I'm a nonmathematical person, but now I have a very basic understanding of trigonometry and is discovering simple mathematics.
For a year ago I did not know the existence of a unit circle, vectors, but now e.g. radians gives some meaning.
Anyone who can explain the basics of inverse functions using BBCBASIC  please bear in mind, I'm new to this topic?
Trigonometry  inverse...
Trigonometry  inverse...
BBC Model B  19841989. 6502 assembler, Unicomal 19881994, Some C and C++, Pascal 19901994. Bought a copy of BBCBASIC 2007, but started to program at a daily basis 2019.
Re: Trigonometry  inverse...
Hi Ivan, not sure what you are looking for, and something like Wikipedia might be a good place to go, but, very simply:
Consider a rightangled triangle with a horizontal baseline and a point on the left. The line sloping up from the bottom left to top right is the hypotenuse, which is the longest side. I will call it C. The horizontal line I will call A, and the vertical line on the right I will call B.
There is an angle t (for theta) at the left "corner". Note that by default this is in radians, but you could put trigfunction(RAD(t)) if you want to use an angle in degrees, where trigfunction is any of the following...
The cosine of theta (COS(t) in BBC BASIC) is the ratio A/C
The sine (SIN(t) ) is the ratio B/C
The tangent (TAN(t) ) is the ratio B/A
The inverse functions take these ratios, and return the angle t (in radians). Again, there is a conversion function: DEG(trigfunction(Ratio)) will give the angle in degrees
arccosine (ACS(A/C) in BBC BASIC) will give the angle t for that ratio (i.e. that cosine value)
arcsine (ASN(B/C) ) gives t given a sin value
arctangent (ATN(B/A) gives the angle for a given tan value.
To give a concrete example, consider a triangle with hypotenuse of length 1 (unit circle!), and an angle t of 30 degrees on the left (=Pi/6 radians)
cos(t) ~ 0.866, sin(t) = 0.5, and tan(t) ~0.577 NB Cos and tan are approximate! Actually sqrt(3)/2 and 1/sqrt(3)
ACS(0.866) = ASN(t) = ATN(0.577) = PI/6, and DEG(ACS(0.866))=30
Things get a little harder to visualise when the angles are outside the range 090 degrees (0  PI/2 radians), but just think of the hypotenuse moving as a radius around your unit circle. It may be helpful to think of the cosine as the projection of that line onto the x axis, and the sine as the projection onto the y axis (which they are). In the quadrant 90  180 degrees, the line projects up and left, so the cosine (and tangent) values become negative. In the range 180  270 degrees the line points down and left, so both cosine and sine are negative (but tan is positive, since the two negatives cancel out). In the last quadrant, 270360 degrees (or, if you prefer, 90  0 degrees), the line points down and right, so cosine is positive, but sine (and tan) are negative.
Note that for the inverse functions, there is a problem: there are two places (for each function) where they can have a particular value. Cos and sin can have values in the range 1 to 1: ACS assumes this refers to the region 0 to 180 degrees, ASN and ATN assume 90 to + 90 degrees. It's then up to you to check the x and y values making up the ratio, and work out which quadrant it was ACTUALLY in....
Does that help?
Consider a rightangled triangle with a horizontal baseline and a point on the left. The line sloping up from the bottom left to top right is the hypotenuse, which is the longest side. I will call it C. The horizontal line I will call A, and the vertical line on the right I will call B.
There is an angle t (for theta) at the left "corner". Note that by default this is in radians, but you could put trigfunction(RAD(t)) if you want to use an angle in degrees, where trigfunction is any of the following...
The cosine of theta (COS(t) in BBC BASIC) is the ratio A/C
The sine (SIN(t) ) is the ratio B/C
The tangent (TAN(t) ) is the ratio B/A
The inverse functions take these ratios, and return the angle t (in radians). Again, there is a conversion function: DEG(trigfunction(Ratio)) will give the angle in degrees
arccosine (ACS(A/C) in BBC BASIC) will give the angle t for that ratio (i.e. that cosine value)
arcsine (ASN(B/C) ) gives t given a sin value
arctangent (ATN(B/A) gives the angle for a given tan value.
To give a concrete example, consider a triangle with hypotenuse of length 1 (unit circle!), and an angle t of 30 degrees on the left (=Pi/6 radians)
cos(t) ~ 0.866, sin(t) = 0.5, and tan(t) ~0.577 NB Cos and tan are approximate! Actually sqrt(3)/2 and 1/sqrt(3)
ACS(0.866) = ASN(t) = ATN(0.577) = PI/6, and DEG(ACS(0.866))=30
Things get a little harder to visualise when the angles are outside the range 090 degrees (0  PI/2 radians), but just think of the hypotenuse moving as a radius around your unit circle. It may be helpful to think of the cosine as the projection of that line onto the x axis, and the sine as the projection onto the y axis (which they are). In the quadrant 90  180 degrees, the line projects up and left, so the cosine (and tangent) values become negative. In the range 180  270 degrees the line points down and left, so both cosine and sine are negative (but tan is positive, since the two negatives cancel out). In the last quadrant, 270360 degrees (or, if you prefer, 90  0 degrees), the line points down and right, so cosine is positive, but sine (and tan) are negative.
Note that for the inverse functions, there is a problem: there are two places (for each function) where they can have a particular value. Cos and sin can have values in the range 1 to 1: ACS assumes this refers to the region 0 to 180 degrees, ASN and ATN assume 90 to + 90 degrees. It's then up to you to check the x and y values making up the ratio, and work out which quadrant it was ACTUALLY in....
Does that help?
Re: Trigonometry  inverse...
Hi DDRM,
I'm afraid I have not been precisely enough:
I have to calculate theta, but I don't know how to use the tan function in BBCBASIC that way.
Thanks for your detailed reply.
I'm afraid I have not been precisely enough:
I have to calculate theta, but I don't know how to use the tan function in BBCBASIC that way.
Thanks for your detailed reply.
 Attachments

 inverse tan.jpg (3.26 KiB) Viewed 878 times
BBC Model B  19841989. 6502 assembler, Unicomal 19881994, Some C and C++, Pascal 19901994. Bought a copy of BBCBASIC 2007, but started to program at a daily basis 2019.
Re: Trigonometry  inverse...
Hi Ivan,
Assuming you know y and x, then you can just use ATN:
theta = ATN(y/x)
This will give you the angle theta, in radians. If you do
theta = DEG(ATN(y/x)) you will get it in degrees.
You will note that you need to ensure x isn't 0, or you will get an error: you can check for this, and simply assign an angle of (+/) 90 degrees/ PI/2 radians.
Note also, as discussed in the last post, if the angle is actually in the range 90 to 270 degrees, you'll need to correct for this (check whether x is negative, and add 180 degrees if it is).
Best wishes,
D
Assuming you know y and x, then you can just use ATN:
theta = ATN(y/x)
This will give you the angle theta, in radians. If you do
theta = DEG(ATN(y/x)) you will get it in degrees.
You will note that you need to ensure x isn't 0, or you will get an error: you can check for this, and simply assign an angle of (+/) 90 degrees/ PI/2 radians.
Note also, as discussed in the last post, if the angle is actually in the range 90 to 270 degrees, you'll need to correct for this (check whether x is negative, and add 180 degrees if it is).
Best wishes,
D
Re: Trigonometry  inverse...
Hi DDRM,
Thanks a lot  you helped me a lot.
Did some exercies and have a basic understanding of trigonometry now.
Next subject will be vectors.
Best wishes,
Ivan
Thanks a lot  you helped me a lot.
Did some exercies and have a basic understanding of trigonometry now.
Next subject will be vectors.
Best wishes,
Ivan
Code: Select all
rem right triangle A, B, C
rem unit circle
mode 12
x = 500
y = 500
Cx = 1500
Cy = 600
Ax = Cx  x
Ay = Cy
Bx = Cx
By = Cy + y
rem draw right triangle
gcol 7
move Ax, Ay
draw Bx, By
draw Cx, Cy
draw Ax, Ay
rem draw unit circle
gcol 3
move Cx, Cy
for i = 0 to 2 * pi step pi / 180
xp = cos(i) * x + Ax
yp = sin(i) * y + Ay
draw xp, yp
next
rem draw a vertical line
gcol 5
xp = cos(pi / 4) * x + Ax
yp = sin(0) * y + Ay
move xp, yp
xp = cos(pi / 4) * x + Ax
yp = sin(pi / 4) * y + Ay
draw xp, yp
BBC Model B  19841989. 6502 assembler, Unicomal 19881994, Some C and C++, Pascal 19901994. Bought a copy of BBCBASIC 2007, but started to program at a daily basis 2019.

 Posts: 593
 Joined: Tue 15 Oct 2019, 09:10
Re: Trigonometry  inverse...
Actually you can't: knowing both y and x means you can determine the angle completely, but ATN takes only one parameter so it gives an ambiguous result. The way to get the angle from y and x is to use the atan2 function, which takes two parameters. A BBC BASIC implementation can be found at the Wiki; I thought (hoped) this was extremely well known.
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Re: Trigonometry  inverse...
Hi Richard,
Fair enough, though you CAN do it with ATN: you just need to check whether the answer is actually "on the right hand side" (i.e. x is positive) and correct. Of course the atan2 version is nice, since it does this for you. The builtin error trapping for x=0 is cool, too.
Fair enough, though you CAN do it with ATN: you just need to check whether the answer is actually "on the right hand side" (i.e. x is positive) and correct. Of course the atan2 version is nice, since it does this for you. The builtin error trapping for x=0 is cool, too.

 Posts: 593
 Joined: Tue 15 Oct 2019, 09:10
Re: Trigonometry  inverse...
Your original phrase was "can just use ATN", which implies that you don't need to know any additional information, which you do.
You could equally have said "in the top half" (i.e. y is positive) because that also allows you to determine the quadrant. But in practice the code is simpler and faster if you use the signs of both x and y because that way you don't also need to store the result from ATN in a temporary, LOCAL, variable.you just need to check whether the answer is actually "on the right hand side" (i.e. x is positive) and correct.
The following example programs, supplied with BBCSDL, all use the FNatan2 function; studying them may be instructive: Rubik.bbc, SkyBaby.bbc, aliens.bbc, bbcowl.bbc, skaters.bbc and world.bbc.
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Re: Trigonometry  inverse...
Ok  atn for angles between (left triangle) 0 and < 90 deg and atan2 for 360 deg...
Topic is also covered by javidx9 here: https://www.youtube.com/watch?v=DPfxjQ6sqrc
Thanks for clarification.
Topic is also covered by javidx9 here: https://www.youtube.com/watch?v=DPfxjQ6sqrc
Thanks for clarification.
Code: Select all
rem right triangle A, B, C
rem unit circle
rem test atn vs atan2
mode 12
x = 500
y = 500
Cx = 1500
Cy = 600
Ax = Cx  x
Ay = Cy
Bx = Cx
By = Cy + y
rem draw unit circle
gcol 3
for i = 0 to 2 * pi step pi / 4
xp = cos(i) * x + Ax
yp = sin(i) * y + Ay
move Ax, Ay
draw xp, yp
r = fn_atan2(ypAy,xpAx) :rem gives correct quadrant
rem r = atn((ypAy)/(xpAx)) :rem division by zero at 90 deg or pi / 2
vdu 5
print "rad: ";r;" deg: ";deg(r)
vdu4
next
end
def fn_atan2(y,x) : on error local = sgn(y)*pi/2
if x>0 then = atn(y/x) else if y>0 then = atn(y/x)+pi else = atn(y/x)pi
BBC Model B  19841989. 6502 assembler, Unicomal 19881994, Some C and C++, Pascal 19901994. Bought a copy of BBCBASIC 2007, but started to program at a daily basis 2019.