I have been asked which of the two quantities $ \sin 28^\circ $ and $ \tan 21^\circ $ is bigger without resorting to a calculator.
I tried taking $ f(x) $ to be
$ f(x) = \sin 4x - \tan 3x $
$ f'(x) = 4\cos 4x - 3\sec^2 3x = \cos 4x(4 - 3\sec^2 3x\sec 4x) $
but to no avail.
I also tried solving $ \tan^2 21^\circ - \sin^2 28^\circ = \tan^2 21^\circ - \sin^2 21^\circ + \sin^2 21^\circ - \sin^2 28^\circ = \tan^2 21^\circ\sin^2 21^\circ + \sin^2 21^\circ - \sin^2 28^\circ $
but again no luck.
There doesn't appear to be a general way of doing this.
Let us say I only know the right angle triangle-based definitions of $ \tan x $ and $ \sin x $.
From right angle triangle I can find that $ \tan 45 = \frac{\text{same length}}{\text{same length}} = 1 $.
Now, I also can manage to find the $ \sin 30 = \frac{1}{2} $ by using the 30-60-90 triangle theorem which can be proved independent of any formulas and using similarities and congruency of triangles.
In brief: $ \tan 45 = 1 $ and $ \sin 30 = \frac{1}{2} $
Now, using half-angle formula I can manage to find the value of $ \tan 22.5 = \sqrt{2} - 1 = 0.4142 $
Solution:
$ t(x) = \tan x \implies t'(x) = \sec^2 x $
Conclusion: The rate of change of values of $ \tan x $ is very high as the value of $ x $ goes on increasing
$ s(x) = \sin x \implies s'(x) = \cos x $
Conclusion: The rate of change of values of $ \sin x $ is very small because the value of $ \cos x $ (over to this $ \cos x $ is decreasing function (as derivative of $ \cos x $ is negative $ \sin $ means the change at $ 30^\circ $ will be less)
Now, We can compare values for
$ \delta t = \tan 22.5 - \tan (22.5-\delta_{deg}) $ and $ \delta s = \sin 30 - \sin (30-\delta_{deg}) $
From their rates of change of values we can conclude the drop of value as $ \frac{\delta t}{\delta s} > 1 $
Now, though it doesn't give you the exact comparison proof but gives the intuitional sense of touch that $ \tan 21^\circ < \sin 28^\circ $ as $ \tan 22.5 $ is already significantly less than $ \sin 30 $.