PureBasic Forums - English

I have:

t*sin(a) = k+cos(a)

I know all but 'a', and i want to know 'a' respect 'k' and 't', but i don't know.
Any help?

My initial working yields:

a = arcsin(k/sqrt(1+t^2)) + arctan (1/t)

I'll do some checking.

Yea looks okay. This will only give valid values though providing
the value of k lies between -sqrt(1+t^2) and +sqrt(1+t^2).

Yes, can explain how did you it please?

Psychophanta wrote:Yes, can explain how did you it please?

Not so easy through the forum. Assuming that your mathematics is at least A-level standard:

Rearrange to get t*sin a - cos a = k.

Now use the standard A-level Mathematics technique to express t*sin a - cos a in the form RSin(a - s) where R and s are to be found.

You should find that R = sqrt(1 + t^2) and s = arctan (1/t).

The equation now becomes sqrt(1 + t^2)sin (a - arctan (1/t)) = k
which means that sin (a - arctan (1/t)) = k / sqrt(1 + t^2)
and thus a - arctan (1/t) = arcsin (k / sqrt(1 + t^2))
from which we get the result.

If you need me to fill in a few more details then I'll need to e-mail you a pdf file or a Word document.

srod wrote:...express t*sin a - cos a in the form RSin(a - s) where R and s are to be found.

You should find that R = sqrt(1 + t^2) and s = arctan (1/t).

Where in teh web can i find these relations?

srod wrote:If you need me to fill in a few more details then I'll need to e-mail you a pdf file or a Word document.

Should be nice

The whole thing hinges on the fact that

sin (A - B) = sin A*cos B - cos A*sin B

(called a compound-angle identity).

If you've come across this before, then I can easily furnish you with a complete proof of the above fact.

srod wrote:The whole thing hinges on the fact that
sin (A - B) = sin A*cos B - cos A*sin B

of course i know, but i don't see the relation of this with the problem :roll:

Give me 10 minutes to hack up a pdf file with a full explanation then. Can you pm me your e-mail address.

**EDIT: all done - just need your e-mail address.

I will try to solve it too and scan in the result if possible (after what I have learned in math)

srod wrote:**EDIT: all done - just need your e-mail address.

Wait, please, i will not defeat, sorry to heve been to loose your time, but i have it all almost solved.

I cant solve it using factorizing and tanA=sinA/cosA

I defeat:
my email is my nick at hotmail.com

e-mail sent.

I see, mhhh, you are in use, i guess, doing these things, don't?