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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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    $60+15$ seems to be closer. – 2017-01-10
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    Okay, I appreciate your comment, but how to be more accurate ? – 2017-01-10
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    $60+15-1.875$ is even closer. – 2017-01-10
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    Hand-held calculators have a number of predefined values (based on *Taylor's formula* ) and use [CORDIC algorithm](https://en.wikipedia.org/wiki/CORDIC). – 2017-01-10
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    Is 1.875 a "standard" angle ? – 2017-01-10
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    what I did on a computer that had no sine, is I used Taylor Series, working in radians for the calcs. Small values of converge more rapidly, so I used your method of sin(A+B) where A was one of the 'known' values. once you've got sine, you can of course use that method to get cos,tan etc Mind you it did do many numerical calculations to do these calcs. – 2017-01-10
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    It is $60/32$, so: yes. – 2017-01-10

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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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    Thanks, the idea of using Taylor expansion looks promising. – 2017-01-10
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    @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. – 2017-01-10
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    @QUANTUM it is beautiful, simple, and good for most everything :D – 2017-01-10
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    @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. – 2017-01-10
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    No probs, at least you explained the method ... – 2017-01-10
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    Wouldn't it be better to use $a=0$ in the standard Taylor Series Notation ? That makes everything a whole lot simpler , allowing to cancel out the odd positioned $\sin$ terms and reducing $\cos(a)$ to 1 . – 2017-01-30
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    @Nirbhay. The convergence would be much slower. – 2017-01-30
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    @ClaudeLeibovici Is there any "way" to actually know how rapidly or slowly the convergence would be there for a particular value of $a$ without actually calculating ??? – 2017-01-31
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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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    How did you arrive at this approximation formula ? – 2017-01-10
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    @QUANTUM it is an approximate version of the answer above. – 2017-01-10