Chapter 5 Review
Dec 7th, 2010 by corricelli
Hello!
Please use this section to post questions/concerns/reflections on Chapter 5, Review (starts on p.418) or Test (p.421).
Happy Blogging,
Mrs. Corricelli
25 Responses to “Chapter 5 Review”
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For number 25 on the review questions, I understand how to get 5cosx, but why isn’t it the absolute value of 5cosx?
Tiana, I looked at the same q and can’t figure out why the answer isn’t in absolute value…the only explanation I can vaguely think of is maybe there are already restrictions to begin with since it’s a square root…I’m honestly not too sure!
Can any one help me with # 21 and 33..including restrictions?
for the ablsolute value question, i’ve come to conclude that because there is a 5 out in front, there can not be an absolute value in your answer.
Like if you think about it, the sq.root of 25 is positive and negative 5,not just positive five, so therefore it can not be the absolute value.
I think that’s right…but im not sure
Hi guys,
I posted this before a 5.3 section was made so please forgive me Mrs. Corricelli!!!
Anyways, I need help with number 35 and 37. how do I factor out the (1/sin) and (1/cos) out of the equation? Do I take an x out with it? please help me!
happy new year’s eve!!!!!!!!!!!!!!
Hey Guys,
For number 85 in the review the problem in the book and the problem on calc chat don’t line up. It may just be me but it didn’t seem right and I have no idea what to do for it.
hey guys anyone do 25? i feel like it should be easy but im not sure im doing the right thing
Hey Rachel,
So for 25in the review I’m pretty sure you just have to plug in 5sin theta into the original equation, and then distribute the squared, take out a 25 and then switch 1-sin squ. theta into cos^2, so your final answer is 5 cos theta when you get rid of the square root.
Hope that makes some sense…Calc Chat should have it too
Hey Guys,
I just wanted to make sure I know what equation we need to know. Its sum and difference, double angle, Power reducing, sum to product and product to sum. Right???
Double angle is on the formula sheet I believe.
the formulas on the “cheat sheet” are sum and diff, half angle, and double angle. the ones that are not on it are pwr reducing and prod-sum/sum-prod
i’m still confused on the sign in the half angle formulas. in 91 on tan, cos(-150) is the same as cos(150), which is the way it’s written, but sin(-150) is not the same as sin(150), because the first is -1/2 and the other is 1/2. so since there are two negatives in there, would the answer not be positive? because it’s not on the answer sheet. from my understanding, sign depends on the quad of u/2 in all the formulas (cos, sin, tan), but what is u/2 then? because i thought it was cos(whatever) but that wouldn’t work for tan because there’s cos and sin.
I’m kinda confused on that too will, but if your talking about the sign outside the absolute value or whatnot, you have to picture it on the circle.
So for 91, your trying to find sin(-75), and you know -75degrees is in quadrant 4, so using the All Students Take Calculus trick, sin is not positive in the 4th quadrant so therefore there is a negative outside the absolutve value of 1-cos150/2.
So the same goes for cos which is positive and tan which is negative.
I think…that was kind of confusing but I hope that answered your question?
so really, you’re looking for the sign of x, y, and y/x for cos, sin, and tan in whatever quadrant the angle falls into, which then goes outside the radical number?
Will:
In general, if you’re given a trig function of u, and are asked to evaluate the 3 trig functions for u/2…
> the quadrant depends on u/2
> whether sine, and cosine are positive or negative on where u/2 is on the unit circle.
I don’t think tangent has that problem.
okay, so mariah, i think you were probably posting about my earlier question, and not my eureka moment, so i’m going to go with what i said unless someone tells me otherwise.
thanks for all the help everyone!!!!
Ooooooo wait, so pretty much I was wrong? The quadrant is actually determined by the u/2 and not the u….?
Mrs. Corricelli,
Does period 7 have to do any other work because we missed class? I just didn’t want to come in on Monday and find out we were supposed to find out the homework for the weekend.
Thanks!
this is a question about 6.1. in example 3, the book explains that sin B = 12(sin 42°/22) which is equal to 0.3649803307. how do you get an answer for B? because the book says B equals about 21.41°, but you can’t divide 0.3659803307 by sin only.
I’m confused on what you are asking, are you asking about example 3 or number 25??
it’s about example 3.
i think my question is related to will’s because example 3 is related to number 25. i found that sin B = .7517 using the formula given but then on calcchat it says that angle C=180-A-B which equals 21.26. we already know that A is equal to 110degrees but idk how to find angle B
Ryan (and everyone stuck on #25),
Since you know sin B= .7517,you can find the measure of angle B by doing arcsin (.7517)= B
Will,
Is that the height- of- the -pole example? You don’t need to find B for that, since it’s given.
Example 3 is finding a (length). You don’t have to use sin B or b, since c is already given and you can find m. angle C by subtracting.
You should end up using the proportion
(a/sin A)= (c/ sin C)
Does this clear up any confusion?