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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.

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    $\begingroup$ Why do you want to compare these values without a calculator? $\endgroup$ Commented Oct 21, 2024 at 21:25
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    $\begingroup$ Have you learned about the Taylor series for $\sin(x)$ and $\tan(x)$? $\endgroup$ Commented Oct 21, 2024 at 21:28
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    $\begingroup$ 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. $\endgroup$ Commented Oct 21, 2024 at 22:10
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    $\begingroup$ 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. $\endgroup$ Commented Oct 21, 2024 at 22:19
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    $\begingroup$ 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. $\endgroup$ Commented Oct 21, 2024 at 22:36

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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 $\tan(35^\circ) \approx 0.71823$.

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  • $\begingroup$ Very nice. Thank you. $\endgroup$ Commented Oct 22, 2024 at 15:49
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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 $\sin(2\pi/9)<0.66$.

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  • $\begingroup$ I understand. Thank you. $\endgroup$ Commented Oct 22, 2024 at 15:50

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