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.