – Niklas Hebe Apr 03 '14 at 15:47

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