How to approximate $\sin x$ without using trigonometry tables?

Question

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 ?

Answer 1

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.

Answer 2

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).