<s>[INST] Which of the two quantities $\sin 28^{\circ}$ and $\tan 21^{\circ}$ is bigger . I have been asked that which of the two quantities $\sin 28^{\circ}$ and $\tan 21^{\circ}$ is bigger without resorting to calculator.
My Attempt:
I tried taking $f(x)$ to be
$f(x)=\sin 4x-\tan 3x$
$f'(x)=4\cos 4x-3\sec^23x=\cos 4x(4-3\sec^23x\sec 4x)$
but to no avail.
I also tried solving $\tan^2 21^{\circ}-\sin^228^{\circ}=\tan^2 21^{\circ}-\sin^221^{\circ}+\sin^221^{\circ}-\sin^228^{\circ}=\tan^2 21^{\circ}\sin^221^{\circ}+\sin^221^{\circ}-\sin^228^{\circ}$
but again no luck.
There doesn't appear to be a general way of doing this
 [/INST] Working with radians, you want to compare
$$\sin \left(\frac{\pi }{6}-\frac{\pi}{90}\right)\qquad \text{to} \qquad \tan \left(\frac{\pi }{8}-\frac{\pi  }{120}\right)$$ Let $\epsilon=\frac \pi {120}$ and we shall compare
$$\sin \left(\frac{\pi }{6}-\frac 4 3\epsilon\right)\qquad \text{to} \qquad \tan \left(\frac{\pi }{8}-\epsilon\right)$$ Using series around $\epsilon=0$
$$\sin \left(\frac{\pi }{6}-\frac 4 3\epsilon\right)=\frac{1}{2}-\frac{2 }{\sqrt{3}}\epsilon +O\left(\epsilon ^2\right)$$
$$\tan \left(\frac{\pi }{8}-\epsilon\right)=\sqrt 2-1+2 \left(\sqrt{2}-2\right) \epsilon +O\left(\epsilon ^2\right)$$
$$\sin \left(\frac{\pi }{6}-\frac{\pi}{90}\right)- \tan \left(\frac{\pi }{8}-\frac{\pi  }{120}\right)=\left(\frac{3}{2}-\sqrt{2}\right)+\left(4-2 \sqrt{2}-\frac{2}{\sqrt{3}}\right)
   \frac \pi {120} +\cdots$$ which is $0.0862$ as an approximation. The true value is $0.0856$.
No calculator used until the last line (I am too lazy and used my phone).
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