Question

when combining sum of 2 trig functions.. like acosx+bsinx

Answer 1

\[=\sqrt{a^2+b^2}(\frac{a}{\sqrt{a^2+b^2}} \cos(x) + \frac{b}{\sqrt{a^2+b^2}} \sin(x) ) \]

Answer 2

what do you want to know

Answer 3

howcome the coefficients before cos(x) and sin(x) when squared and added together becomes 1 ??

Answer 4

what is that a result of

Answer 5

Is this to do with QM, normalization to 1?

Answer 6

Or, say you have a vector (3,4) and you want a unit vector, then you divide by sqrt(3^+4^2) = 5 -> (3/5, 4/5) and this is unit, 3/5 squared + 4/5 squared = 1

Answer 7

so we can let \[\sin\phi=\frac{a}{\sqrt{a^2+b^2}}\]\[\cos\phi=\frac{b}{\sqrt{a^2+b^2}}\]and finally\[a\cos x+b\cos x=\sqrt{a^2+b^2}(\frac{a}{\sqrt{a^2+b^2}} \cos(x) + \frac{b}{\sqrt{a^2+b^2}} \sin(x) )\]\[=\sqrt{a^2+b^2}(\sin \phi\cos(x) + \cos \phi \sin(x) )=\sqrt{a^2+b^2} \ \sin(x+\phi )\]which helps us finding range of original function\[-\sqrt{a^2+b^2}\le a\cos x+b\sin x \le\sqrt{a^2+b^2}\]

Answer 8

\[(\frac{a}{\sqrt{a^2+b^2}})^2 + (\frac{b}{\sqrt{a^2+b^2}})^2 =1 \] just kinda wondering why .. and yeah like the unit vector thingo

Answer 9

It is often the case that we want some coefficients normalized to yield 1 somewhere (probability, quantum mechanics, vector algebra, etc).

So the coefficients have been deliberately calculated so as to achieve this condition.