Mr. O'Brien's 2009-10 Red Pre-AP Calculus: 4/7/2010

Today we started off with a little quiz on 5.1 and 5.2. The back page was a little tricky.

Then we got to look at all the Quarter 3 projects from R3 and W2. Mr. OB was right, Andy's was very entertaining.

We got to go over some problems from last class's homework on page 396: 5, 21, 33, 35, 56, 68, & 75.

The goal of last night's homework was to get us to practice solving them only with technology. Then we will practice solving them entirely without technology.

$[0,2\pi )$

33) csc(x)+cot(x)=1 (enter into calc as $\frac{1}{sin(x)} + \frac{cos(x)}{sin(x)}$ and set scale to pi/12, domain )

Be careful when entering cotangent into the calculator, it should be entered as cos(x)/sin(x), not 1/tan(x) which is undefined to the calculator.

Answer: $\frac{\pi}{2}+2\pi k$ , $k\epsilon Z$ (funny looking Z that wasn't on the equation editor)

75) $h(t)=53+50sin(\frac{\pi}{16}t-\frac{\pi}{2})$ Factor....

$h(t)=53+50sin(\frac{\pi}{16}(t-8))$

a) So, at 8 seconds and 24 seconds it is 53 ft above the ground.

b) During 160 seconds, the person will be at the top after 16, 48, 80, 112, and 114 seconds.

We can then confirm this on a calculator and set limits/scale.

Solving with Algebraic Methods:

5) Use the unit circle when solving algebraically, it's quite helpful to check your work.

If you have to solve this... begin by factoring:

$2sin^{2}x-sinx-1=0$

Let sinx=u

$2u^{2}-u-1=0$

$u = -\frac{1}{2}, 1$

This is when you would use the Unit Circle:

$sinx=-\frac{1}{2}$ at $\frac{7\pi}{6}$ and $\frac{11\pi}{6}$

at $\frac{\pi}{2}$

$x= \frac{\pi}{2}+2\pi k$

$\frac{7\pi}{6}+2\pi k$

$\frac{11\pi}{6}+2\pi k$ , $k\epsilon Z$

21) $cos^{3}x=cosx$

$cos^{3}x-cosx=0$

Factor out a cosx...

$cosx(cos^{2}-1)=0$

and

$cos^{2}x-1=0$

$cos^{2}x=1$

$cosx=\pm 1$

You can than look at the unit circle to see where cos is equal to 0, 1, and -1.

Or, if you prefer a graphical analysis, you can look at the graph of a cosine wave and see where the graph is at 0, 1, and -1.

$x=0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}$

**28) This was not on the homework but Mr. OB thought it would be a good one to go over in class.

Separate the sec and the tan on separate sides and square...

Factor...

$sec^{2}x=1-2tanx+tan^{2}x$

and since we know that $sec^{2}x=1+tan^{2}x$ through the pythagorean identity, they cancel on both sides to get...

and tan equals 0 at...

$x=0, \pi$

However, if we evaluate this on a calculator and check the values on the table, we can see that pi does not work - it is an EXTRANEOUS SOLUTION, that comes from squaring both sides in the beginning of the process. We must always remember to check our answers before moving on. As Mr. OB says, "It's like brushing your teeth and flossing. You can't get away without it"...

35) $cos2x=\frac{1}{2}$ This is considered a multiple angle function because the 2 shrinks the period of the function to pi, meaning there are more revolutions in a 2pi segment. We know that cos is equal to 1/2 at pi/3 and 5pi/3, so: