Ask Uncle Colin: A Cosec Proof | Colin

Dear Uncle Colin

I’m stuck on a trigonometry proof: I need to show that $\cosec(x) – \sin(x) \ge 0$ for $0 < x < \pi$. How would you go about it?

– Coming Out Short of Expected Conclusion

Hi, COSEC, and thank you for your message! As is so often the case, there are several ways to approach this.
The most obvious one

The first approach I would try would be to turn the left hand side into a single fraction: $\frac{1}{\sin(x)} – \sin(x) \equiv \frac{1 – \sin^2(x)}{\sin(x)}$.

The top of that is $\cos^2(x)$, so you have $\frac{\cos^2(x)}{\sin(x)}$.

In the specified region,…

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Tessellations and cuboids | Colin

On a recent1 episode of Wrong, But Useful, Dave mentioned something interesting2: if you take three regular shapes that meet neatly at a point – for example, three hexagons, or a square and two octagons – and make a cuboid whose edges are in the same ratio as the number of sides on each shape (e.g., 6 by 6 by 6 or 4 by 8 by 8), the resulting cuboid has a volume that’s (numerically) the same as its surface area (here, 216 or 256).
Let’s prove it!

The interior angle of a regular $n$-gon is $\pi – \frac{2n}{\pi}$, or $180 – \frac{360}{n}$ if you insist on silly angle measures. In fact, it’s…

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Ask Uncle Colin: Simultaneous Trigonometry | Colin

Dear Uncle Colin,

I’m normally pretty good at simultaneous equations, but I can’t figure out how to solve this for $a$ and $b$.

$\cos(a)-\cos(b) = x$

$\sin(a)-\sin(b) = y$

– Any Random Circle

Hi, ARC, and thanks for your message!

This is, it turns out, a bit trickier than it looks at first glance – and (in most cases) unlikely to lead to exact answers.

So, rather than get to an explicit answer, I’ll talk through a strategy.
Draw a picture

Before anything else, we should draw a picture of what we’re looking at, using the geometric definitions of sine and cosine:


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Are you sure that’s a right angle? | Colin

What’s that, @pickover? Shiver in ecstasy, you say? Just for a change.

Shiver in ecstasy. The sides of a pentagon, hexagon, & decagon, inscribed in congruent circles, form [a] right triangle.

— Cliff Pickover (@pickover) May 20, 2017
That’s neat. But why?

Let’s suppose the circles all have radius 1, without loss of generality. Then the triangle’s side lengths are (in decreasing order) $2\sin\left( \piby 5 \right)$, $2\sin\left( \piby 6 \right)$ and $2\sin\left( \piby {10} \right)$.

Those will form a right-angled triangle if $\sin^2\left( \piby 5…

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Ask Uncle Colin: Shouldn’t this be simple? | Colin

Dear Uncle Colin,

I’ve got a funny square and I can’t find $x$. Can you help?

– Oughta Be Simple, Can’t Unravel Resulting Equations

Hi, OBSCURE, and thanks for your message! You’re right, it ought to be simple… but it turns out not to be.

It is simple enough to set up some equations.

The sine of the ‘big’ triangle’s bottom-right angle is clearly $\frac{1}{x+1}$.

The cosine of the top-left angle of the medium triangle (the one with the right-angle in the top-right, and which is similar to the big triangle) is $\frac{1}{x}$.

Those two angles are the same, which means…

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Lines and squares | Colin

This puzzle presumably came to me by way of @ajk44, some time ago. Thanks, Alison!

The problem, given here, is to find the equations of two lines that complete a square, given:
Two of the lines are $y=ax+b$ and $y=ax+c$
One of the vertices is at $(0,b)$.

The example given has $a=2$, $b=1$ and $c=4$, so that the given lines are $y=2x+1$ and $y=2x+4$.

A perpendicular line to either, passing through $(0,1)$, would be $y= -\frac{1}{2}x + 1$, using the negative reciprocal property. Another possible line, given to us in the question, is $y= -\frac{1}{2}x + \frac{5}{2}$.1
But why does that…

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Ask Uncle Colin: touching cubics | Colin

Dear Uncle Colin,

I’m told that the graphs of the functions $f(x) = x^3 + (a+b)x^2 + 3x – 4$ and $g(x) = (x-3)^3 + 1$ touch, and I have to determine $a$ in terms of $b$. Where would I even start?

– Touching A New Graph Except Numerically Troubling

Hi, TANGENT, and thanks for your message!

For two graphs to touch, there needs to be a point that lies on both curves, and for the derivatives of the two curves at that point to be equal.
The naive way

A naive approach, then, is to differentiate and find out where the gradients are the same, then use that information to find the common…

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