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I've recently picked up the book Mathematics and it's History by John Stillwell due to a recent curiosity in the history of math.

I started doing one of the exercises in the book and got a little stumped (I'll admit, I'm quite rusty at math).

I'm asked to use this figure ...

enter image description here

... to show that $\cos \theta = \dfrac{1 - t^2}{1 + t^2}$.

I labeled the bottom side of the smaller triangle $a$, and the right side of the smaller triangle as $b$, which makes the bottom side of the bigger triangle $1 + a$.

In these terms, $\cos \theta = a$, since the hypotenuse of the smaller triangle is $1$.

I also came up with the following equations:

$$ a^2 + b^2 = 1 $$ $$ (a + 1)^2 + b^2 = t^2 $$

Solving the first equation for $b$ gives: $$ b = \sqrt{1 - a^2} $$

Substituting this equation into the second equation above gives: $$ (a + 1)^2 + (\sqrt{1 - a^2})^2 = t^2 $$ $$ a^2 + 2a + 1 + 1 - a^2 = t^2 $$ $$ 2a + 2 = t^2 $$ $$ a = \frac{t^2 - 2}{2} $$

As you can tell, $\frac{t^2 - 2}{2}$ is not equal to the given $\frac{1 - t^2}{1 + t^2}$.

Where am I going wrong in my logic? Thanks!

Michael Hardy
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MrDiddly
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  • I think there may actually be some problem: given the figure, at $t=\sqrt{2}$ presumably the segment of length $t$ goes from $(-1,0)$ to $(0,1)$, so that $\theta=\pi/2$ and $\cos(\theta)=0$. I think the problem is probably more with the figure than anything else because the relations $\cos(\theta)=\frac{1-t^2}{1+t^2}$ and hence $\sin(\theta)=\frac{2t}{1+t^2}$ do come up, for instance in https://en.wikipedia.org/wiki/Tangent_half-angle_substitution – Ian Sep 06 '17 at 23:45
  • If $t$ is the length of that chord, then, for non-obtuse $\theta$ (in fact, for all $\theta < 120^\circ$, but that's beside the point), that length is greater than $1$, so that $1-t^2< 0$. But the cosine of a non-obtuse angle is non-negative, so there's something wrong with this problem. – Blue Sep 06 '17 at 23:46
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    Looking at the picture, I thought $t$ stood for the vertical coordinate of the intersection between the $y$ axis and the ´segment between $(-1,0)$ and $(a,b)$, not for the length of the latter segment, as implied by OP's second equation. It complies with the given solution for $ \theta = 0, \pi/2$. – An aedonist Sep 06 '17 at 23:48
  • @Anaedonist Your suggestion is consistent with the analogous figure in the Wikipedia article I linked above (which is much clearer by virtue of its scale). – Ian Sep 06 '17 at 23:50
  • Yes, the answer that was just provided confirms it. – An aedonist Sep 06 '17 at 23:52

1 Answers1

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You never say what $t$ is. I suppose that it is the slope of the line. Since it is a line passing through $(-1,0)$ and $(\cos\theta,\sin\theta)$, you have $t=\frac{\sin\theta}{1+\cos\theta}$. So,$$t^2=\frac{1-\cos^2\theta}{(1+\cos\theta)^2}$$and therefore\begin{align}\frac{1-t^2}{1+t^2}&=\frac{(1+\cos\theta)^2-1+\cos^2\theta}{(1+\cos\theta)^2+1-\cos^2\theta}\\&=\frac{2\cos^2\theta+2\cos\theta}{2+2\cos\theta}\\&=\cos\theta.\end{align}

José Carlos Santos
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    The OP did say what $t$ is in their solution, just not in words, but they apparently interpreted it as the length of that line segment. But in fact it is the y-intercept of that line, which is also necessarily its slope since the x-intercept is -1. – Ian Sep 06 '17 at 23:51
  • It'd be consistent with the diagram's labeling of the radius as $1$, if the $t$ refers to the $y$-intercept. – fleablood Sep 06 '17 at 23:54
  • Yes, I seem to have misinterpreted $t$ as the length of the line, not the slope of the line at that point. That makes a lot more sense. Thank you for the insight! – MrDiddly Sep 06 '17 at 23:55
  • The only way for this to make sense is that $t$ is the $y$-intercept of the line. – Michael Hardy Sep 07 '17 at 01:38