8 methods for three by one

Proof

From the diagram, $a =\tan^{-1}(1/3), b =\tan^{-1}(1/2)$, and $c = \tan^{-1}(1)$.

We have to prove $a+b=c$, which is the same as proving that $\tan^{-1}(1/3)+\tan^{-1}(1/2)=\tan^{-1}(1)$.

Note that

$$

\begin{eqnarray}

\tan\left(\tan^{-1}\left(\frac{1}{3}\right)+\tan^{-1}\left(\frac{1}{2}\right)\right)

&=&\frac{\tan\left(\tan^{-1}\left(\frac{1}{3}\right)\right)+\tan\left(\tan^{-1}\left(\frac{1}{2}\right)\right)}

{1-\tan\left(\tan^{-1}\left(\frac{1}{3}\right)\right)\tan\left(\tan^{-1}\left(\frac{1}{2}\right)\right)}\cr &=&\frac{\frac{1}{3}+\frac{1}{2}}{1-\frac{1}{3}\cdot\frac{1}{2}}\cr &=&1 

\end{eqnarray}

$$

Proof

 

Using Pythagoras we can calculate the lengths of the diagonal lines:

 

Image

 

 $$

\sin a = \frac{1}{\sqrt{10}}\quad \cos a = \frac{3}{\sqrt{10}}\quad \sin c = \frac{1}{\sqrt{2}}\quad \sin b = \frac{1}{\sqrt{5}}\quad \cos b = \frac{2}{\sqrt{5}}

$$

 

 

Using the identity $\sin(a+b) = \sin a \cos b + \sin b \cos a$ we see that

$$

\begin{eqnarray}

\sin(a+b) &=&   \frac{1}{\sqrt{10}}\cdot\frac{2}{\sqrt{5}}+\frac{1}{\sqrt{5}}\cdot \frac{3}{\sqrt{10}}\cr

&=& \frac{2+3}{\sqrt{10\cdot 5}}\cr

&=& \frac{5}{\sqrt{50}}\cr

&=&\frac{1}{\sqrt{2}}\cr

&=& \sin c

\end{eqnarray}

$$

 Hence the result is proved.

 This method required us to extend the diagram as follows:

 

Image

 

From the altered diagram it can be seen that $x=3+2=5$. 

 

Next, $d$ can be found using the cosine rule $c^2= a^2+b^2-2ab \cos C$.

 

Substitution of these values gives

$$

\cos d = \frac{-1}{\sqrt{2}}

$$

Thus $d=135^\circ$. Therefore $a+b = 180^\circ - d = 45^\circ = c$

 

 

Hence the result is proved.

Image

 

 

 

Image

 

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Hence the result is proved.

 

Let $\bf{d} = (x, y)$ be any point in the $x-y$ plane. Let $\bf{d}_1$ be the point obtained by rotating $a^\circ$ about the origin.

  

Image

 

 

 Let $\bf{d}_2$ be the point obtained by rotating $\bf{d}_1$ by $b^\circ$ around the origin. 

 

 

Image

Finally, let $\bf{d}_2$ be the point obtained by rotating $\bf{d}$ by $c^\circ$ around the origin.

 

 

Image

 

Hence a rotation by $a+b$ is the same as a rotation by $c$ degrees. Hence, $a+b=c$, as none of the angles is greater than $90^\circ$.  

 

 

Image

By looking at the leftmost unit square this diagram can be drawn

 

 

Image

 

 $a+b=c \Leftrightarrow x+x+y  = x+y+z \Leftrightarrow x=z$

Hence it must be proven than $x=z$:

 

 

 

 Split triangle ADE into two right angled triangles ADF and EDF

 

 

Image

From this it can be seen that

$$

EF = DF = \frac{\sqrt{2}}{4}

$$

and

$$

AF = AE-FE = \sqrt{2} -\frac{\sqrt{2}}{4} = \frac{3\sqrt{2}}{4}

$$

  

Thus

$$

AF:AB = \frac{3\sqrt{2}}{4}:1

$$

and

$$

DF:BC = \frac{\sqrt{2}}{4}:\frac{1}{3} = \frac{3\sqrt{2}}{4}:1

$$

Image

 

 

 Hence ADF and ABC are similar triangles and, therefore, $x=z$. Thus, $a+b=c$.

 The coordinate geometry proof is based on the following diagram; the gradients are easy to read off the original image.

Image

  

 Let $B$ be the point $( x_B ,y_B )$ on the line $y = -\frac{1}{2} x$  at a distance of $1$ from the origin. Since $\sqrt{x^2_B+y^2_B} = 1$ and $y_B = -\frac{1}{2}x_B$ we have

$$

\begin{eqnarray}

\sqrt{x^2_B+\frac{1}{4}x^2_B} &=& \sqrt{\frac{5}{4}x^2_B} = 1\cr

\Rightarrow x_B = \frac{2}{\sqrt{5}}\cr

\Leftrightarrow y_B = \frac{-1}{\sqrt{5}}

\end{eqnarray}

$$

Thus,

$$

B=\left(\frac{2}{\sqrt{5}}, \frac{-1}{\sqrt{5}}\right)

$$

We can now determine the equation of the line through $B$ perpendicular to the line $y = -\frac{1}{2} x$ to be

$$

y = 2x-\sqrt{5}.

$$

This line intersects $y=\frac{1}{3} x$ at $A$ which we can calculate to be

$$

A= \left(\frac{3}{\sqrt{5}}, \frac{1}{\sqrt{5}}\right)

$$

Now that we konw points $A$ and $B$ we can calculate the distance between them as

$$

|AB| = \sqrt{\left(\frac{3}{\sqrt{5}}-\frac{2}{\sqrt{5}}\right)^2 +\left(\frac{1}{\sqrt{5}}+\frac{1}{\sqrt{5}}\right)^2}= \sqrt{\frac{1}{5}+\frac{4}{5}}=1

$$

Referring back to the diagram shows that $a+b=45^\circ$ and the result is proved.

This method uses aspects of other proofs presented (and therefore could be shortened) but we include it because it gives a nice example of how complex numbers, matrices and all other methods fit nicely together.

 

 

 Let $r=\sqrt{x^2+y^2}$.

 

Now, the complex number corresponding to the point with coordinates $(x, y)$ in the argand diagram can be written as

$$

z= x+iy = r\exp\left(i\tan^{-1} \left(\frac{y}{x}\right)\right)

$$

Let $z'$ be the complex number obtained by rotating $z$ by $2(a+b)$ degrees. Then we can use the identities $\sin(a+b) = \sin a \cos b +\cos a\sin b$ and $\cos(a+b) = \cos a \cos b-\sin a \sin b$ to see that:

$$

\begin{eqnarray}

z' &=& r\exp\left(i\tan^{-1}\left(\frac{y}{x}\right)+2i(a+b)\right)\cr

&=& r\left[\cos\left(\tan^{-1}\left(\frac{y}{x}\right)+2(a+b)\right)+i\sin\left(\tan^{-1}\left(\frac{y}{x}\right)+2(a+b)\right)\right]\cr

&=&r\left[\cos\left(\tan^{-1}\left(\frac{y}{x}\right)\right)\cos\left(2(a+b)\right)-\sin\left(\tan^{-1}\left(\frac{y}{x}\right)\right)\sin\left(2(a+b)\right)\right]\cr

&& +ir\left[ \sin\left(\tan^{-1}\left(\frac{y}{x}\right)\right)\cos\left(2(a+b)\right)+\cos\left(\tan^{-1}\left(\frac{y}{x}\right)\right)\sin\left(2(a+b)\right) \right]

\end{eqnarray}

$$

 

We  can simplify the arctangents by considering the following diagram:

 

 

 

 

Image

From this we can see that $\cos\left(\tan^{-1}\left(\frac{y}{x}\right) \right)= \frac{x}{r}$ and $\sin\left(\tan^{-1}\left(\frac{y}{x}\right) \right)= \frac{y}{r}$

This simplifies the expression for $z'$ to

$$

z' = x\cos\left(2(a+b)\right)-y\sin\left(2(a+b)\right)+i\left(y\cos\left(2(a+b)\right)+x\sin\left(2(a+b)\right)\right)

$$

 

 Alex and Neil then used some double angle formulae and the fact that $\sin(a+b)=\cos(a+b)=\frac{1}{\sqrt{2}}$ to determine that 

$$

z' = -y+ix

$$

Since this is a complex number rotated through $90^\circ$ we can conclude that $a+b = 45^\circ$