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Mrs. Johnson's 20152016 BC Calculus Students ROCK!
Jose R
1/31/2011 06:10:36 am
I have no clue how to even start number 56...help please?
Anika Gupta
1/31/2011 09:30:30 am
For that one, I started by finding a common denominator for both fractions, so i had ((1+sinx)squared)/1sinsquaredx))((1sinx)squared/1sinsquaredx)). Then i distributed that and got 1+2sinx+sinsquaredx (12sinx+sinsquaredx)/1sinsqured x. Then, I combined like terms on the top, which got rid of both ones and both sinsquaredx, leaving me with 4sinx/cossquaredx (since 1sinsquaredx is equal to cossquaredx). Then i split it apart, so i got four times sin/cos times 1/cos, which is equal to 4tanxsecx.
Kelly Mathesius
1/31/2011 11:22:03 am
For 68, I started by taking the 1 in the numerator and replacing with the trig identity cos^2 X + sin ^2 X. After combining the like terms, you get sin^2 X – cos^2 X in the numerator. Then, separate the single fraction into two separate ones: sin^2 X/sinXcosX – cos^2 X/sinXcosX. When you cancel one of the sin’s in the first fraction and one of the cos’s in the second fraction, you get sinX/cosX – cosX/sinX, which equals tan X – cot X. And sorry I couldn’t get 38 either.
melina moussetis
1/31/2011 01:11:41 pm
Could someone please show me how to do number 62? Or maybe just how to start it?
Kayla Simon
2/1/2011 02:57:14 am
For #4 on 7.2 I keep getting 1+root3/1root3 instead of 2root3. I used tangent (pi/4+pi/3) and plugged it into the formula, so I have no idea what I'm doing wrong
Sejzelle ErastusObilo
2/2/2011 02:28:33 pm
Can someone help me with problems 1319 on 7.2 please? It seems like a review of what I already know, but I just can't remember. I tried the 20 * pi/180 but it didn't work.
Alex masiak
2/3/2011 04:45:39 am
for 1319 the problems are like the reverse of the problems before that. For example Sin 20 cos 10 + cos 20 sin 10 Goes to Sin (20 + 10) or sin 30 Which is a point on the unit circle. So instead of expanding you're condensing
Sejzelle ErastusObilo
2/3/2011 09:19:59 am
Ohhhhhhhh thank you, Alex!
Alex Tazic
2/3/2011 01:02:53 pm
what section are we going to in the homework?
Chandan Yashraj
2/3/2011 02:53:02 pm
Alex: As of today, Thursday, we are supposed to be doing 7.3 homework...but I don't know if there will be a quiz tomorrow. So I would prep for it just in case.
Ratuja Reddy
2/6/2011 04:15:48 am
How do you do numbers 11 and 12 on page 459?
Sohaila Mali
2/6/2011 08:29:26 am
I was a little confused on how to go about solving number 16 from 7.3..If anyone could show me how to set it up that would be great. Thanks :)
Mihir Surati
2/6/2011 02:59:47 pm
For 16, I multiplied 9pi/8 by 2 so that we can then use the half angle formula for tangent. Therefore, theta would now be 9pi/4. Now, you plug 9pi/4, for theta, in the half angle formula for tangent. Then solve exactly like the ones before.
Ayo Adewole
2/7/2011 11:07:26 am
can someone explain to me how they did number 16, i dont get the book explanation
Chandan Yashraj
2/7/2011 01:28:37 pm
In 7.5, for the numbers like 4552...is it really that simply that the answer is the value that is being defined by the trig functions? For for number 45, it will be .54, number 46, 7.4, etc.? Ayo: Is this from what Mihir explained in the comment before yours, or is it the one that we had to do for for 7.5 homework? If you are referring to the one from tonight, then all you have to do is FIRST picture which angle's sine will result in root(2), and once you know that, then just imagine what the csc, or opposite, of that would be. I don't think that's the best explanation...but do you sort of get it?
Sejzelle ErastusObilo
2/7/2011 02:45:22 pm
Does anyone know how to do numbers 16 and 19 from 7.5?
Sejzelle ErastusObilo
2/7/2011 02:46:00 pm
and 55?
Cody Essling
2/8/2011 01:20:36 pm
How are you supposed to do the first question we did in class today. I distributed the tan and got the wrong answer. tan((tan^1(3/4)+(sin^1(1/2)))
Lexy Neville
2/9/2011 05:25:11 am
how do you do problems like # 25 on 7.6?
Kayla Simon
2/9/2011 09:12:24 am
Lexy,
Kayla Simon
2/9/2011 09:13:03 am
What are you supposed to do for problem 37 on 7.6? I'm trying to use the cotangent formula but what is coty?
Chandan Yashraj
2/9/2011 10:12:16 am
Cody,
Chandan Yashraj
2/9/2011 10:13:22 am
That's Mrs. Johnson...with no "s" at the end. My bad.
Vish Patel
2/9/2011 10:28:04 am
Hey guys I don't really know how to figure out number 15 on the homework tonightt. It's probably pretty easy but i just can't figure it outt
Shivani D
2/9/2011 11:09:17 am
Hi Vish! I can help you with 15.
Meghan I
2/10/2011 08:31:37 am
I just started the third section of 7.6 and I am already super confused on number 42. Do you use the double angle formula? Can you cancel anything? Where can I start?
Anika Gupta
2/10/2011 09:28:04 am
For number 42, i used the double angle formula, and because of the inverse of sign, it can only go from 0pi. since the cosine is positive, that means that the sin is too. so since its the cos of 4/5 then that means that the sin would be 3/5 because of drawing the triangles and everything. so you plug that in to the formula cossquaredxsinsquaredx, and you end up with 7/25. i think at least :)
Nithya Sridhar
2/10/2011 09:51:30 am
I had the same question as Anika, on problem #66 in 7.6.
Madeline Zehnal
2/10/2011 11:51:07 am
Not only was I having difficuly with number 42, I was wondering if someone could explain number 68... I dont know if you are supposed to do it like the one we did in class or not. I know it is establishing identities but i was confused becasue it is cot^1e^v = tan^1ev and i didnt know what to do with the negative v.
Sivani Aluru
2/10/2011 12:07:39 pm
Could someone help me with number 61 from Day 2 of 7.6?? I'm having trouble incorporating the trig identities.
Madeline Zehnal
2/10/2011 12:30:31 pm
Ok also on last nights homework, we had problems like sec(tan^1 1/2). I do not understand these types of problems. in the book it gives you an example but when i follow the example it does not help to to get the answer i am looking for. any help?
Sivani Aluru
2/10/2011 12:38:59 pm
never mind! I got it already.
Madeline Zehnal
2/10/2011 01:08:57 pm
I actually figued them out except for number 25.
Alex Tazic
2/10/2011 01:31:37 pm
For number 38 you can use the angle addition formula for cos(ab) and i just thought that since csc is 5/3 that its the same as sin 3/5 so i use that instead of csc in the formula which made it easier, i assumed that csc^1(5/3) is the same as sin^1(3/5)
Chandan Yashraj
2/12/2011 02:41:59 am
For section 7.7, when finding all the possible values that correlate with the problem with the given domain, then how would you know when to add the "pi"? This is confusing me, because for a number like 13, which is both 3pi/4 AND 7pi/4, how would you do number 15 then? I got 11pi/6, and that is just the only answer.
Shivani D
2/12/2011 01:13:03 pm
I have the same question as Chandan. Since for 15, if you did subtract pi, you would get 5pi/6, which is still in the domain, so it should work...
Shivani D
2/12/2011 01:16:26 pm
I also had a question on 24 for 7.6. Can somebody help me out?
sohaila and chandan
2/13/2011 02:53:32 am
in problem 69 from the review section of the chapter why cant the answer be in both quadrant 2 (3pi/4) or quadrant 4 (7pi/4).
Bethany F
2/13/2011 06:27:35 am
Sejzelle (regarding problems 1619 from 7.5)
Bethany F
2/13/2011 06:30:41 am
I dont understand #37 or #41 or #36 from section 7.6.
Sejzelle ErastusObilo
2/13/2011 11:54:53 am
Thanks, Bethany!
maddie strick
2/13/2011 12:50:09 pm
can someone help with 7.7 #13?
Kelly Mathesius
2/13/2011 01:05:09 pm
for #13 from 7.7, since cos (2x – pi/2) = 1, then inverse cos 1 = 2x – pi/2. using the unit circle, you can determine that inverse cos 1 = pi. if you substitute this in, then you get pi = 2x – pi/2. Add pi/2 to each side: 2x = 3pi/2. Then divide each side by 2: x = 3pi/4.
sohaila m
2/13/2011 01:09:43 pm
in section 7.7 i understand how to get pi/2 as an answer...but i am not sure how they got 7pi/6 and 11pi/6 as solutions too.
Jamez Hunter Yo
2/13/2011 01:27:05 pm
Hey Bethany, if you still need it.
The Doctor (Also Known As James Hunter)
2/13/2011 01:36:30 pm
For number 41 from 7.6
Tarzan (Also Known As James Hunter)
2/13/2011 01:44:21 pm
For number 36 from 7.6 you need to draw two triangles (Double the fun!)
Alec Kestler
2/13/2011 02:05:58 pm
Hey can somebody help me with problem 22 7.7 ?
Alec KEstler
2/13/2011 02:14:16 pm
Nevermind i figured it out myself. I forgot to copy the 5 down at the front of the equation.
Cody Essling
2/13/2011 03:17:38 pm
Does anyone know if we can use the laminated sheet for the quiz tomorrow.
Chammy
2/14/2011 09:30:38 am
ok i'm not sure whether or not we're allowed to do this, but for number one 1 on Form D of the quiz, does anyone know how to do it? every way i do it i get like 2x((1x^2)^.5)=1...
Shivani D
2/14/2011 12:35:31 pm
I'm also not sure if this is allowed, so if not, feel free to disregard! But for number 1 on form c, is the question wrong? As no matter how many ways I try, I can't get the left side to equal 1.
Ayo Adewole
2/14/2011 01:31:01 pm
i have form a shivani! but i think it might be the same problem, theres no way to make it work. i was having the same problem and i went to the ARC center, and the teachers there told me either it has to equal a 1 or it has to be an addition problem. so im pretty sure its a typo
Ayo Adewole
2/14/2011 01:39:52 pm
can someone explain to me how they did number 22?
Julianne DiLeo
2/14/2011 03:18:41 pm
Yeah i have form D too!! i dont
Anika Gupta
2/15/2011 11:39:13 am
Hi i was doing tonight's homework and i'm really confused on numbers 23 and 32. could someone please help me?
Ashwin Chakilum
2/15/2011 12:12:49 pm
@ Anika G.
Sohaila M
2/15/2011 12:14:36 pm
Hi, i was going over some problems from yesturdays hw and i was just wondering how you would solve #16..
Chandan Yashraj
2/15/2011 01:49:05 pm
Cody: Yes, we always get the laminated sheet that Mrs. Johnson gave us.
Chandan Yashraj
2/15/2011 01:57:32 pm
if someone can please explain numbers 8 and 19 from 7.8, that would be extremely helpful. the problem with these for me is that I just don't know how to start then.
Sohaila Mali
2/15/2011 02:10:11 pm
Thanks chandan
Chandan Yashraj
2/16/2011 08:51:41 am
Thanks Sohaila. 2/16/2011 09:15:17 am
Chandan i agree with you for the sin(4x) however for sin(6x) i would make it sin(2x + 4x) and then when you would expand to make it sin2xcos4x + cos2xsin4x epand the cos4x and sin4x as you said. In the end sin6x should look like this:
Alex
2/16/2011 09:16:13 am
Does anyone get how to do number 48 on the chapter review section? Also, i would really appreciate it if someone could help me with 72, 94, and 98. Thanks.
Chandan Yashraj
2/16/2011 09:26:16 am
I am currently on number 39 for extra practice...and am confused as to how you would go about solving that. The way that I would do it is use the halfangle formula based off of the pi/4...because half of that is pi/8. is this the right way?
Chandan Yashraj
2/16/2011 09:29:32 am
I have another question: can someone properly explain how to solve tan^1(tan4pi/7)...? We did this in class, but I got lost and the bell rang.
Chandan Yashraj
2/16/2011 09:36:27 am
Alex: Maybe you are making the problem too hard...for number 48 it's asking for csc(alpha)=2, and this is nothing else than sin(alpha)=1/2, and same for the sec(beta), where cos(beta)=1/3. Then you solve accordingly.
Chandan Yashraj
2/16/2011 09:42:14 am
Another question: On number 55 from the math review at the end of the chapter , for the question that asks cos^1(root3/2), how is the answer inly 5pi/6...and not 7pi/6...?? I am getting confused on these problems.
Alyssa Pezan
2/16/2011 09:42:43 am
Chandan: First you start with figuring out what the tangent of 4pi/7 is. Since this is not on the unit circle, you will normally need to use a calculator. But since you will be taking the inverse tangent, you will end up with 4pi/7.
Chandan Yashraj
2/16/2011 09:44:01 am
Same goes for number 53 in the same section...how is is pi/4 and not 5pi/4 as well...? The tangent of both is 1...please help!
Alex
2/16/2011 09:46:15 am
@Chandan
Chandan Yashraj
2/16/2011 09:50:01 am
Oh ok I understand now...
Nithya Sridhar
2/16/2011 10:04:35 am
no, chandan
Chandan Yashraj
2/16/2011 10:30:29 am
How would you calculate numbers like 85 and 86 in the review section with a calculator...?
Anika Gupta
2/16/2011 11:21:01 am
Hi I was doing the review for the test tomorrow and i really needed help on problem 33b and 81.
Cody Essling +Chandan Yashraj
2/16/2011 11:23:12 am
Does anyone know how to do problem 15 in the review?
Beatrice Koka!
2/16/2011 11:53:30 am
For #15 on the review basically you start with the left side.
Mihir Surati
2/16/2011 12:19:20 pm
Anika, is the 33b supposed to be just a 33 because i don't see a "b" in the review section. Anyways, if you want to solve 33, you have to use the sine half angle formula. You do this by multiplying the 165 by 2. Therefore, by doing this, you will be able to say that sin (330/2). Then just look up the sin double angle formula and plug in 330 for x or theta.
Kyle Cluver
2/16/2011 12:49:02 pm
Can someone help me with 71?
Christian Noblett
2/16/2011 01:15:18 pm
For # 19 you manipulate the left side so:
Julianne DiLeo
2/16/2011 01:43:29 pm
I dont understand how you spit it up in the second step:( do u think u can explain that in a different way?
Justin Temple
2/16/2011 02:08:19 pm
Julianne
Shivani D
2/16/2011 02:15:47 pm
For 71:
Shivani D
2/16/2011 02:20:26 pm
Alec Kestler
2/16/2011 02:22:48 pm
Kyle:
Shivani D
2/16/2011 02:23:23 pm
Sorry! My message got deleted:)
Alec Kestler
2/16/2011 02:24:01 pm
Dear Shivani, i was typing my answer in and did not refresh the page so i did not see yours come up. Sorry for stealing your answer.
Shivani D
2/16/2011 02:33:51 pm
Dear Alec,
Connor Allison
2/16/2011 02:48:13 pm
For #15 in the review, i don't
Connor Allison
2/16/2011 02:51:52 pm
Oh wait it's already been answered
Muhammed Alikhan
2/16/2011 08:43:46 pm
Connor, for 97 it says you can use a calculator, so I guess you don't have to manipulate either side of the equation at all. I would just graph each as Y1 and Y2, then find the point(s) of intersection.
Alex
2/17/2011 07:26:10 am
Oh man that test was hard! Comments are closed.

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