Figuring out angle from trigonometric function without calculator

Question

Consider $\tan^{-1}(2+\sqrt3)$. From this one can figure out that

$$\sin(\theta)=\frac{1+\sqrt3}{2\sqrt2}$$ and

$$\cos(\theta)=\frac{\sqrt3-1}{2\sqrt2}$$

So we now know the sides of the right angle. However, if we try to use the laws of sines or cosines we will run into an obstacle, since we will, again, need to determine the values of trigonometric expressions.

So how can we determine, without using a calculator, the value of $\theta$ in this case?

Answer 1

Look at $$ \sin ( 2\theta ) = 2\sin (\theta ) \cos(\theta)$$

$$ \sin ( 2 \theta ) = 2\frac{1+\sqrt3}{2\sqrt2}\frac{\sqrt3-1}{2\sqrt2}= 1/2$$

We can find $2\theta $ from the equation

$$\sin ( 2\theta ) =1/2$$

Considering the values for $\sin (\theta )$ and $\cos (\ theta )$ we get $$ 2\theta =5\pi /6$$ Thus $$ \theta =5\pi /{12}$$

Answer 2

Consider the complex number $$z = \cos(\theta) + i \sin(\theta) = \frac{\sqrt{3} - 1 + i(1 + \sqrt{3})}{2\sqrt{2}}.$$ Then \begin{align*} z^2 &= \cos(2\theta) + i \sin(2\theta) \\ &= \frac{(\sqrt{3} - 1)^2 - (1 + \sqrt{3})^2 + 2i(\sqrt{3} - 1)(\sqrt{3} + 1)}{8} \\ &= \frac{-4\sqrt{3} + 4i}{8} = \frac{-\sqrt{3} + i}{2} \\ &= \cos\left(\frac{5\pi}{6}\right) + i\sin\left(\frac{5\pi}{6}\right). \end{align*} From here, we conclude that $z = \cos\left(\frac{5\pi}{12}\right) + i\sin\left(\frac{5\pi}{12}\right)$ or $z = \cos\left(-\frac{7\pi}{12}\right) + i\sin\left(-\frac{7\pi}{12}\right)$. Clearly $z$ lies in the first quadrant, so we have $$z = \cos\left(\frac{5\pi}{12}\right) + i\sin\left(\frac{5\pi}{12}\right).$$ In other words, $\theta = \frac{5\pi}{12} + 2\pi k$ for some $k \in \mathbb{Z}$.

Answer 3

There are several ways to proceed. Here is just one. From a standard isosceles right triangle we know that $\sin 45^0 = \cos 45^0=\frac 1 {\sqrt 2}, and, \tan 45^0=1$. By considering a equilateral triangle we find $\sin 60^0=\frac {\sqrt 3} 2, and, \cos 60^0=\frac 1 2, so, \tan 60^0=\sqrt 3$. Now we can also write that $$\tan(A+B)=\frac{\tan A +\tan B}{1-\tan A\tan B}$$.

If we let $A=45^0, and, B=30^0 $, then we find that, after some algebra, $\tan 75=2+\sqrt 3$.

Answer 4

$$2+\sqrt3=\csc2x+\cot2x=\cot x=\tan(90^\circ-x)$$

Here $2x=360^\circ n+30^\circ$ where $n$ is any integer

Use this

Answer 5

Use the Law of tangents: $$\frac{a-b}{a+b}=\frac{\tan \frac{\alpha-\beta}{2}}{\tan \frac{\alpha+\beta}{2}}.$$ Consider the right angled triangle:

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The Law of tangents: $$\begin{align}\frac{\frac{\sqrt{3}+1}{\sqrt{2}}-\frac{\sqrt{3}-1}{\sqrt{2}}}{\frac{\sqrt{3}+1}{\sqrt{2}}+\frac{\sqrt{3}-1}{\sqrt{2}}}&=\frac{\tan \frac{\theta-(90^\circ-\theta)}{2}}{\tan \frac{\theta+(90^\circ-\theta)}{2}} \Rightarrow \\ \frac{1}{\sqrt{3}}&=\frac{\tan (\theta-45^\circ)}{1}\Rightarrow \\ \tan(\theta-45^\circ)&=\frac{1}{\sqrt{3}} \Rightarrow \\ \theta -45^\circ&=30^\circ \Rightarrow \\ \theta &= 75^\circ.\end{align}$$

Answer 6

$\arctan(2+\sqrt{3})=75^\circ$. Consider the following configuration:

If the length of each red segment is $1$, the length of the blue segment is $\sqrt{3}$. It follows that $\arctan\left(\frac{1}{2+\sqrt{3}}\right)$ is the orange angle, which is one fourth of $60^\circ$ by symmetry and the relation between angle at the center/angle at the circumference.