All sinusoidal graphs could be represented like this:
$$r\sin(cx + \alpha)+d$$
Where "r" increases the amplitude, "$\alpha$" shifts the function left or right, "c" increases or decreases the period, and "d" increases/decreases the y-intercept.
It is also seen that $r\sin(x + \alpha)$ can be transformed into $a\sin x + b\cos x$, with these relations:
Relation 1: $$r = \sqrt{a^2+b^2}$$
Relation 2: $$cos \alpha = \frac{a}{r}$$
Relation 3: $$sin \alpha = \frac{b}{r}$$
Using this logic, how can you change $r\sin(cx + \alpha)+d$ into a similar cosine and sine function form? What relations would be seen there? If you were to use the same steps as seen with $r\sin(x + \alpha)$, then you would get $\sin cx$ and $\cos cx$ terms. I would like for it to just have $\sin x$ and $\cos x$ terms and the additional "d" value.
Thanks!
Edit: You can use the same logic (compound angle formulae), but for that you would need to expand $\sin cx$ and $\cos cx$ afterwards.
The sine function $ f(x)= r\sin(cx + \alpha)+d$ is signal with wavelength ( $x$ is time)
$$ \lambda = \dfrac{2 \pi}{c}$$
Different $c$ frequency waves cannot in general be analytically compounded.
Using Fourier series not in a single but employing an infinite ( finite if frequency is multiple, $c$ is integer ) series of sine functions $f(x)$ can be summed up.
To find sum of an odd function
$$ f(x)= \Sigma_{n=1} ^{\infty} b_n \sin nx $$
coefficients of harmonics are evaluated as
$$ ~~b_n= \frac{1}{\pi}\int f(x) \sin nx ~dx $$
When function is not zero at $x=0,$ even series coefficients $a_n$ should be also bincluded/evaluated.