# If $\cos A=\tan B$, $\cos B=\tan C$ …

If $\cos A=\tan B$, $\cos B=\tan C$ and $\cos C=\tan A$, prove that $\sin A=\sin B=\sin C$.

My Attempt. Let us consider $x$, $y$ and $z$ as:.

$$x = \tan^2A$$ $$y = \tan^2B$$ $$z = \tan^2C$$ $$\cos^2A = \tan^2B$$ $$\frac {1}{\sec^2A}= \tan^2B$$ $$\frac {1}{1 + \tan^2A} = \tan^2B$$ $$\frac {1}{1 + x} = y$$ $$(1 + x)y = 1\tag{1}$$ Similarly, $$(1 + y)z = 1\tag{2}$$ $$(1 + z)x = 1\tag{3}$$

• See quora.com/… – lab bhattacharjee Sep 21 '16 at 16:36
• @Lab, Is the same process to be applied to get $sinB$ and $sinC$? – pi-π Sep 21 '16 at 16:41

$$\tan^2C=\cos^2B=\dfrac1{1+\tan^2B}=\dfrac1{1+\cos^2A}=\dfrac1{2-\sin^2A}$$

$$\cos^2C=\tan^2A$$ $$\implies\tan^2C=\sec^2C-1=\dfrac{1-\cos^2C}{\cos^2C}=\dfrac{1-\tan^2A}{\tan^2A}=\dfrac{1-2\sin^2A}{\sin^2A}$$

Equating the values of $\tan^2C$ and writing $\sin^2A=x$

$$\dfrac1{2-x}=\dfrac{1-2x}x\implies x^2-3x+1=0\implies x=\dfrac{3-\sqrt5}2$$

As $0\le x\le1$

• So $x = \varphi^2? ...$ the Golden Ratio? – Narasimham Sep 21 '16 at 17:00
• @Narasimham, Yes. From this, we can find $$\sin^2B,\sin^2C$$ to be same – lab bhattacharjee Sep 21 '16 at 17:03
• Interesting,.. would love to see this special "Golden" configuration ! – Narasimham Sep 21 '16 at 17:17
• @Narasimham, just found answers.yahoo.com/question/index?qid=20100812174010AAlk9pP – lab bhattacharjee Sep 21 '16 at 17:36
• Yes fine. The side question is, if they are not of a triangle, than what and how to draw these straight lines ? That should be may be another related question. – Narasimham Sep 21 '16 at 17:47

Define $a := \cos^2 A$, and $b := \cos^2 B$, and $c := \cos^2 C$. Then $$\cos A = \tan B \quad\to\quad \cos^2 A = \tan^2 B = \sec^2 B - 1 \quad\to\quad a = \frac{1}{b} - 1 \quad\to\quad b = \frac{1}{1+a}$$ Likewise, $$c = \frac{1}{1+b} \qquad\text{and}\qquad a = \frac{1}{1+c}$$ so that, adding a slightly-gratuitous $1$, $$1+a \;=\; 1+\frac{1}{1+\frac{1}{1+\frac{1}{1+a}}} \\ \tag{\star}$$ The evident recursion reveals that $1+a$ (and also $1+b$ and $1+c$) is the "$1$s all the way down" continued fraction, which some will recognize as representing the Golden Ratio, $\phi := \frac{1}{2}(1+\sqrt{5}) = 1.618\ldots$. Consequently, $$a = b = c = \phi - 1 = \phi^{-1}$$

We finally calculate $$\sin^2 A = \sin^2 B = \sin^2 C = 1 - \phi^{-1} = \phi^{-2}$$ so that

$$|\sin A| = |\sin B| = |\sin C| = \phi^{-1} = 0.618\ldots$$

That the signs of the sines match is left as an exercise to the reader.

Observe that we'd reach the same conclusion no matter the length of the "loop" of equations. Once $1+a$ appears, at any stage, in the right-hand-side of the counterpart of $(\star)$, the implied continued fraction collapses to the simpler form $$1 + a = 1 + \frac{1}{1+a} \quad\text{, giving}\quad a = \frac{1}{1+a}$$ which represents the one-equation loop, $\cos A = \tan A$.

If no two of $x,y,z$ are equal then all of them are unequal. Assume $x>y>z.$ From $(1) and (2)$ $y - z + y(x - z) = 0$. This implies $y < 0$(since $y - z > 0, x - z > 0)$.....$(4)$. From $(2) and (3)$:$z -x + z(y - x) = 0$ This implies $z < 0$ (since $z - x < 0, y - x < 0)$ $......(5)$. From $(3) and (1)$: $x - y + x(z - y) = 0$ This implies $x > 0$(since $x - y > 0, z - y < 0$)$.........(6)$ (4) and (6) contradicts (1), which has (1 + x)y = 1 → our assumption that $x, y, z$ are all unequal is incorrect and some two of them are equal. If $x = y$, $(2)$ becomes $(1 + x)z = 1 = (1 + z)x$ (by $(3)$) →$1 - z = xz = 1 - x → z = x→ x = y = z$. If $y = z$, it similarly follows that $x = y = z$. So we have: $tan²A = tan²B = tan²C$ →$A = B = C →sinA = sinB = sinC$.

From $\cos A=\tan B, \cos B=\tan C, \cos C=\tan A$,

$\begin{array}\\ \cos A &= \sin B/\cos B\\ &= \sin B/\tan C\\ &= \sin B/(\sin C/\cos C)\\ &= \sin B \cos C/\sin C\\ &= \sin B (\sin A/\cos A)/\sin C\\ \end{array}$

or

$\cos^2 A = \sin A\sin B/\sin C$.

Similarly,

$\begin{array}\\ \cos B &= \sin C/\cos C\\ &= \sin C/\tan A\\ &= \sin C/(\sin A/\cos A)\\ &= \sin C \cos A/\sin A\\ &= \sin C (\sin B/\cos B)/\sin A\\ \end{array}$

or

$\cos^2 B = \sin B\sin C/\sin A$.

Also,

$\begin{array}\\ \cos C &= \sin A/\cos A\\ &= \sin A/\tan B\\ &= \sin A/(\sin B/\cos B)\\ &= \sin A \cos B/\sin B\\ &= \sin A (\sin C/\cos C)/\sin B\\ \end{array}$

or

$\cos^2 C = \sin A\sin C/\sin B$.

Letting $x=\sin A, y = \sin B, z = \sin C$, these become $1-x^2 = xy/z, 1-y^2 = yz/x, 1-z^2 = zx/y$.

Dividing the first two, $\frac{1-x^2}{1-y^2} =\frac{x^2}{z^2}$.

From the last one, $x = \frac{y(1-z^2)}{z}$ so $1-y^2 =\frac{zy}{\frac{y(1-z^2)}{z}} =\frac{z^2}{1-z^2}$ or $y^2 =1-\frac{z^2}{1-z^2} =\frac{1-2z^2}{1-z^2}$.

Similarly, $\frac{1-x^2}{\frac{z^2}{1-z^2}} =\frac{x^2}{z^2}$ or $(1-x^2)(1-z^2) =x^2$ or $1-z^2 =\frac{x^2}{1-x^2}$ or $z^2 = 1-\frac{x^2}{1-x^2} = \frac{1-2x^2}{1-x^2}$.

Similarly, again, we get

$\begin{array}\\ x^2 &= \frac{1-2y^2}{1-y^2}\\ &= \frac{1-2\frac{1-2z^2}{1-z^2}}{1-\frac{1-2z^2}{1-z^2}}\\ &= \frac{1-z^2-2(1-2z^2)}{1-z^2-(1-2z^2)}\\ &= \frac{3z^2-1}{z^2}\\ &= \frac{3\frac{1-2x^2}{1-x^2}-1}{\frac{1-2x^2}{1-x^2}}\\ &= \frac{3(1-2x^2)-(1-x^2)}{1-2x^2}\\ &= \frac{2-5x^2}{1-2x^2}\\ \text{or}\\ x^2-2x^4 &=2-5x^2\\ \text{or}\\ 0 &=2x^4-6x^2+2\\ \text{or}\\ 0 &=x^4-3x^2+1\\ \text{or}\\ x^2 &=\dfrac{3\pm\sqrt{9-4}}{2}\\ &=\dfrac{3\pm\sqrt{5}}{2}\\ \end{array}$

Since $0 \le x^2 \le 1$, we must have $x^2 =\dfrac{3-\sqrt{5}}{2}$, so $x =\pm\dfrac{\sqrt{5}-1}{2}$.

Going through the same thing for $y$ and $z$, we get $x = y = z =\pm\dfrac{\sqrt{5}-1}{2}$.

To determine the signs, multiply $1-x^2 = xy/z, 1-y^2 = yz/x, 1-z^2 = zx/y$ together. We get $(1-x^2)(1-y^2)(1-z^2) =xyz$. Therefore $xyz > 0$ so that 0 or 2 of them are negative and the other(s) are positive.