For acosA +(-) bsinA +(-) c
a= rcosA & b= rsinA ( a not = A^o)
Then a^2+b^2 = r^2-cos^2-A + r^2-sin^2-A = r^2 that= r is sqrt(a^2+b^2)..................................... .......(I could not understand this method. could you explain me about it)
For acosA +(-) bsinA +(-) c
a= rcosA & b= rsinA ( a not = A^o)
Then a^2+b^2 = r^2-cos^2-A + r^2-sin^2-A = r^2 that= r is sqrt(a^2+b^2)..................................... .......(I could not understand this method. could you explain me about it)
I may not understand your question.
However, to start, r is the hypotenuse of a triangle formed by the axes and an angle.
$\displaystyle a = r cos(A)$ is true because cosine is the adjacent side (a) over the hypotenuse (r), so this is just solving for a. This is similar with solving for b as well, but it is the side opposite the angle, so we use sine instead of cosine.
Pythagorean theorem says that for a right triangle $\displaystyle a^2 + b^2 = c^2$, so when we substitute the above this gives us $\displaystyle r^2 cos^2(A) + r^2 sin^2(A)$.
If you elaborate more on your question, that would help me.
Patrick
"Extreme values of acosA +(-) bcosA +(-) c(anot=A^o) is
[c - sqrt(a^2+b^2) , c + sqrt(a^2+b^2)]" (I think now it is better) (I edited the qestion could you again read it)