Isolating sin in an equation

• Apr 8th 2010, 11:28 PM
florx
Isolating sin in an equation
Solve the equation with 0 ≤ α ≤ 2pi. Give exact answers if possible.

sin(2α) + 3 = 4

I need help isolating the sin so that it is by itself. How would I do that? I get stuck at this point:

sin(2α) + 3 = 4
sin(2α) = 1

What would I need to do next in order to isolate sin in order to solve this problem? Thank you.(Nod)
• Apr 9th 2010, 12:21 AM
Sudharaka
Quote:

Originally Posted by florx
Solve the equation with 0 ≤ α ≤ 2pi. Give exact answers if possible.

sin(2α) + 3 = 4

I need help isolating the sin so that it is by itself. How would I do that? I get stuck at this point:

sin(2α) + 3 = 4
sin(2α) = 1

What would I need to do next in order to isolate sin in order to solve this problem? Thank you.(Nod)

Dear florx,

$\displaystyle Sin(2a)=1$

Now it is given that, $\displaystyle 0\leq{a}\leq{2\pi}$

Therefore, $\displaystyle 0\leq{2a}\leq{4\pi}$

Now can you imagine what values 2a could take if Sin(2a)=1 given that, $\displaystyle 0\leq{2a}\leq{4\pi}$ ????

Hope you could persevere from here!!
• Apr 9th 2010, 02:48 AM
mathemagister
Quote:

Originally Posted by florx
Solve the equation with 0 ≤ α ≤ 2pi. Give exact answers if possible.

sin(2α) + 3 = 4

I need help isolating the sin so that it is by itself. How would I do that? I get stuck at this point:

sin(2α) + 3 = 4
sin(2α) = 1

What would I need to do next in order to isolate sin in order to solve this problem? Thank you.(Nod)

Sorry if I'm misundertanding, but do you mean you really want to "isolate sin" as in end up with sin = something? Just in case you were thinking of that, I think you are a bit confused about what sin means. "sin" is just a trigonometric function that can't stand on its own.

Having $\displaystyle \sin = ...$ is like having $\displaystyle \sqrt{\ \ } = ...$

To solve for whatever is inside the sin function (e.g. sin(x)), you take the arcsin of both sides [just like you would square both sides to solve for what's in the square root].

Hope that clears things up :)

Mathemagister

PS If you already knew everything I just said and I misinterpreted your words, no harm no foul!
• Apr 9th 2010, 12:48 PM
florx
Thank you so much for your input. I have figured out how to do it already. To cancel out that sin I just put arcsin on the right side so it would look like:
2α = arcsin 1 and just solve it for alpha from there.
• Apr 10th 2010, 04:35 AM
mathemagister
Quote:

Originally Posted by florx
Thank you so much for your input. I have figured out how to do it already. To cancel out that sin I just put arcsin on the right side so it would look like:
2α = arcsin 1 and just solve it for alpha from there.

Exactly. To work out arcsin 1, you could have used the unit circle. At what angle is the y-value (sin) equal to 1?

As Sudharaka said: $\displaystyle 0\leq{2\alpha}\leq{4\pi}$

NOTE: you will have 2 possible values for alpha.

$\displaystyle 2\alpha = \frac{\pi}{2} \quad \text{or} \quad \frac{5\pi}{2} \quad \implies \quad \alpha = \frac{\pi}{4} \quad \text{or} \quad \frac{5\pi}{4}$

If so, good job!

If not, hope I helped ;)

Mathemagister