Just started studying trigonometry, with my highschool 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-

$Sin((π/2)-x)=cos(x) $






$Sin(π+x)=-sin(x) $


This graphical interpretation of transformations does not work the following-


Where $n$ is a whole number and $n>0$.

For example By my method,


But, by solving using

$sin(a-b)= sin(a)cos(b)-cos(a)sin(b)$

Result comes out-


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.


graphical transformations for a given function $f{(x)}$ are as follows;

$f{(x)}$ $\to$ $f{(-x)}$ refers to rotating the graph about the y-axis.

$f{(-x)}$ $\to$ $f{(a-x)}$ refers to shifting the graph $a$ units to the right along the x-axis.

p.s: to determine in which direction to shift the graph think;

"if $f{(0)}$ for $x=0$ in $f{(x)}$ for which value of x does $f{(0)}$ occur in $f{(a-x)}$"

if you follow this process for $sin{(x)}$ $\to$ $sin{(\pi - x)}$ you would get $$sin{(\pi - x)} = sin{(x)}$$ generalizing for any positive integer $n > 0$ ; $$sin{(n\pi - x)} =(-1)^{n-1} sin{(x)} $$


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 $x=\pi/2$

  • $\begingroup$ "f(a−x)f(a−x) is a reflection of f(x)f(x) through the line x=a/2x=a/2. " This I already assumed, and please elaborate how proving sin(x) is symmetric about x=x/2 will explain the linear shift of origin graphically? $\endgroup$ – Tanishq Jaiswal Nov 7 '17 at 23:20
  • $\begingroup$ If it's symmetric, then reflection will result in the same function as before $\endgroup$ – Dylan Nov 7 '17 at 23:51

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