Question

guys I really need help with part c) I literally have no idea how to this. given that tan(pi/8)=sqrt(2)-1. I am really looking forward to step-by-step explanation.

Answer 1

Answer:

$a \geq \frac{1}{\sqrt{2} -1}$

Step-by-step explanation:

This equation is more intimidating than the problem you have to solve.

You know that the sine of everything is always between -1 and +1. So for the entire expression to be >= 0, the a*tan(pi/8) bit has to be 1 at least. Given this, we can forget about the sin(...) term of the equation for the remainder of solving it.

You already figured out that tan(pi/8) is sqrt(2)-1.

So what we're saying is a * (sqrt(2) - 1) has to be 1 at least.

If we solve a(sqrt(2)-1) >= 1 for a we get:

a = 1/(sqrt(2)-1)

Answer 2

$c)\\\tan\left(\dfrac{\pi}{8}\right)=\tan\left(\pi-\dfrac{7\pi}{8}\right)=\tan\left(-\dfrac{7\pi}{8}\right)=-\tan\left(\dfrac{7\pi}{8}\right)\\\\=-(1-\sqrt2)=\sqrt2-1\\\\y=\sin(2x-1)+a\tan\dfrac{\pi}{8}\\\\\text{We know}\ -1\leq\sin(2x-1)\leq1.\\\\y\geq0\ \text{therefore}\ a\tan\dfrac{\pi}{8}\geq1\\\\\text{We have to move the graph at least one unit up}\\\\a(\sqrt2-1)\geq1\qquad\text{divide both sides by}\ (\sqrt2-1)>0\\\\a\geq\dfrac{1}{\sqrt2-1}\cdot\dfrac{\sqrt2+1}{\sqrt2+1}\\\\a\geq\dfrac{\sqrt2+1}{(\sqrt2)^2-1^2}$

$a\geq\dfrac{\sqrt2+1}{2-1}\\\\a\geq\dfrac{\sqrt2+1}{1}\\\\a\geq\sqrt2+1\\\\Answer:\ \boxed{a=\sqrt2+1}$