1. ## H/W help urgent

find the minimum value of $\sqrt{2} sin (x - 40)$, hence find smallest positive value of x where the minimun value occurs.

2. You know the range for the graph y = sin(x) is [-1, 1]. For your question, you should know that the -40 inside the brackets will not affect the vertical shifting of the graph. So:

$-1 \leq \sin (x - 40) \leq 1$

$-\sqrt{2} \quad \leq \quad \underbrace{\sqrt{2}\sin(x - 40)}_{\mbox{original equation}} \quad \leq \quad \sqrt{2} \quad \quad \mbox{Multiplied both sides by } \sqrt{2}$

To find what x values occur at the max and min, just set your expression equal to them

3. Are you aware of stretches of translations of graphs ?

first can you draw $\sin x$ ? with reasonable accuracy showing where it is zero and the min/max points between -360 to 360. If you draw this graph well the rest of the problem is very easy.

If you can do that then, you need to understand that $\sqrt{2} \sin x$ is $\sin x$ stretched vertically. and that $\sqrt{2} \sin (x - 40)$ is a horizontal shift.

Bobak.

4. Thank youuuuuuuu
o_O and bobak

Got the answer for the question