questionsss be4 regional

[datnguyen160]

I don't know how to read a trigonometry graph and their alterations (phase shift, amplitude, period, etc etc)... They'll give you a graph, and expect you to know y=_______

Also, how do I solve problems that gives an x and y equation, asks the # of pairs of integers that will solve the equation. Here: [unparseable or potentially dangerous latex formula]

Area of any polygons?

How can I increase my speed? I wanna do 50 problems in 40 minutes. In practice I can, but in tests, time beats me. What strategy should I use?

I might add more questions. Thanks ahead!

[Kenon]

Alright, area of a polygon is a very fun one. Want a long drawn out proof?


Ok, so let's take an n-gon of side lengths s. This is just a generalization image btw.
By definition, we know the center angles are going to be equal to [unparseable or potentially dangerous latex formula]. By using this little piece of information alone, we can find out all of it.
Let us partition our n-gon into n isosceles triangles. One of the angles in it is going to be the central angle, and the side opposite is the edge of length s.
so we have [unparseable or potentially dangerous latex formula].

Now we know that the area of a triangle is [unparseable or potentially dangerous latex formula], so we need to find our respectable variables. b is simply s, we know that. The key comes down to what h is. Let us construct a perpendicular bisector of this triangle from the side with s length. Due to it being an isosceles triangle, the resulting opposite angle for the created triangles is now divided by two, or [unparseable or potentially dangerous latex formula]. We then must use some trig. If we have a side length of (now) [unparseable or potentially dangerous latex formula], the the resulting length for the side h is [unparseable or potentially dangerous latex formula]. Thus, we get for the area of the triangle, [unparseable or potentially dangerous latex formula]. Thus, the final area is [unparseable or potentially dangerous latex formula].

[88bobcat]

[sinkokuyou]

[Kenon]

Oh wow, I didn't even notice that, the text had all the substance in that.

Updated.

Trig graphs are not a problem anyways. I'll just use sin for an example.
[unparseable or potentially dangerous latex formula]
A is the amplitude iirc (or is it half of it? I don't remember right now, it's 5 am and I just woke up)
B is the frequency shift. Higher the number, faster it oscillates.
C is the phase shift. Shifts the graph more to the left.
D is the vertical shift. Shifts the graph up.

The period is [unparseable or potentially dangerous latex formula] for sin/cos/sec/csc and [unparseable or potentially dangerous latex formula] for tan/cot.

The way to catch it is to identify the lowest Y value and the highest Y value on the graph. The average is D and the difference is A. Then find areas where the graph is equal to D. Find the separation between it and the nearest one of that. That divided by pi gets you B for tan, 1/2 B for sin/cos. At that point I can generally get the answer very simply, because I can generally infer what C is based on the graph.