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If the ratios sine, cosine and tangent are applicable to right angled triangles, then how can they be applicable for the angles of 0, 180 and 360 degrees?
They don't seem to have a right angle, since there more like a flat line.

For example, the cosine of 0° is 1, but doesn't this contradict the required 90% straight angle to make the rule valid (?).
The formula adjacent side / hypothenuse would ultimately "appear" to result into 1 if they would fall together, but what is the proof they actually get there in the scope of trigonometry?
Why can't there be assumed an "infinite approximation" instead, without it ever getting to 1?

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  • $\begingroup$ Your question should be the other way round:- sine/cosine/tangent are used to handle angles from $0^0$ to $90^0$ only; what should I do if I want to find the sine/cosine/tangent of an angle beyond that range. $\endgroup$ – Mick Jan 19 '16 at 9:50
  • $\begingroup$ I assume in that case it just becomes the opposite, negative angle below the 90°. $\endgroup$ – Trace Jan 19 '16 at 10:12
  • $\begingroup$ No. An angle beyond that range could be $100^0$, say. $\endgroup$ – Mick Jan 19 '16 at 10:16
  • $\begingroup$ Of course, but I would use the opposite angle for practical calculations. $\endgroup$ – Trace Jan 19 '16 at 10:27
  • $\begingroup$ If we have a thorough study of the sine/.. /… of the non-normal angles, wouldn’t it be more direct than handling their “opposite” angles? For $\theta$ in the non-normal range, I guess your “opposite” angle is $180^0 - \theta$. How do you handle $\sin (210^0)$ and $\tan (210^0)$ by “opposite” angles? $\endgroup$ – Mick Jan 19 '16 at 10:44
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The $\sin$ function has a definition that goes beyond simply right angled triangles.

The most useful definition for high-school level math is this:

  • $\sin x$ is defined as the $y$-coordinate of a point $P$ on the unit circle for which the length of the arc between $(1,0)$ and $P$ is equal to $x$
  • $\cos x$ is the $x$-coordinate of the same point.
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In spite of it's name, Trigonometry wasn't originally (and still isn't) about triangles (trigonon, in Greek). It was about the relationship between the length of chords of circles and angles of subtension, as the Greeks were primarily interested in Astronomy applications. Right angled triangles were useful in the calculation of these lengths of chords, as reference triangles, if you will.

Later, in India, the notion of the Half-Chord was developed. The Half-Chord is one half of the length of the chord of twice the angle, which is similar to our modern $\sin$ function, although the Indians (and Greeks) used a circle with a more convenient radius (chosen such that the circumference was about the same as the number of minutes in a full revolution).

enter image description here

Today, our Trigonometry is interested in vertical and horizontal offsets on the unit circle, which are analyzed with right angled triangles. For example, the $\sin$ function may refer to the height of the crank pin on a locomotive wheel relative to the center, and similarly, the $\cos$ may refer to the horizontal offset. You can see how triangles may be useful in analyzing these offsets, but shouldn't be considered as "how trig is defined".

enter image description here

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  • $\begingroup$ Thanks, but even your answer appears to suggest that sin, cos are applicable to right angled triangles because of their practical usefulness. So if there is no right angled triangle (as formulated in the question), are these ratios theoretical? What would be the proof that the cos of 0* actually exists as 1, opposed to the thought that the closest would be 0,99999..., an infinite approximation to 1 (for as far as a right triangle exists)? $\endgroup$ – Trace Jan 19 '16 at 18:12
  • $\begingroup$ Who says that a right angled triangle has to have all sides of nonzero length? $\endgroup$ – John Joy Jan 20 '16 at 17:00
  • $\begingroup$ I don't know, it's a part of the question I guess. Can a right angle exist with a nonzero length side? $\endgroup$ – Trace Jan 20 '16 at 18:11
  • $\begingroup$ Try this, draw a line of length 1 cm, then draw a line perpendicular to it of length 0 cm. $\endgroup$ – John Joy Jan 21 '16 at 0:00
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    $\begingroup$ see this image that I just uploaded to my website. cheers :) mathuprising.comlu.com/images/trivial-right-triangle.png $\endgroup$ – John Joy Jan 22 '16 at 14:22

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