# Solve the equation (sin and cos problems)

• jegues
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
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?

Last edited:
jegues said:
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?

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

Last edited:
jegues said:
$$\Rightarrow$$ $$-10(sin(\alpha) + 2\sqrt{3}cos(\alpha)) = 21$$
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)

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

Last edited:
jegues said:
$$a sin \theta + b cos \theta \equiv R cos \alpha sin \theta + R sin \alpha cos \theta$$ 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=?

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?

jegues said:
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)

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

jegues said:
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

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.

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?

Solve for $\alpha+73.9^\circ$.

jegues said:
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.

vela said:
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)?

$\alpha+73.9^\circ$

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.

Last edited:
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...

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

## What is the difference between sine and cosine?

Sine and cosine are both trigonometric functions that relate the angles of a triangle to the lengths of its sides. The main difference between them is that sine relates the opposite side of the triangle to the hypotenuse, while cosine relates the adjacent side to the hypotenuse.

## How do I solve an equation involving sine and cosine?

To solve an equation involving sine and cosine, you can use trigonometric identities and algebraic manipulation. For example, you can use the Pythagorean identity (sin^2x + cos^2x = 1) or the double angle formula (sin2x = 2sinx*cosx) to simplify the equation and find the solution.

## What is a unit circle and how does it relate to sine and cosine?

A unit circle is a circle with a radius of 1 centered at the origin on the Cartesian plane. It is commonly used in trigonometry to visualize and understand the relationships between sine and cosine. The x-coordinate on the unit circle represents the cosine value and the y-coordinate represents the sine value for a given angle.

## What are the possible solutions to a sine or cosine equation?

The possible solutions to a sine or cosine equation depend on the interval and range specified by the problem. In general, there are an infinite number of solutions since sine and cosine are periodic functions. The solutions can be expressed as a general solution using the inverse functions (arcsine and arccosine) or as specific solutions within a given interval.

## Can I use a calculator to solve a sine or cosine equation?

Yes, most scientific calculators have built-in functions for sine and cosine. You can use these functions to find the value of an angle or a side length in a triangle, or to solve an equation involving sine or cosine. However, it is important to understand the principles behind these functions in order to use them accurately and interpret the results correctly.

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