A trigonometric equation

Albi said:
Why not is the same thing
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It works with multiplication and division, but not with functions.
E.g. (ab)/2 = a(b/2).
But sin(180°)/2 = 0 and sin(180°/2) = 1.
You need to review this before working on trig equations.
 
lev888 said:
It works with multiplication and division, but not with functions.
E.g. (ab)/2 = a(b/2).
But sin(180°)/2 = 0 and sin(180°/2) = 1.
You need to review this before working on trig equations.
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What should I do then?
 
Albi said:
What should I do then?
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The first you should do is learn to correctly learn function notation.
The sine is a function. So use function notation. We don't write fx for \(f(x)\) do we?
Thus we don't write \(\sin x\) for \(\sin(x)\).
 
pka said:
The first you should do is learn to correctly learn function notation.
The sine is a function. So use function notation. We don't write fx for \(f(x)\) do we?
Thus we don't write \(\sin x\) for \(\sin(x)\).
Click to expand...
So I should write it this way sin (x)+sin (x/2)=0
 
Albi said:
So I should write it this way sin (x)+sin (x/2)=0
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Thank you, thank you. Yes indeed, I wish the rest of our community would be so accommodating.
As head of a division of mathematical sciences, I told any textbook committee not to bring me any recommendation on a text that use that notation: \(\sin x, \cos x, \tan x\) or use \(\ln(x)\text{ for }\log(x).\)
 
pka said:
Thank you, thank you. Yes indeed, I wish the rest of our community would be so accommodating.
As head of a division of mathematical sciences, I told any textbook committee not to bring me any recommendation on a text that use that notation: \(\sin x, \cos x, \tan x\) or use \(\ln(x)\text{ for }\log(x).\)
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I accept that a function should be written in the form that you said but anyways let's get to the main point do you know what should I do to solve the equation?
 
Albi said:
I accept that a function should be written in the form that you said but anyways let's get to the main point do you know what should I do to solve the equation?
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I suggested double angle formulas, did you look them up?
 
pka said:
Thank you, thank you. Yes indeed, I wish the rest of our community would be so accommodating.
As head of a division of mathematical sciences, I told any textbook committee not to bring me any recommendation on a text that use that notation: \(\sin x, \cos x, \tan x\) or use \(\ln(x)\text{ for }\log(x).\)
Click to expand...
ln(x) is still preferred in Physics. So thhhhpppttt! :)

-Dan
 
lev888 said:
I suggested double angle formulas, did you look them up?
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I'll add one more piece of detail:
[math]sin(x) + sin \left ( \dfrac{x}{2} \right ) = 0[/math]
Let y = 2x. Then
[math]sin(2y) + sin(y) = 0[/math]
Now use the double angle formulas.

-Dan
 
20200511_034732.jpg
Is it correct?
Also instead of 3pi/2 it should be 2pi/3 but I made e mistake
 
lev888 said:
How did -1/2 become 3pi/2?
Also y is not 2x.
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If you know the unit circle you know that -1/2 can be written as 2pi/3.
As for the second part I took y as 2x to help me to solve the equation easier you can see that in the end I replaced y with x/2.
 
Albi said:
If you know the unit circle you know that -1/2 can be written as 2pi/3.
As for the second part I took y as 2x to help me to solve the equation easier you can see that in the end I replaced y with x/2.
Click to expand...

I think these were both mere typos! At the bottom of post #16 you corrected the 3 pi/2 to 2 pi/3, and while you wrote y = 2x, you clearly replaced x/2, not 2x, with y, and vice versa at the end. That is, you let x = 2y. But it is important to write what you mean.

Of course, what you meant here is not "-1/2 can be written as 2pi/3", but "the inverse cosine of -1/2 is 2pi/3 ".
 
Dr.Peterson said:
I think these were both mere typos! At the bottom of post #16 you corrected the 3 pi/2 to 2 pi/3, and while you wrote y = 2x, you clearly replaced x/2, not 2x, with y, and vice versa at the end. That is, you let x = 2y. But it is important to write what you mean.

Of course, what you meant here is not "-1/2 can be written as 2pi/3", but "the inverse cosine of -1/2 is 2pi/3 ".
Click to expand...
If I would let 2y=x that would've made more sense.
 
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