# Trig equation

1. Jan 1, 2015

### ciubba

I simplified a trig equation to

$$\theta=\Big\{sin^{-1}(\frac{1}{2})+2\pi k\Big\}\; and \; \Big\{\pi-sin^{-1}(\frac{1}{2})+2\pi k\Big\}$$

Whereas the book simplified it to

$$\theta=\pm\frac{\pi}{6}+\pi k$$

Obviously these answers encompass the same set of values, with the book's answer being much simpler. How would one simplify my answer into the book's form?

2. Jan 1, 2015

### Staff: Mentor

The two answers are different, so you can't go from what you have to the book's answer.
The book's answers include $7\pi/6$ and $11\pi/6$ and your answers don't include these and a bunch more.

It would help if you showed us the equation you started from.

Also, please don't delete the three parts of the template...

3. Jan 1, 2015

### ciubba

Sorry, I wrote down my answer incorrectly. The one halves should be plus or minus.

Here is the question:

solve the following trig equation (in radians):
$$2sin^2(\theta)-cos(2\theta)=0$$

I translated everything into sin via the double angle formula, so

$$2sin^2\theta-(1-2sin^2\theta)=0$$

$$4sin^2\theta-1=0$$

$$sin\theta=\pm \sqrt {\frac{1}{4}}$$

$$sin^{-1}\bigg(sin(\theta)\bigg)=sin^{-1}\Big(\pm \sqrt{\frac{1}{4}}\Big)$$

$$\theta=\Big\{sin^{-1}(\pm\frac{1}{2})+2\pi k\Big\}\; and \; \Big\{\pi-sin^{-1}(\pm\frac{1}{2})+2\pi k\Big\}$$

My guess is that the book factored $$4sin^2\theta-1$$ and then solved from there.

Edit: Actually, factoring gave me $$\theta=\Big\{\frac{\pi}{6}+2\pi k \Big\} \; and \; \Big\{\frac{5\pi}{6}+2\pi k \Big\}$$

Last edited: Jan 1, 2015
4. Jan 1, 2015

### Staff: Mentor

You're better off working with this -- $sin\theta=\pm \sqrt {\frac{1}{4}} = \pm \frac{1}{2}$ directly, rather than taking the inverse sine of both sides. The inverse trig functions are one-to-one, so you tend to lose solutions when you use them.

If $sin(\theta) = 1/2$, then $\theta = \pi/6 + 2k\pi$ or $\theta = 5\pi/6 + 2k\pi$, with similar work for the negative value of $sin(\theta)$.

5. Jan 1, 2015

### ciubba

For $$sin\theta=\frac{1}{2}$$, don't I still need to take the inverse sin function in order to determine the value of theta? Also, that still doesn't really answer my question of how to facilitate the transfer of $$\theta=\Big\{\frac{\pi}{6}+2\pi k \Big\} \; and \; \Big\{\frac{5\pi}{6}+2\pi k \Big\}$$ to
$$\theta=\pm\frac{\pi}{6}+\pi k$$

This one wasn't too confusing, but I'd like a general analytic approach to that simplification, if one exists.

6. Jan 1, 2015

### Staff: Mentor

If you apply the inverse sine to both sides of the above, you get one value: $\theta = \pi/6$. But $\theta = 5\pi/6$ is also a solution. In fact, there are an infinite number of solutions to your equation above; namely, $\theta = \pi/6 + 2k\pi# and$\theta = 5\pi/6 + 2k\pi#, where k is an integer.
What you show for $\theta$ above are only the solutions for $sin(\theta) = 1/2$. You need to combine these solutions with those of the other equation, $sin(\theta) = -1/2$, and then list four or five members of each set of solutions, and you'll see how this can be simplified to what the book has.

7. Jan 1, 2015

### lurflurf

I don't understand any of Mark44's inverse sine talk.
$$\dfrac{5\pi}{6}-\left(-\dfrac{\pi}{6}\right)=\pi$$
so they generate the same set.
to see this more clearly write
$$-\dfrac{\pi}{6}+n\, \pi=\dfrac{5\pi}{6}+(n-1)\, \pi$$

8. Jan 1, 2015

### Staff: Mentor

I'm not sure what the above has to do with this problem. In the first post of this thread, ciubba had this as his solution set (slightly simplified):
$\theta = \pi/6 + 2k\pi$ or $\theta = 5\pi/6 + 2k\pi$
He claimed that the above was the same as the book's solution, which was ##\theta =\pm \pi/6 + k\pi. I said that ciubba's solution and the book's solution are not the same, and then he/she admitted that a sign was missing.

9. Jan 1, 2015

### lurflurf

yes post#1 had a typo which he/she corrected in post #3 after you pointed it out.
Then you started talking about inverse sine
the question was how do we know that
$$-\dfrac{\pi}{6}+n\, \pi \text{ and } \dfrac{5\pi}{6}+n\, \pi$$
are the same set? we know because
$$\dfrac{5\pi}{6}-\left(-\dfrac{\pi}{6}\right)=\pi$$
if we look for solutions in one period say $$(-\pi,\pi]$$
we find $$-\dfrac{5\pi}{6} \\ -\dfrac{\pi}{6} \\ \phantom{-}\dfrac{\pi}{6} \\ \phantom{-}\dfrac{5\pi}{6}$$
we can give all solutions as
$$-\dfrac{5\pi}{6} +2k\pi \\ -\dfrac{\pi}{6}+2k\pi \\ \phantom{-}\dfrac{\pi}{6}+2k\pi \\ \phantom{-}\dfrac{5\pi}{6}+2k\pi$$
we can combine these into two using plusminus
$$\pm\dfrac{5\pi}{6} +2k\pi \\ \pm\dfrac{\pi}{6}+2k\pi$$
after we notice
$$\dfrac{5\pi}{6}-\left(-\dfrac{\pi}{6}\right)=\pi$$
we combine into one
$$\pm\dfrac{\pi}{6} +k\pi$$
or
$$\pm\dfrac{5\pi}{6} +k\pi$$
these are the same since since
$$\dfrac{5\pi}{6}-\left(-\dfrac{\pi}{6}\right)=\pi$$

10. Jan 1, 2015

### Staff: Mentor

I was talking about the inverse sine only because the OP was talking about it, and I said that this wasn't the way to go.

11. Jan 2, 2015

### lurflurf

$$\sin^{-1}\left(\dfrac{1}{2}\right) \text{ and } \frac{\pi}{6}$$ are two names for the same number, they mean the same thing.