Sin, Cos, and 360°

In My Humble Opinion
1

I’ve been playing around with traversing using horizontal distances and compass azimuths. I’ve been finding latitudes and departures using Sin and Cos. For example given an azimuth of 32° the Sin of the angle is equal to the opposite over the hypotenuse. When I plug in my HD for the hypotenuse and solve for the opposite I get the departure. Doing the same thing with Cos would get me my latitude.

Ok so far

The problem I’m having trouble visualizing is when I move into different quadrants. To me the Sin of a 110° azimuth should be the same as a 20°, as 110°s is 20° past 90°. The world seems to think differently. The Sin of 110° is the same as 70° as it is 70°s from South.

Does anyone have a rule or way to simplify this for my brain. On the one hand it seems to simplify things because the Sin will always be my latitude and the Cos my departure but it seems counterintuitive to my compass using brain, if that makes sense.

2

Should the sin of 90° be the same as the sin of 0°, since 90° is 0° past 90°?

3

I’m not sure what you mean by “horizontal distance” and “departure.”

Can you explain what you’re trying to do?

4

Sorry I didn’t explain. Traversing is a simple land surveying technique using distance and direction from point to point. Distance is measured in horizontal distance as opposed to slope distance but in rereading my OP I realize it doesn’t matter for my question.

Latitudes are how far North or South I traveled and departures are how far East or West. They are calculated from the distance and azimuth and are the opposite and adjacent sides for each.

5

The familiar sine wave is symmetrical about 90°. The sine of 90 + x is the same as the sine of 90 - x.

You seem to be defining a different function entirely, having a different purpose.

I’m not actually having any success envisioning what you are trying to describe: geometry without diagrams is a wet row to hoe!

6

Surveying is geometry using 18th century terminology. Lots of ways to fail to communicate between surveyors and people schooled in modern trig.

7

You can construct your own diagram as follows:

Take a piece of paper, mark a point, and use a compass to construct a circle around it. Draw a horizontal diameter to be your x-axis. Now any ray from the centre making an angle a with the axis (measured anti-clockwise) will intersect the circle at the point (cos a, sin a) in units of the radius. Identities like sin(90 deg+a) = sin(90 deg - a) should now become completely obvious.

But, now you should be aware that while everybody uses the same trigonometric functions, in applications like navigation more often than not 0 degrees represents a course due north and compass headings are taken clockwise. The different conventions could be confusing, so practice with some charts.

8

With trig functions quite often sin is considered the ratio of the y component of a unit vector with a unit vector. So sin 0° is 0 and sin 90° is 1 etc. what you are doing is choosing a particular side to be considered opposite the angle and doing the ratio of opposite/hypotenuse.

That side you are choosing would be the x projection of the unit vector and would be the cos.

9

Just to add to what I said, if you rotate the diagram I described by a right angle you will observe that sin(90° ± x) = cos x. That should clear up any confusion between the two functions.

The thing is, in actual surveying, the surface of the Earth is curved. So there are various identities and tables and approximations to deal with that. I don’t know what method you are using, but I reiterate my suggestion to check your work on some practice problems.