## Monday, February 12, 2007

### Scribe for today

First of all, I would like to remind all of you guys and gals that we'll be having our first pre-test on Wednesday. For those new to Mr. K's class, the pre-test starts off with us doing the pre-test first, then after a certain amount of time, Mr. K puts all of us in groups and we discuss all of our answers in our group and we hand in one paper for the group.

Anyway,
I'm the scribe for today. Hopefully mine could be as good as the previous scribes.

Today was pretty much just a review of what we did last week, and near the end of the class we started talking about graphing trigonometric functions. First, we started off the class with a couple of questions:

The point (-2, 5) is on the terminal arm of angle θ. Find the exact values of:
a) sin θ b) sec θ

This question is pretty straightforward. First we draw a circle, then put in the coordinates:

Then we solve for the hypotenuse so we can get the sin of the angle:
a2+b2=c2
(-2)22+(5)2=c2

4+25=c2
29=c2

c=
29

Now that we know the length of the hypotenuse, we can solve sin θ and sec θ
a) sin θ = opp/hyp
sin θ = 5 / √29

b) sec θ = reciprocal of cos θ
cos θ = -2 / √29
little note: We don't have to rationalize the denominators.

Onto the next problem(s):

a) cot x = -2.3
tan x = 1/-2.3
arctan(tan x) = arctan (1/-2.3)
x = -0.4101, in quadrants II and IV

b) csc x = -2.3

sin x = 1/-2.3

arcsin(sin x) = arcsin (1/-2.3)

x = -0.4498
, in quadrants III and IV

But those aren't the answers that we want. We want the related angles. Unless the question gives us a restricted domain, we must solve for all real numbers.

Just to help us in this problem, I made a diagram showing where the two angles are:

I recall Mr. K telling us to store those angles in our calculator's memory. To store it, press sto> (store), then alpha, then a.
We want the positive value though, so we press (-), alpha, a, then sto> (store), alpha, a.

We add/subtract π or 2π to know where exactly the angle is in 0 <x<2π or to know where exactly it is in the circle.

So, to solve:
a) cot x = -2.3
x = -0.4101

a = .4101 (answer converted to positive via calculator [(-), 2nd, ans, sto>, alpha, a])

2π - a = 5.8731 To get the related angle in quadrant II:
π - a = 2.7314
therefore:
x = 5.8731 + 2kπ ;k ∈ I
x = 2.7314 + 2kπ ;k ∈ I
Since the angles are only 180 degrees apart, we could leave the answer as:
x = 2.7314 + kπ ; k ∈ I

For csc:
b) csc x = -2.3
sin x = 1/-2.3

arcsin(sin x) = arcsin (1/-2.3)

x = -0.4498
, in quadrants III and IV
a = 0.4498
2π - a = 5.8334
This time we add π because the angle is past 180°.
π + a = 3.5914
therefore: x = 3.5914 + 2kπ ; k ∈ I
x = 5.8334 + 2kπ ; k ∈ I

Whew.

After the questions, we jumped in to today's lesson, graphing trigonometrical functions in our graphing calculator, using the values of pi. It's nothing really new to us since we learned this in Grade 11 besides that we're using the values of pi..

On y = sinx, the sinusoidal axis a.k.a vertical shift is 1, which is the maximum value of the function. On y = 2sinx, the vertical shift is 2. Those also are called amplitudes.

To see π values:
1)Press 2nd function, window. Table Setup should appear.
2)Scroll down using your arrow keys to the line that says ΔTBL and put π/2 in it's place.
3)Now press y=, then punch in sin(x).
4)Press 2nd function, table. On the left hand side (X table) you should see different values of π, and on the right hand side (y table) you should see outputs.

I think that's pretty much it. Homework for tonight is ex.6, all of it, and another one that should be posted by Mr. K. The scribe for tomorrow is Tim_MATH_y.
Just comment if there are any mistakes, and I'll change them as soon as I can.

Grey-M said...

Your kidding me! I did that drawing to give you the general idea, it is nowhere near how well that question should be explained! It is not of Grey-M quality for posting! Grrrrr.....

Grey-M said...

maybe edit that out and just use your text explanation?

aichelle s. said...

haha lol yeah change it haha!!! but your post is very detailed ! nice

Grey-M said...

and besides my one objection the post is done very well.

Ricky said...

LMAO!!!! HA HA HA yes... It's not of Grey-M quality... Nice one Grey-M. Thanks for giving John a general Idea but your drawing is so not Grey-M quality. Man... That's funny. But seriously John... Fix that image up yo.

John D. said...

Yes Mr. Grey-M I've removed the image, lol I never thought you'd react that way. That was funny though, "It's not of Grey-M quality for posting" haha