Solve the equation (sin and cos problems)

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In summary: Can I simply use sin^{-1}, to move the sin to the other side of the equation?If so,\alpha = -109.52^{o} +360^{o}k,\forall k Z ,and the equation becomes\alpha=-109.52+360+k=0
  • #1
jegues
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Homework Statement


Solve the equations:
[tex]100sin(\alpha) -400cos(30^{o}-\alpha) = 210[/tex]


Homework Equations


[tex]cos(A-B)=cos A cos B + sin A sin B[/tex]


The Attempt at a Solution



I threw the identity into the original equation, simplified a bit now I'm stuck at the following:

[tex]200sin(\alpha) -400\sqrt{3}cos(\alpha) - 400 sin(\alpha) = 420[/tex]

[tex] \Rightarrow[/tex] [tex] 10sin(\alpha) -20\sqrt{3}cos(\alpha) - 20 sin(\alpha) = 21 [/tex]

Any ideas?
 
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  • #2
jegues said:
I threw the identity into the original equation, simplified a bit now I'm stuck at the following:

[tex]200sin(\alpha) -400\sqrt{3}cos(\alpha) - 400 sin(\alpha) = 420[/tex]

Any ideas?

200sinα-400sinα simplifies to?
 
  • #3
[tex] \Rightarrow[/tex] [tex] -10(sin(\alpha) + 2\sqrt{3}cos(\alpha)) = 21 [/tex]
Slowly getting there... Anymore nudges in the right direction?

I can make things into an uglier equation ;),

[tex] \Rightarrow[/tex] [tex]sin(\alpha) + 2\sqrt{3}cos(\alpha) = \frac{-21}{10} [/tex]

but that really doesn't get me anywhere either... I have a feeling I may have taken the wrong route from the get go.
 
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  • #4
jegues said:
[tex] \Rightarrow[/tex] [tex] -10(sin(\alpha) + 2\sqrt{3}cos(\alpha)) = 21 [/tex]
Slowly getting there... Anymore nudges in the right direction?

I can make things into an uglier equation ;),

[tex] \Rightarrow[/tex] [tex]sin(\alpha) + 2\sqrt{3}cos(\alpha) = \frac{-21}{10} [/tex]

but that really doesn't get me anywhere either... I have a feeling I may have taken the wrong route from the get go.

Put the ledt side in the form Rsin(α+A) or Rcos(α-A)
 
  • #5
[tex] a sin \theta + b cos \theta \equiv R cos \alpha sin \theta + R sin \alpha cos \theta[/tex] So,

For [tex] sin\theta, 1=Rcos\alpha[/tex]

For [tex] cos\theta, 2\sqrt{3}=Rsin\alpha[/tex]

So,

[tex]2\sqrt{3} = tan\alpha[/tex]

So,

[tex]\alpha = 49.1^{o}[/tex]

Is this correct?

EDIT: I'm lost in how to use the form [tex]Rsin(\theta\pm\alpha)[/tex], could you show me how to apply it ?
 
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  • #6
jegues said:
[tex] a sin \theta + b cos \theta \equiv R cos \alpha sin \theta + R sin \alpha cos \theta[/tex] So,

For [tex] sin\theta, 1=Rcos\alpha[/tex]

For [tex] cos\theta, 2\sqrt{3}=Rsin\alpha[/tex]

So,

[tex]2\sqrt{3} = tan\alpha[/tex]

So,

[tex]\alpha = 49.1^{o}[/tex]

Is this correct?

EDIT: I'm lost in how to use the form [tex]Rsin(\theta\pm\alpha)[/tex], could you show me how to apply it ?

Yes that looks correct, and what is R=?
 
  • #7
R would simply be as follows,

[tex] R = \sqrt{a^{2} + b^{2}} = \sqrt{13} [/tex]

But,

[tex] \alpha = 49.1^{o}[/tex], doesn't seem to be satisfying my original equation? Thoughts?
 
  • #8
jegues said:
R would simply be as follows,

[tex] R = \sqrt{a^{2} + b^{2}} = \sqrt{13} [/tex]

But,

[tex] \alpha = 49.1^{o}[/tex], doesn't seem to be satisfying my original equation? Thoughts?

tanα=2√3 ⇒α=tan-1(2√3)=73.9° (re-check that)
 
  • #9
Even with [tex] \alpha = 73.9^{o} [/tex], my original equation still isn't being satisfied.
 
  • #10
jegues said:
Even with [tex] \alpha = 73.9^{o} [/tex], my original equation still isn't being satisfied.

Sorry your notation confused me. I used 'A' for a reason. A=73.9°. Your equation is now

√13sin(α+73.9°)= -21/10
 
  • #11
You need to be more careful. You have

[tex]sin(\alpha) + 2\sqrt{3}cos(\alpha) = \frac{-21}{10}[/tex]

and the identity

[tex]a sin \theta + b cos \theta \equiv R cos \alpha sin \theta + R sin \alpha cos \theta[/tex]

But the [itex]\alpha[/itex] in the first line isn't the same as the [itex]\alpha[/itex] in the identity. Compare what you have to the LHS of the identity to see how the various quantities match up.
 
  • #12
Sorry your notation confused me. I used 'A' for a reason. A=73.9°. Your equation is now

√13sin(α+73.9°)= -21/10

Sorry if I'm wrong, but won't this lead me in circles?

[tex]sin(A+B)=sin A cos B + cos A sin B[/tex]

Ill be back to having an equation with a
[tex] sin\alpha,cos\alpha[/tex],
making it just as hard as the original to solve no?
 
  • #13
Solve for [itex]\alpha+73.9^\circ[/itex].
 
  • #14
jegues said:
Sorry if I'm wrong, but won't this lead me in circles?

[tex]sin(A+B)=sin A cos B + cos A sin B[/tex]

Ill be back to having an equation with a
[tex] sin\alpha,cos\alpha[/tex],
making it just as hard as the original to solve no?

If you have

√13sin(α+73.9°)= -21/10 and then divide by √13, then you will have


sin(α+73.9°)= -21/10√13

Now just do as vela suggests and solve for α+73.9° and get α.

vela said:
Solve for [itex]\alpha+73.9^\circ[/itex].


Also, your initial question did not state the domain of α. Are you to find a general solution or did you leave out the the range for your answer(s)?
 
  • #15
[itex]\alpha+73.9^\circ[/itex]

Can I simply use [tex]sin^{-1}[/tex], to move the sin to the other side of the equation?

If so,

[tex] \alpha = -109.52^{o}[/tex]

... Turns out this works! Pefect! Thank you both for all your help it's appreciated!

Also, your initial question did not state the domain of α. Are you to find a general solution or did you leave out the the range for your answer(s)?

The question only states "Solve the equations:" and nothing else, so I'm assuming it's a general solution.
 
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  • #16
Is it valid to write this as an answer?

[tex] \alpha = -109.5^{o} +360^{o}k,\forall k Z [/tex],

I'm just not sure how to state the "for all integers k" part...
 
  • #17
Bump, just looking for some final clarification on my notation.
 

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