Do trigonometry mentally

Question

I know that sound a bit stupid, but is there a way to calculate basic trigonometric functions mentally like: $$ \sin(x) $$ $$ \cos(x) $$ $$ \tan(x) $$ $$ \arcsin(x) $$ $$ \arccos(x) $$ $$ \arctan(x) $$ I know it's a lot of functions, but if you can give me a simple and nice writen answer, that would be very nice!

Answer 1

You can use the Taylor Series expansion. For example, $$\sin x= x - \frac{x^3}{3!}+\frac{x^5}{5!}+\dots$$ So as $x \approx 0$ $\sin x \approx x-\frac{x^3}{6}$. Since the error grows a bit larger as $x$ grows, you may also want to use the fact that $\sin x$ is a periodic function. Thus $$\sin(x+2 \pi)=\sin x$$ You can calculate $\cos$ and other trigonometric functions in similar ways. For $\cos$, the expansion is $$\cos x = 1-\frac{x^2}{2!}+\frac{x^4}{4!}-\dots$$ And since $\tan x =\frac{\sin x}{\cos x}$, you can calculate the tangent as well. Other tools you could use include the double angle formula, and you can approximate using well known values such as $$\sin \frac{\pi}{4}=\frac{\sqrt{2}}{2}, \quad \sin \frac{\pi}{6}=\frac{1}{2}, \quad \sin \frac{\pi}{3}=\frac{\sqrt{3}}{2}$$ Or, somewhat less known but still famous $$\sin \frac{\pi}{12}=\frac{\sqrt{6}-\sqrt{2}}{4}$$ For the inverse trigonometric functions, the Taylor series for $\arcsin$ is as follows $$\arcsin x=x+\frac{1!!}{2!!}\frac{x^3}{3}+\frac{3!!}{4!!}\frac{x^5}{5}+\dots$$Where $n!!$ is the double factorial.Then, we have that $\arccos x=\frac{\pi}{2}-\arcsin x$. Using these, we can approximate the values required.

Answer 2

Sorry, but it can't be done mentally. And I don't know about inverse functions but I can give you my simple method to calculate sin of any angle upto at least one digit of accuracy. First, convert your angle to radians if it's in degrees.

If you want to calculate $sin\theta$ where $\theta$ is in degrees, then first calculate $x=\frac{\pi\theta}{180}$

Now, if $\theta$ lies between 0 and 22.5 degrees, then $sin\theta=x$ approximately. The accuracy decreases as the angle increases.

If $\theta$ lies between 22.5 and 45 degrees, then $sin\theta=x\sqrt{1-(\frac{x}{2})^2}$ approximately. The closer $\theta$ is to 45 degrees, the lesser the accuracy.

If $\theta$ lies between 45 and 67.5 degrees, then use $sin\theta=\frac{1}{\sqrt{2}}(x-\frac{\pi}{4}+\sqrt{1-(x-\frac{\pi}{4})^2})$. Accuracy decreases as $\theta$ gets closer to 67.5 degrees.

If $\theta$ lies between 67.5 and 90 degrees, use $sin\theta=\sqrt{1-(\frac{\pi}{2}-x)^2}$. Accuracy increases as $\theta$ gets closer to 90 degrees.

Answer 3

Convert the argument to a multiple of $\pi/2\approx1.57$, i.e. express the angle in "quadrants". It is an easy matter to reduce to the first quadrant.

Then

$$\cos x^*\approx 1-x^2$$ is a resonable approximation ($<5\%$ maximum error, green curve).

You can get better with an extra term,

$$\cos x^*\approx(1-x^2)\left(1-\frac{\pi^2-8}8x^2\right)\approx(1-x^2)\left(1-\frac7{30}x^2\right).$$

($<0.3\%$, blue curve.)

The sine is obtained from $\sin x^*=\cos(1-x^*)$ and the tangent from the ratio. The arc sine/cosine can be drawn form the first approximation, but require a square root extraction, which is uneasy.