# Trig problem help (Sin(2x))

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1. Feb 8, 2014

### xtrubambinoxpr

1. The problem statement, all variables and given/known data

Sin(2x)=.98

2. Relevant equations

I know you take arcsin of both sides and divide by 2 to get the answer which is 39, but where does this guy get 50 from? I would have assumed it occurred again at 2*39.. not 50. it is an online lecture.

3. The attempt at a solution

Arcsin of both sides solving give me 39 degrees

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2. Feb 8, 2014

### haruspex

Wrong assumption. Look at a graph of the sine function. Draw a line across just below y=1. You should see pairs of intercepts close together. How are they related?

3. Feb 8, 2014

### xtrubambinoxpr

right it crosses the graph twice. just before the peak like he drew it, but how is that figured out? i know it is between half a period, or (pi) because it repeats itself after that. and in radians its .68.. help me figure this out without telling me! lol i really want to deduce it on my own

4. Feb 8, 2014

### Mentallic

With the graph $y=\sin(x)$ for $0<x<\pi$ what is the line of symmetry?

5. Feb 8, 2014

### xtrubambinoxpr

what do you mean by line of symmetry ? its in respect to the origin

6. Feb 8, 2014

### Mentallic

Do you know what a line of symmetry is? It'll just take a second to understand if you look it up, it's a simple concept. And no, it's not the origin (the origin is also not a line, it's a point).

7. Feb 8, 2014

### xtrubambinoxpr

well you mean that between 0 to pi you get sin of a value twice. like sin(60)=sin(120)..?

8. Feb 8, 2014

### Mentallic

Yes. Basically, the line of symmetry is at $x=90^o$ which means that the sine of an angle 2 degrees less than 90 is the same as the sine of the angle 2 degrees more, etc. In general, $\sin(90^o+x)=\sin(90^o-x)$.

Now, at what x value does $y=\sin(2x)$ have a maximum value? This is also the same as the line of symmetry problem I asked earlier.

9. Feb 8, 2014

### xtrubambinoxpr

well it would be where sinx = 1 thats as high as it can go so x = 1/2

10. Feb 8, 2014

### Mentallic

Right!

Also, remember that if you're referring to degrees, you should add the sup tags (which is the x2 button located just above your post if you pick advanced edit) and put an 'o' in to represent degrees.

11. Feb 8, 2014

### xtrubambinoxpr

well its at ∏/2 or 90°..

12. Feb 8, 2014

### Mentallic

Nope. if $x=\pi / 2$ then $\sin{2x}=\sin{\pi}=0$.

The 2x as opposed to just x makes a difference to where the min and max are located. $\sin(ax)$ for some constant a will either squish or expand the sine graph.

13. Feb 8, 2014

### xtrubambinoxpr

sin(2x)=1

arcsin(sin(2x))=arcsin(1)

x=arcsin(1)/2

45°.. sin(90°-x) = sin (90°+x)

90-45 = 45 ????

14. Feb 8, 2014

### Mentallic

Up to this point, the maths tells you that the maximum is located at $x=45^o$, right?

That equality is only valid for the graph $y=\sin{x}$ since the line of symmetry (the line that passes through the maximum) is located at $x=90^o$. Now we're dealing with $y=\sin(2x)$ where you've found that the maximum is located at $x=45^o$ so in this case, for $y=\sin(2x)$, we have that $\sin(45^o+x)=\sin(45^o-x)$.

So if 39.3o is one solution, then what's the other?

It would also help if you drew the graphs of sin(x) and sin(2x).

15. Feb 8, 2014

### xtrubambinoxpr

ahhh i see so the first value i get for x will either be added or subtracted from the half period or however you want to phrase it. so if its 39.3° then 90°-39.3° = 50.7°

16. Feb 8, 2014

### Mentallic

Yep, exactly! For future reference, you should always draw your graphs, label any important info, and try to deduce any facts you may know about the symmetry or special features of the graph. It will help you immensely.

17. Feb 8, 2014

### xtrubambinoxpr

thanks man! really appreciate it!

18. Feb 8, 2014

### Mentallic

Happy to help