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February 27th, 2017, 04:27 AM   #1
Bee
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A math course lost

Hello

I switched schools at a bad time and now when I'm about to graduate I have one course uncompleted from last grade and the teacher said I can do the test in one go if I study independently. But I don't quite understand everything

Here's a sample task:

Solve the equation in the given region:
sin x = -½ ; [-π;2π]

I have no idea what to do here. Should I draw some kind of graphics too?
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February 27th, 2017, 05:08 AM   #2
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You should know the identities sin(A) ≡ sin(A + 2$\pi$k), where k is any integer, and sin(A) ≡ sin($\pi$ - A).

You should also know that sin(-$\pi$) = -1/2.

It follows that sin(-5$\pi$/6) = sin(-$\pi$/6) = sin(7$\pi$/6) = sin($11\pi$/6) = -1/2.

Other values of x for which sin x = -1/2 do not lie in the given interval [-$\pi$, 2$\pi$].
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February 28th, 2017, 05:51 PM   #3
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sin(-π) = -1/2 (wrong) should be sin(-π) = 0.
I think he meant.
sin(-π/2) = -1 or sin(-π/6) = -1/2
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March 1st, 2017, 04:16 AM   #4
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Quote:
Originally Posted by Bee View Post
Hello

I switched schools at a bad time and now when I'm about to graduate I have one course uncompleted from last grade and the teacher said I can do the test in one go if I study independently. But I don't quite understand everything

Here's a sample task:

Solve the equation in the given region:
sin x = -½ ; [-π;2π]

I have no idea what to do here. Should I draw some kind of graphics too?

Here's a step-by-step method for getting the answer to questions like $\displaystyle \sin x = -1/2$ on a calculator paper:

1. Make sure your calculator is in the right angle (degrees or radians)
2. Make a note of the domain that your answers need to be in (in this case it is between $\displaystyle -\pi$ and $\displaystyle 2\pi$)
3. Calculate the inverse sine of -1/2 to get the angle. The answer will be returned somewhere between $\displaystyle -\pi/2$ and $\displaystyle \pi/2$
4. Use your answer to get the other answers by remembering the quadrant formulae for sine, cosine and tangent.

You'll need to remember the top-part of this diagram

https://image.slidesharecdn.com/trig...?cb=1279119098

Have a go

Also... here's a top tip: whenever you get a problem about sine, cosine or tangent, draw a little diagram somewhere to remind you what it looks like. Remember to draw these from memory:

http://img.sparknotes.com/content/te...triggraphs.gif

Practise! You can draw a little one in a margin or on some scrap paper for each question. Don't worry about the numbers to begin with, just remember the shape and where those graphs cross the axes. For example, the sine wave starts at zero, then goes up and round. The cosine starts at 1 and then goes down. Remembering the shape of these curves can be crucial for answering trigonometry questions, even on calculator papers.


-------------------------------------
Some extra info:
-------------------------------------

You're probably used to questions having only one or two answers. That's because if you draw a straight line like ($\displaystyle y = 2x -3$) or a quadratic (like $\displaystyle y = x^2 + 2x - 5$), those function only cross the x-axis once or twice (at most). Some quadratics don't cross the x-axis at all.

However... the trigonometric functions are perioidic. That's basically a fancy word for "repeating over and over and over again". Those functions will continue forever, following those same shapes over and over again.

That means that if you want to find the answer to, say, $\displaystyle \sin x = 1$ (the peaks of the sine wave), you don't just have one solution, you have many, many solutions. In fact, you have infinite solutions. One solution is at an angle of $\displaystyle \pi/2$ radians, another at $\displaystyle 5\pi/2$ radians, another at $\displaystyle 9\pi/2$ degrees, ..... you can also go backwards... there is a solution at $\displaystyle -3\pi/2$, one at $\displaystyle - 7\pi/2$, etc.

There is a formula that describes all of those solutions, which is:

$\displaystyle x = 2\pi k + \pi/2$

where $\displaystyle k$ is any positive or negative whole number (including 0). For example, if you choose that k=0 (first peak) than $\displaystyle x = \pi/2$. If you want the second peak, choose k = 1, so the answer is $\displaystyle x = 2 \pi + \pi/2 = 5\pi/2$

The above only works for $\displaystyle \sin x = 1$, but there is a formula for any solution. That's the one in SkipJack's post.

Knowing the formulas is a really good idea, but imho it is much, much, much more important to remember the two diagrams given in the links above, so remember those first



Let us know how it goes!
Thanks from Denis

Last edited by Benit13; March 1st, 2017 at 04:22 AM.
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