Formula for finding $\cos(\theta)$?

Question

Suppose we have that $\cos(\theta) = \sqrt{3} / 2$. Now I know that I can find the value of $\theta$ easily by referring to a trigonometric table but I want to ask is there any other way to find it's value without the table ? A sort of relationship or formula etc.

Answer 1

Without using the table you have to derive the values.

In general, you can build a construction of geometric shapes that demonstrate that the angle is a certain proportion of the angle swept out by a full revolution. Keeping in mind that the angle of a full revolution is known to be 360 degrees or 2π radians you can obtain the angle.

The exact value of $\text{arccos}( \sqrt{3 }/ 2 )$ is $\frac{\pi}{6}$ radians.

So take a look at the construction of the Dodecagon.

If that still isn't very intuitive there is an even nicer construction which demonstrates this particular angle which comes from the bisection of an equilateral triangle.

In an equilateral triangle all of the sides have the same length and so all of the angles are the same. Call their value A radians. We know the interior angles of a triangle add to be $\pi$ radians, and so $A+A+A = \pi$. Therefore the angle on an equilateral triangle is $\frac{\pi}{3}$ radians. When you bisect an equilateral triangle the angle is $\frac{\pi}{6}$ radians. If you bisect an equilateral triangle and flip the pieces to make a rectangle, the diagonal across opposite corners has length equal to the original side length. Let's give that original side a value of 1 for simplicity. The diagonal is length 1, The short side of the rectangle is length $\frac{1}{2}$ since we bisected the triangle. So, by the Pythagorean theorem, the length of the long side of the rectangle $L$ satisfies the equation:

$(1/2)^2 + L^2 = 1^2 $

So $L$ is $\sqrt{3 }/ 2$. We can then use the relationship between the sin and cos of angles and the length of the sides of triangles to find theta so that

$\cos( \theta ) = \sqrt{3 }/ 2$

Answer 2

I think for your purposes,series expansion of $\cos^{-1}(x)$ would do : \begin{align} \cos^{-1}(x) &= \frac{\pi}{2} - \sin^{-1}(x)\\ &=\frac{\pi}{2} - \sum_{n=0}^{\infty} \frac{2n!}{(2^nn!)^2}\frac{x^{2n+1}}{2n+1}; |x|\leq 1 \end{align} Obviously the formula is much more complicated than the values. But I reckon, given that $|x|\leq1$, and the exponent grows "exponentially" quick,the first few terms should do.

Answer 3

https://en.wikipedia.org/wiki/Special_right_triangle

This link basically has everything you are asking about. I can clarify things if you like, if you find anything unclear.