## Trig Trouble and Triumph

Young Tirivanhu was trying to do his homework, which included mathematics (otherwise we wouldn't print this story in Zimaths). Uncle Temba sat on a sofa and read the newspaper; he was waiting for Tirivanhu's father Precious to return from work.

Mother Shamiso had just come in to serve Uncle Temba a drink of Mazoe orange juice when she noticed that Tirivanhu's face looked as if he was in pain. ``Hey, Tiri, what's the problem?" - ``I have problems with the sines." - ``So why don't you go to the doctor?" She had understood that he had problems with his sinuses - which wouldn't have been surprising because he also had a cold. A little absent-mindedly, he replied: ``But I have problems with the cosines as well!" - ``Then go to two doctors." Mother Shamiso was preoccupied with balancing the tray on which was a bottle of Mazoe, a jug of water, and a glass. She took them across the room towards the side table next to Uncle Temba. He folded up the newspaper so that his grinning face became visible. ``So, where's the problem, Tiri?" Temba, being an engineer in the construction business, thought he might be of some help. And Tirivanhu had hoped, indeed, that his uncle would. So he began explaining his problem.

``Look, I have two right-angled triangles ABC and A1B1C1 with sides of lengths a,b,c and a1,b1,c1 respectively. The sides c and b as well as c1 and b1 form the same angle x. c and c1 are the hypotenuses of the two triangles.'' - ``So the two triangles are similar; but where's the problem?" Uncle Temba was eager to get to the point. ``Well, they (he meant his teacher and what was written in the book), say that sinx is the ratio of the opposite side of x divided by the hypotenuse. But how do I know that
 ac = a1c1 ?
Is that really so? Is it really true?'' - ``Draw any two similar triangles, measure the corresponding sides and ratios, and you'll see that they'll be roughly the same,'' Uncle Temba was a very practical man. ``Just roughly the same is not good enough for an equation,'' objected Tiri. ``You got a point here, young Tiri, but didn't they teach you that in similar triangles, these ratios are always the same, even if those triangles were not right-angled?'' - ``Yes, they did last year. We were taught that it can be said that the larger triangle is a scale drawing of the smaller one, so [a/(a1)] = [b/(b1)] = [c/(c1)].'' - ``So, [a/c] = [(a1)/(c1)]; where's the problem, Tiri?'' Uncle Temba thought he had solved Tiri's problem. ``Something's fishy about this similarity business - `it can be said that...' , this doesn't sound too convincing to me... I WANT AN EXACT PROOF THAT
 aa1 = bb1 = cc1 ;
THAT'S WHAT I WANT.'' Now they were at the heart of the problem!

Uncle Temba thought for a little while, then he said: ``Okay, let's start by putting the smaller triangle on top of the other, so we get this -'' and he drew the following picture:

``So we assume A1B1C1 is the larger triangle.'' Tiri tried to follow Uncle Temba's attempt to come up with a proof. - ``Now let me ask you this, Tiri: Do you know what's the area of a triangle, and have you learned Pythagoras' Theorem already?'' ``Sure, half base times height, and a2+b2 = c2 !'' Tiri's memory functioned well. - ``Okay, so let's apply this to the following figure which we get from the first one by adding a line parallel to BB1 and starting at C.'' While saying this he drew the figure which finally looked like this:
``Now, let's determine the area of the big triangle in two ways: First by using its short side, and then by adding the areas of the two right angled triangles and the rectangle.'' - ``Okay, okay, let me try to do it, I think I can do it, Uncle Temba, please.'' Tiri needed to actively take part in proving what he wanted. It was like cooking some dish together: one person throws the veggies into the pot, then another pours in some oil and some water, then come the spices, etc. ``So,'' Tiri continued, ``we have":
 b1a12 = ba2 + (b1-b)(a1-a)2 +(b1-b)a
``Now what?'' Tiri seemed to be stuck at this equation. ``Try to simplify the right hand side of the equation,'' Uncle Temba encouraged Tiri. ``Oh yes! so we have'' - (Tiri tried to copy his teacher), ``- after multiplying the whole equation by 2 - '' (he tried to get rid of the fraction):
 b1a1 = ba+b1a1-ba1-b1a+ba+2b1a-2ba
which yields finally ba1 = b1a and thus
 bb1 = aa1
``Now that explains what they meant by scale drawing!'' - ``Not fully,'' Uncle Temba wanted to caution Tiri, ``for we need to show that [b/(b1)] = [a/(a1)] = [c/(c1)]. Now let's apply Pythagoras' Theorem to the original right-angled triangle. To get what we want, just set r: = [b/(b1)] = [a/(a1)], with r standing for ratio. ''

``Oh yes, Uncle Temba, look!"
 a2+b2 = c2          and         a12+b12 = c12
and using rb1 = b, ra1 = a gives c2 = (ra1)2+(rb1)2 = r2(a12+b12) = r2c12 and finally
 c = rc1
``Now we have'' - and Tiri smiled all over his face -
 bb1 = aa1 = cc1
``and sinx = [a/c] = [(a1)/(c1)] and,'' Tiri continued, ``we also have cosx = [b/c] = [(b1)/(c1)], tanx = [a/b] = [(a1)/(b1)], cotx = [b/a] = [(b1)/(a1)], so, really, the definition of the trigonometric functions does not depend on the side lengths of the right-angled triangles, as long as they have the same angle x, that is, as long as they are similar.'' ``In other words, the trigonometric functions are well defined.'' Uncle Temba cited the shorter mathematical way of expressing what Tiri described in his long sentence. ``Now, see whether there is a relationship between sinx and cosx, by using Pythagoras' Theorem.'' And Tirivanhu wrote:
 a2+b2 = c2 Þ ( ac )2+( bc )2 = 1,

 hence        sin2 x+cos2 x = 1
At that moment, Tirivanhu's father Precious arrived home from work, to find his son and his friend celebrating their victory with Mazoe, and his wife looking at them in amazement.

Tirivanhu was satisfied for the moment. But, like any good mathematician, he would soon be asking questions again! This all works well if we operate with acute-angled triangles, that is, if x lies between 0° and 90°. But what if x is an angle greater than 90°? How can we then define the trigonometric functions?

And there is another question: if you take any two (not necessarily right-angled) similar triangles with the respective side lengths a,b,c and a1,b1,c1, is it then true also that
 aa1 = bb1 = cc1 ?
Try to answer this question by proceeding analogously to what Tirivanhu did.

- Herbert Fleischner

File translated from TEX by TTH, version 2.78.
On 18 Mar 2001, 11:12.