Trig Problem: Where Did 50 Degrees Come From in Sin(2x)=.98?

[xtrubambinoxpr]

Homework Statement

Sin(2x)=.98

Homework 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.

The Attempt at a Solution

Arcsin of both sides solving give me 39 degrees

 

[haruspex]

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.

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?

 

[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

 

[Mentallic]

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

 

[xtrubambinoxpr]

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

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

 

[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).

 

[xtrubambinoxpr]

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).

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

 

[Mentallic]

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

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.

 

[xtrubambinoxpr]

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.

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

 

[Mentallic]

well it would be where sinx = 1 that's as high as it can go

Right!

so x = 1/2

Umm... Think about this again.

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' into represent degrees.

 

[xtrubambinoxpr]

Right!



Umm... Think about this again.

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' into represent degrees.

well its at ∏/2 or 90^°..

 

[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.

 

[xtrubambinoxpr]

sin(2x)=1

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

x=arcsin(1)/2

45^°.. sin(90°-x) = sin (90^°+x)

90-45 = 45 ?

 

[Mentallic]

sin(2x)=1

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

x=arcsin(1)/2

45^°

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

sin(90°-x) = sin (90^°+x)

90-45 = 45 ?

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

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

 

[xtrubambinoxpr]

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

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

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°

 

[Mentallic]

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°

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.

 

[xtrubambinoxpr]

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.

thanks man! really appreciate it!

 

[Mentallic]

Happy to help