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

We can use some trigonometric identities to help us here. We need to know three things here:

*

*$\sin 2\theta = \frac{2\tan \theta}{1+\tan^2\theta}$

*$\tan 2\theta = \frac{2\tan \theta}{1-\tan^2\theta}$

*$\tan 3\theta = \frac{3\tan\theta-3\tan^3\theta}{1-3\tan^2\theta}$
You were onto something when looking at $\sin4\theta$ and $\tan3\theta$. We can use these definitions to find that $\sin4\theta = \frac{2\tan2\theta}{1+\tan^22\theta} = \frac{2\frac{2\tan \theta}{1-\tan^2\theta}}{1+(\frac{2\tan \theta}{1-\tan^2\theta})^2}$. Simplifying this expression gives us the more manageable $\sin4\theta=\frac{4\tan\theta(1-\tan^2\theta)}{(1+\tan^2\theta)^2}$.
Now we take $\theta$ to be equal to $7^\circ$ or $\frac{7\pi}{180}$. Define $x=\tan\frac{7\pi}{180}$, and look at the following inequality:

*

*$\frac{4x(1-x^2)}{(1+x^2)^2}>\frac{3x-x^3}{1-3x^2}$
This is actually the case assuming that $\sin 28^\circ > \tan 28^\circ$. This expression is equivalent to

*

*$4x(1-x^2)(1-3x^2) > (3x-x^3)(1+x^2)^2$
which in turn is the same as

*

*$12x^5-16x^3+4x>-x^7+x^5+5x^3+3x$
Subtracting the right hand side from the left gives us

*

*$x^7+11x^5-21x^3+x>0$
We can pull out a factor of $x$ to get

*

*$x^6+11x^4-21x^2+1>0$
Something interesting worth noting is that for small $\theta$, $\tan\theta\approx\theta$, and in fact, if $\theta<9^\circ$, the percentage error between the two is less than one percent (https://en.wikipedia.org/wiki/Small-angle_approximation#Error_of_the_approximations). So, we can simply evaluate this expression at $\frac{7\pi}{180}$ instead of $\tan\frac{7\pi}{180}$. If you wish to remain truly calculator-less, I suppose you could further approximate this to $\frac{217}{1800}$ and work it out by hand, but I will be using the expression with pi. Swapping it out for $x$ gives us

*

*$(\frac{7\pi}{180})^6+11(\frac{7\pi}{180})^4-21(\frac{7\pi}{180})^2+1>0$
and running this through WolframAlpha gives us roughly

*

*$0.689>0$
Since this expression is true, it follows that $\sin 28^\circ > \tan 21^\circ$.
