Cosine Curve (1 Viewer)

Skillz17 · May 29, 2016

Hey

How come when i try to graph sqrt(2) cos(2x+pi/4) I can't simply draw sqrt(2) cos(2x) and shift the curve to the left by pi/4. I think the curve is being shifted to the left by pi/8 but I don't properly understand why.

Thanks

Shuuya · May 29, 2016

Skillz17 said:
Hey

How come when i try to graph sqrt(2) cos(2x+pi/4) I can't simply draw sqrt(2) cos(2x) and shift the curve to the left by pi/4. I think the curve is being shifted to the left by pi/8 but I don't properly understand why.

Thanks

You have to factorise out the 2 first. So it becomes √2cos2(x+pi/8), which is why it is shifted left pi/8 units.

(remember that c identifies the shift when it is f(x+c), so the coefficient of x needs to be made into 1 before shifting)

InteGrand · May 29, 2016

Skillz17 said:
Hey

How come when i try to graph sqrt(2) cos(2x+pi/4) I can't simply draw sqrt(2) cos(2x) and shift the curve to the left by pi/4. I think the curve is being shifted to the left by pi/8 but I don't properly understand why.

Thanks

$\noindent (We'll ignore the influence of the $\sqrt{2}$ factor here since this doesn't affect the query.) This is because the function is of the form $f\left(2x + a\right)$, rather than $f(x+a)$ (where here, the function $f$ is the cosine function, i.e. $f(u) = \cos u$). If it was $f(x+a)$, you could just shift the graph of $y=f(x)$ and it right by $a$, but since its $f(2x+a)$, if you want to shift the graph of $y=f(2x)$ (which is what you were asking about), the reason the shift amount is $\frac{a}{2}$ instead is that $f(2x+a) = f\left(2\left(x+\frac{a}{2}\right)\right)$. So to get the graph we want, which is $y=f\left(2 \left(x+\frac{a}{2} \right)\right)$, we can sketch first $y=f(2x)$, and then shift this left by $\frac{a}{2}$ (because to get from $f(2x)$ to $f\left(2 \left (x+\frac{a}{2}\right) \right)$, we have replaced $x$ by $x+\frac{a}{2}$, which we know means we shift the graph left by $\frac{a}{2}$).$

$\noindent If you want to make it easier to see perhaps, just write $g(x) = \cos \left(2x\right)$. Then $\cos \left(2x + a\right) = \cos \left(2\left(x+\frac{a}{2}\right)\right) = g\left(x+\frac{a}{2}\right)$, so to sketch $y=\cos \left(2x+a\right)=\cos \left(2\left(x+\frac{a}{2}\right)\right) = g\left(x+\frac{a}{2}\right)$, we shift the graph of $y=g(x)=\cos \left(2x\right)$ left by $\frac{a}{2}$.$

davidgoes4wce · May 29, 2016

Skillz17 said:
Hey

How come when i try to graph sqrt(2) cos(2x+pi/4) I can't simply draw sqrt(2) cos(2x) and shift the curve to the left by pi/4. I think the curve is being shifted to the left by pi/8 but I don't properly understand why.

Thanks

Here is my working out:

davidgoes4wce · May 29, 2016

Here is my graph on Desmos:

Skillz17 · May 29, 2016

Thank you all so much I understand it now

Cosine Curve (1 Viewer)

Skillz17

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