sin-1

[qbasicking]

For my GUIs scripting language I want it to be able to do sine-1, does anybody know how to do this? I can't figure out how to do it.
sin-1 is unclear so here's what I mean. I want it to do sine backwards:
sine(1) = .0174
sine-1(.0174) = 1
I also want to do this with tangent and cosine

 

[austim]

Hi, qbasicking.

Sin-1(x) [or asn(x) as it is known in some versions of Basic] can be calculated from the following series (originally formulated by Sir Isaac Newton more than 300 years ago) :

sin-1(x)=x+1/2*x^3/3+1/2*3/4*x^5/5+1/2*3/4*5/6*x^7/7 ...etc
(where x is in radians)

This may not be the most efficient method of calculating sin-1(x) (there are probably other series that converge more quickly), but it will get you there.

Once you can calculate sin-1(x), you can get the others quite easily:

cos-1(x)=pi/2-sin-1(x), or sin-1[sqr(1-x^2)]

tan-1(x)=sin-1[x/sqr(1+x^2)]

 

[qbasicking]

thanks a lot

 

[logiclrd]

You can also hack something up using ATN, the arctangent function. Remember that tan(x) = sin(x) / cos(x), and that cos²(x) + sin²(x) = 1. Then,

cos²(x) = 1 - sin²(x)

cos(x) = (1 - sin²(x))^1/2

So,

tan(x) = sin(x) / ((1 - sin²(x))^1/2)

Therefore, knowing just sin(x), you can get the value to pass into the arctangent function. Assuming you want to get the arcsine x of a, a is equal to sin(x). Replacing through, we get that:

tan(x) = a / ((1 - a²)^1/2)

And therefore:

x = tan^-1(a / ((1 - a²)^1/2))

The tan^-1 function is defined in QB as ATN, so the final code in QB form is:

x = ATN(a / SQR(1 - a ^ 2))

Of course, this only gives you one of the two possible values (when a <> +1). Subtract x from pi to get the other value.

 

[qbasicking]

also I was wondering, is it possible for qbasic to do logs with bases other than 10?

 

[logiclrd]

Logarithm identity:

log_b a = (log a) / (log b)

Any base at all may be used on the right hand side, as long as it is the same for both logs.

 

[qbasicking]

thanx