Solve the equation (sin and cos problems)

[jegues]

Homework Statement

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

Homework Equations

cos(A-B)=cos A cos B + sin A sin B

The Attempt at a Solution

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

200sin(\alpha) -400\sqrt{3}cos(\alpha) - 400 sin(\alpha) = 420

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

Any ideas?

 

[rock.freak667]

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

200sin(\alpha) -400\sqrt{3}cos(\alpha) - 400 sin(\alpha) = 420

Any ideas?

200sinα-400sinα simplifies to?

 

[jegues]

\Rightarrow -10(sin(\alpha) + 2\sqrt{3}cos(\alpha)) = 21
Slowly getting there... Anymore nudges in the right direction?

I can make things into an uglier equation ;),

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

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

 

[rock.freak667]

\Rightarrow -10(sin(\alpha) + 2\sqrt{3}cos(\alpha)) = 21
Slowly getting there... Anymore nudges in the right direction?

I can make things into an uglier equation ;),

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

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)

 

[jegues]

a sin \theta + b cos \theta \equiv R cos \alpha sin \theta + R sin \alpha cos \theta So,

For sin\theta, 1=Rcos\alpha

For cos\theta, 2\sqrt{3}=Rsin\alpha

So,

2\sqrt{3} = tan\alpha

So,

\alpha = 49.1^{o}

Is this correct?

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

 

[rock.freak667]

a sin \theta + b cos \theta \equiv R cos \alpha sin \theta + R sin \alpha cos \theta So,

For sin\theta, 1=Rcos\alpha

For cos\theta, 2\sqrt{3}=Rsin\alpha

So,

2\sqrt{3} = tan\alpha

So,

\alpha = 49.1^{o}

Is this correct?

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

Yes that looks correct, and what is R=?

 

[jegues]

R would simply be as follows,

R = \sqrt{a^{2} + b^{2}} = \sqrt{13}

But,

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

 

[rock.freak667]

R would simply be as follows,

R = \sqrt{a^{2} + b^{2}} = \sqrt{13}

But,

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

tanα=2√3 ⇒α=tan^-1(2√3)=73.9° (re-check that)

 

[jegues]

Even with \alpha = 73.9^{o}, my original equation still isn't being satisfied.

 

[rock.freak667]

Even with \alpha = 73.9^{o}, 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

 

[vela]

You need to be more careful. You have

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

and the identity

a sin \theta + b cos \theta \equiv R cos \alpha sin \theta + R sin \alpha cos \theta

But the \alpha in the first line isn't the same as the \alpha in the identity. Compare what you have to the LHS of the identity to see how the various quantities match up.

 

[jegues]

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?

sin(A+B)=sin A cos B + cos A sin B

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

 

[vela]

Solve for \alpha+73.9^\circ.

 

[rock.freak667]

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

sin(A+B)=sin A cos B + cos A sin B

Ill be back to having an equation with a
sin\alpha,cos\alpha,
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.

Solve for \alpha+73.9^\circ.

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)?

 

[jegues]

\alpha+73.9^\circ

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

If so,

\alpha = -109.52^{o}

... 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.

 

[jegues]

Is it valid to write this as an answer?

\alpha = -109.5^{o} +360^{o}k,\forall k Z,

I'm just not sure how to state the "for all integers k" part...

 

[jegues]

Bump, just looking for some final clarification on my notation.