Just started studying trigonometry, with my high school math book and finally reached the part where Identities are proved with only using basic algebra, I tried to develop my own method of understanding transformations graphically because there aren't any in the book.
So in quest to interpret/transform trigonometric identities graphically, I made my own method, which is explained below.
I assumed that any function of the form $\sin(nπ ± x)$ or $\cos(nπ ± x)$is basically a transformation of $\sin(x)$ and $\cos(x)$ respectively. Hence to obtain any function of those forms we only have to shift the origin of base function linearly on x-axis.
Graphical interpretation/transformation of $\cos((π/2)-x)$ to $\sin(x)$ with my method is as follows-
First I observed the graph of $\cos(x$) function and the value at $π/2$ in it, when we take slope of the left hand limit of $π/2$ and multiply it with $-1$,(if the value of x would've been +ve we should've taken slope of right hand limit and multiplied it with $+1)$, which comes out positive, so when we shift $\cos()$ by $π/2$ towards right or the positive $x$-axis, we get the transformed function which is $\sin(x)$
I used this and interpreted successfully the transformations following functions- \begin{align} \sin\left(\frac{\pi}{2} - x \right) &= \cos(x) \\ \cos\left(\frac{\pi}{2} - x \right) &= \sin(x) \\ \sin\left(\frac{\pi}{2} + x \right) &= \cos(x) \\ \cos\left(\frac{\pi}{2} + x \right) &= -\sin(x) \\ \cos(π-x) &= -\cos(x) \\ \cos(π+x) &= -\cos(x) \\ \sin(π+x) &= -\sin(x) \\ \cos(2π-x) &= \cos(x). \end{align}
This graphical interpretation of transformations does not work the following- $$\sin(nπ-x)$$ Where $n$ is a whole number and $n>0$.
For example: By my method, $$ \sin(π-x) = - \sin(x)$$ But, by solving using $$\sin(a-b) = \sin(a) \cos(b) - \cos(a) \sin(b)$$
Result comes out- $$ \sin(π-x) = \sin(x)$$
For sure I know that my method is wrong but, why it works except one pattern?
I'm keen to know how to interpret trigonometric identities graphically.
Please pardon my grammar, English is not my first language.
Hint: $f(a-x)$ is a reflection of $f(x)$ through the line $x=a/2$. Once you know this, just show that $f(x)=\sin x$ is symmetric about $x=\pi/2$