Why does the graph of $\sin(x+{45}^\circ$) shifts to left instead of right comparing with $\sin x$. With respect to point at which line passed from origin.
Mathematically i know , I can fill in values and plot to verify , but is there any other easy to understand explanation for this ?
Sorry if this sounds foolish question.
Thanks
See graph here
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The basic difference between sin(x) and sin(x+45) is the difference in the argument of the function. In case of sin(x+45), notice that the values of the function sin(x) which we ought to observe at x=45, which is far right with respect to x=0, are actually observed at x=0 instead. Similarly, it's easy to see that the values you'd expect at sin(46), in case of sin(x+45), are observed at x=1, instead of x=46 and since, we can see that the newer position on the x axis is actually always to the left of the older position (by 45 units), it is safe to conclude that the function of sin(x) shifts to the left when 45 is added to it's argument.
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As a child I learned to think about functions as "little machines that take in a number and put out a number." For instance $\sin(\cdot)$ takes in a value $x$ and outputs a value $\sin(x)$.
When poser's question tripped me up (many decades ago) I helped myself by asking: "At what input value of $x$ will I get the same output from the function $\sin(x+45)$ as I would from $\sin(0)$?" Oh.... I have to put in $x = -45$... that way I'll get the same output as $\sin(0)$.
In short, the function will appear shifted to the left.
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I don't know if this is a very good way to visualize shifts, but this is how I think about them.
Vertical shifts are of the form $f(x)+k$ and the addition happens after evaluating $f$, so the shift is applied to the function values themselves. So if you think of the graph, you basically pick up the curve and move it up or down depending on the sign of $k$.
Horizontal shifts are of the form $f(x+k)$. The addition is applied to the input and then passed into the function for evaluation. I think of applying this shift to the coordinate grid. In terms of the graph, you hold the function in place and slide the grid right or left depending on the sign of $k$. If $k=45$, then you slide the grid to the right by 45 units underneath the curve. It's like pulling a tablecloth underneath a plate that you hold in place.
Then when you step back and look at the picture, the $y$ axis is to the right of you graph. So your function went left.
Actually, I think of all transformations this way. Anything happening outside of the parentheses, like $af(x)+b$ is a transformation of the curve. Anything inside the parentheses $f(ax+b)$ is a transformation of the grid, where I hold the curve in place.
This used to bug me too and I've come to just accept that it's counterintuitive. Now I get it right each time by remembering that it's the opposite of what I think it should be. But here's an explanation that could possibly help you understand:
Instead of the variable $x$, let's use $t$ so that it reminds you of time. So we're comparing $\sin(t)$ and $\sin(t+45)$. To simplify our thinking, let's just say we're at $t=0$. By adding $45$ to $t=0$ we're actually asking what's happening to $sin$ at a "future" time (since $45$ is positive hence future). To answer this question, we look at the original $\sin(t)$ function and look at this future time i.e.$t=45$. This obviously graphically occurs to the right of the origin. Now we want this behavior of $\sin$ at $t=45$ to correspond to the behavior of $\sin(t+45)$ at $t=0$. So what must we "physically" do? We need to pull it "back" (i.e. to the left) so that the graphs line up.
A similar explanation would also work if you were, say, subtracting 45 instead.
Hope this helped.