3

I know that for the same angle, $\tan$ is greater than $\sin$ in the first quarter of unit circle. But for different angles, there is uncertainty. For example, $\sin 40^\circ \gt \tan 32^\circ$, but $\sin 40^\circ \lt \tan 33^\circ$. Can you explain with calculations? Thanks.

J. W. Tanner
  • 63,683
  • 4
  • 43
  • 88
Hüseyin Aygül
  • 47
  • 5
    Why do you want to compare these values without a calculator? – Amogh Oct 21 '24 at 21:25
  • 1
    Have you learned about the Taylor series for $\sin(x)$ and $\tan(x)$? – Ghoster Oct 21 '24 at 21:28
  • 1
    Does expressing it in terms of radicals count as a valid "without calculator" solution? This would make sense as an exercise in a high school trig class. – TurlocTheRed Oct 21 '24 at 22:10
  • @TurlocTheRed: Does expressing it in terms of radicals count as a valid "without calculator" solution? --- The only integer degree angles whose trig values can be expressed in terms of radicals (here I'm interpreting this as "real radicals"; see also here and here) are those divisible by $3,$ and neither $40$ nor $35$ are divisible by $3.$ – Dave L. Renfro Oct 21 '24 at 22:15
  • 2
    Was thinking $\sin(120^\circ)=\sin(3\cdot 40^\circ)=\sqrt{3}/2$. Then an identity converts that into a cubic equation in terms of $\sin 40^\circ$. Then the cubic formula can be solved in terms of radicals. Specifically $x=\sin(40^\circ). -4x^3+3x-\sqrt{3}/2=0$ Then I think that has solutions in radicals. – TurlocTheRed Oct 21 '24 at 22:19
  • The trouble is that the radical solution for this cubic will involve cube roots of (non-real) complex numbers that provably cannot be expressed in terms of arithmetic operations and radicals of real numbers applied to rational number inputs. I'm pretty sure that the radicals the cubic formula gives you in this case don't lead you to anything that can be algebraically rewritten in a way that allows you to make an ordering comparison. – Dave L. Renfro Oct 21 '24 at 22:23
  • 2
    If this arose in a math competition, the fact that $30 + 45 = 75$ (an angle whose trig values are simple quadratic irrationals) might be relevant -- I'm thinking of sum-to-product formulas for products of sine and cosine (one such product shows up when expressing $\frac{\sin 35}{\cos 35} - \sin 40$ as a single fraction). Of course there's the issue of the $5$ degree difference between these two angles, among other things. This is a good example of why giving the context background for a question here (which you haven't done) can be important. – Dave L. Renfro Oct 21 '24 at 22:36
  • I need to explain it with a mathematical solution without Taylor Series (which i know how to do it). But you know we can explain how tan values are always larger than sin values for the same angle in first quarter with a unit circle. It is obvious. I wonder if there is an explanation like that. I see maybe there is none except using Taylor Series or a calculator. Thank you for your thoughts. – Hüseyin Aygül Oct 22 '24 at 15:47

2 Answers2

9

You can use the Taylor series for $\sin(x)$ and $\tan(x)$ if you really have to do this without a calculator.

$$40^\circ = \frac{40\pi}{180} \approx 0.7 \text{ radians}$$ $$35^\circ = \frac{35\pi}{180} \approx 0.6 \text{ radians}$$

A rough estimate gives you $\sin(0.7) \approx 0.7 - \frac{0.7^3}{6} \approx 0.64283$ and $\tan(0.6) \approx 0.6+\frac{0.6^3}{3} = 0.672$

This is not too far off from the actual values of $\sin(40^\circ) \approx 0.64279$ and $\tan(35^\circ) \approx 0.70021$, which is quite remarkable for a pen-and-paper calculation!

In fact, even at $x=90^\circ=\frac{\pi}{2} \text{ radians}$, the value of $x-\frac{x^3}{6}$ is only $7.5\%$ off from $\sin(x)$; on the other hand, errors in $\tan(x)$ become very significant even at $x=45^\circ=\frac{\pi}{4} \text{ radians}$ ($33.3\%$ error).

In such cases you can use the transformation formula $\tan(90^\circ-x) = \cot(x)$ and use the corresponding Taylor series. If you want to go really old-school you can mark a unit circle with a protractor and measure lengths to calculate trigonometric ratios, else I'd say you should stick to the calculator.

Addendum: You could also use linear interpolation between $30^\circ$ and $45^\circ$ to approximate the values of trigonometric functions. I believe this is more efficient than computing Taylor series by hand.

$$\sin(40^\circ) \approx \sin(30^\circ) + \frac{\sin(45^\circ)-\sin(30^\circ)}{45^\circ-30^\circ}(40^\circ-30^\circ) \approx 0.63807$$

Similarly, we get $\tan(35^\circ) \approx 0.71823$.

Amogh
  • 1,718
1

Tangent is concave up in the region with these angles. So using the tangent line at $\pi/6$, we have $\tan(7\pi/36)>\tan(\pi/6)+\frac{\pi}{36}\cdot\sec^2(\pi/6)=\frac{1}{\sqrt3}+\frac{\pi}{27}>0.57+0.11=0.68$.

Sine is concave down in the same region. A similar computation based at $\pi/6$ for sine shows $\sin(2\pi/9)<0.66$.

2'5 9'2
  • 56,991