# 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

• Reading the "Handbook of Mathematical Functions" on page 194 it has $\,\sin 28.0^\circ = 0.46947\,15627\,85891\,$, and on page 198 it has $\,\tan 21.0^\circ = 0.38386\,40350\,35416.\,$ Does this answer your question? If not, then why? Commented Jan 20, 2022 at 22:20
• @Somos: In fact, there is no need to even get up from one's computer. Googling "sin(28 degrees)" gives $0.46947156278$ and googling "tan(21 degrees)" gives $0.38386403503.$ Commented Jan 22, 2022 at 21:56
• @DaveL.Renfro Yes, but notice that this results page shows the answer appearing above the keyboard of a Scientific Calculator. That is, this simulates the result of typing "sin(28 degrees)" in such a calculator and therefore this conflicts with the question which requires "without resorting to calculator." Of course, there are links to other pages which give the numeric result. My method does not need the use of any electric or mechanical devices. Commented Jan 22, 2022 at 22:21
• @Somos Well I don’t have the book Commented Jan 23, 2022 at 4:05
• @Somos: Oops, I didn't see that. I assumed from your comment that the OP forgot to indicate the "rules of engagement" for the problem. Maverick, as for not having the book, there are hundreds of freely available digitized algebra and trig books from the 1800s that have basic trig tables at the back of the book, not to mention extensive "tables of values" books (some running to many volumes). I said 1800s, rather than before early 1980s (rough cut-off at which pretty much every earlier such book had tables at the back), because most everything from 1800s is not behind a paywall. (continued) Commented Jan 23, 2022 at 11:06

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 $$\sin 30 = \frac 12$$ by using $$30^0-60^0-90^0$$ theorem which can be proved independent of any formulas and using similarities and congruency of triangles

• In brief: $$\tan 45 = 1$$ and $$\color{blue}{\sin 30 = \frac 12}$$

Now, using half-angle formula I can mange to find the value of $$\color{blue}{\tan 22.5 = \sqrt 2 -1 = 0.4142}$$

• 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 $$30^0$$ 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 $$\color{blue}{\tan 21^0 < \sin 28^0}$$ as $$\tan 22.5$$ is already significantly less than $$\sin 30$$

You may ask that $$\delta$$ is significantly large as compared to the infinitesimal small deltas that we use in calculus then yes it's true but here you are checking for degrees so when you convert to radians they are as significantly small as $$\delta_{rad} = \frac \pi{180}\delta_{degree}$$

• I don't know if this answer will match the exact need of your solution but I just tried to prove that $\tan 21^0 < \sin 28^0$🤞 Commented Jan 20, 2022 at 13:38

I could only think of using approximations.

$$\Delta y\approx\frac{dy}{dx}\Delta x$$

Therefore $$\sin(\frac{28\pi}{180})\approx\sin(\frac{30\pi}{180})-\frac{d}{dx}\sin x\bigg|_{x=\frac{\pi}{6}}\frac{2\pi}{180}=\frac{1}{2}-\frac{3\pi}{180}$$

Now using $$\tan 22.5^{\circ}=\sqrt2-1=0.414$$

It is clear that $$\sin 28^{\circ}>\tan 22.5^{\circ}$$ and since $$\tan x^{\circ}$$ is an increasing function , therefore $$\tan 22.5^{\circ}>\tan 21^{\circ}$$.

Hence $$\sin 28^{\circ}>\tan 21^{\circ}$$

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 $$\sin 28^\circ > \tan 21^\circ$$.

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).

For every sine, the bigger the angle, the bigger the value. Sine 1 is 0.0174... Sine 2 is 0.034... And sine 3 is 0.052... . Sine 45 is 0.7071... Sine 46 is 0.7193... Sine 47 is 0.7313... Sine 48 is 7431... Sine 49 is 0.7547... Sine 50 is 7660... Sine 51 is 0.7771... And finally Sine 52 is 0.7880... (All values are in degrees mode) Used for reference only, and these values are from a modern scientific calculator. Also, just to tell you, sine 0 is 0 and sine 90 is 1. We also know that it applies to the tangent ratio. For example, tan 1 is 0.0174... and 0.0349... Now, Tan 21 is a little bigger than sin 22, because tan = sinθ / √(1-sin2θ). So sin 22 is smaller than sin 28 or in other words, sin 28 is greater than tan 21.