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An Opening Note : First of all, I want to make this very clear that by the phrase "without using trigonometry tables", I mean without using them to find $\sin$ values of the "non-standard angles" (For example $73.5^\circ$).

Now, its obviously easy to find a somewhat "broader" range for the answer. Taking the example of $73.5^\circ$, its obvious that $\sin 73.5^\circ$ will lie between $\sin 60^\circ$ and $\sin 90^\circ$. But, how can this range be narrowed ?

One answer I think would be between $sin 60^\circ$ and $sin(60^\circ+18^\circ)$.

What to do next ?

Is there a way to find an even better approximation without using "much" calculations ?

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  • $\begingroup$ $60+15$ seems to be closer. $\endgroup$ Jan 10 '17 at 10:29
  • $\begingroup$ Okay, I appreciate your comment, but how to be more accurate ? $\endgroup$
    – user399078
    Jan 10 '17 at 10:30
  • $\begingroup$ $60+15-1.875$ is even closer. $\endgroup$ Jan 10 '17 at 10:42
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    $\begingroup$ Hand-held calculators have a number of predefined values (based on Taylor's formula ) and use CORDIC algorithm. $\endgroup$
    – Bernard
    Jan 10 '17 at 10:55
  • $\begingroup$ Is 1.875 a "standard" angle ? $\endgroup$
    – user399078
    Jan 10 '17 at 11:10
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Suppose that you know the closest angle for which you know the value of trigonometric function. Let us name it $a$. Now develop, around $x=a$, the since function as a Taylor series. You should get

$$\sin(x)=\sin (a)+(x-a) \cos (a)-\frac{1}{2} (x-a)^2 \sin (a)+O\left((x-a)^3\right)$$

Let us apply the formula for $x=\frac{5 \pi }{12}$ using $a=\frac \pi 3$. We should get $$\sin\left( \frac{5 \pi }{12}\right)=\frac{\sqrt{3}}{2}+\frac{\pi }{12}\times\frac 12-\frac 12 \left( \frac{ \pi }{12}\right)^2\frac{\sqrt{3}}{2}\approx 0.967247$$ while the exact value would be $\approx 0.965926$.

For sure, using more terms in Taylor expansion will make the result better and better.

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  • $\begingroup$ Thanks, the idea of using Taylor expansion looks promising. $\endgroup$
    – user399078
    Jan 10 '17 at 11:09
  • $\begingroup$ @QUANTUM. Taylor series is really the simplest approximation of a function close to a given point. There are better things such as Pade approximants (don't worry : you will learn about them). Cheers. $\endgroup$ Jan 10 '17 at 11:12
  • $\begingroup$ @QUANTUM it is beautiful, simple, and good for most everything :D $\endgroup$ Jan 10 '17 at 12:23
  • $\begingroup$ @QUANTUM. I apologize : I did read $75$ instead of $73.5$ (I am almost blind). Use $x-a=\frac{3 \pi }{40}$ in the formula. $\endgroup$ Jan 10 '17 at 12:37
  • $\begingroup$ No probs, at least you explained the method ... $\endgroup$
    – user399078
    Jan 10 '17 at 14:53
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If you want a mental cool approximation, here it is: for $\sin(x)$ you can state that

$$\sin x \approx \frac{x}{60} ~~~~~~~~~~~ x \leq 30^{\circ}$$ $$\sin x \approx \frac{30 + x}{120} ~~~~~~~~~~~ x > 30^{\circ}$$

From this you can derive

$$\cos x = \sin (90-x)$$

and

$$\tan x \approx 0.017\ x$$

With an error from $5$ to $8$ percent (for the tangent one).

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  • $\begingroup$ How did you arrive at this approximation formula ? $\endgroup$
    – user399078
    Jan 10 '17 at 11:11
  • $\begingroup$ @QUANTUM it is an approximate version of the answer above. $\endgroup$ Jan 10 '17 at 12:24

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