H Precalculus Ch. 7

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H Precalculus Ch. 7

1/31/2011

Homework Help! Have fun blogging!

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)/1-sinsquaredx))-((1-sinx)squared/1-sinsquaredx)). Then i distributed that and got 1+2sinx+sinsquaredx- (1-2sinx+sinsquaredx)/1-sinsqured x. Then, I combined like terms on the top, which got rid of both ones and both sinsquaredx, leaving me with 4sinx/cossquaredx (since 1-sinsquaredx is equal to cossquaredx). Then i split it apart, so i got four times sin/cos times 1/cos, which is equal to 4tanxsecx.
sorry for the squareds and stuff i don't really know how to do those on my computer!
Also, i need help on numbers 38 and 68. could someone help me please?

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+root3/1-root3 instead of -2-root3. I used tangent (pi/4+pi/3) and plugged it into the formula, so I have no idea what I'm doing wrong

Sejzelle Erastus-Obilo

2/2/2011 02:28:33 pm

Can someone help me with problems 13-19 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 13-19 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 Erastus-Obilo

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 45-52...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 Erastus-Obilo

2/7/2011 02:45:22 pm

Does anyone know how to do numbers 16 and 19 from 7.5?

Sejzelle Erastus-Obilo

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,
I would first start off converting the 2sqrt5/5 to 2/sqrt5 so you can see what numbers go on what sides of the triangle. Draw a triangle for quadrant #1 (because sin's domain is from -pi/2 to pi/2) and label the hypotenuse sqrt 5, the opposite side 2, and the adjacent 1. Then find the cosine from that triangle (1/sqrt 5) and flip it (sqrt 5/1) to find the secant. Hope this makes sense...

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,

It's funny how we worked together on this and got it wrong at first. Mrs. Johnsons said that this is actually just the angle addition formula tangent; you can't distribute tangent like that into the equation. so look at the identity sheet and follow the tangent angle addition formula, and you should get the right answer.

I hope this helped. I guess we took the wrong approach to this.

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.
The problem states sin^(-1)[sin(-7/pi)].
First we can start with finding the sin of (-7/pi). -7/pi is equal to -pi-(-pi/6). This is where you use the sin difference formula:
sin(a-b)=sinAcosB-cosAsinB
sin(-pi-(-pi/6))= (0 x root3/2) - (-1 x (-1/2))
= 0 - 1/2
= -1/2
then, is it is the inverse sin, you have to find the angle with a sin of -1/2. These angles include 210 degrees and 330 degrees.
I hope that helps!

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 0-pi. 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 cossquaredx-sinsquaredx, and you end up with 7/25. i think at least :)
Also, i needed help with number 66 in section 7.6. could someone pelase help me?

Nithya Sridhar

2/10/2011 09:51:30 am

I had the same question as Anika, on problem #66 in 7.6.
I'm not sure if this is correct but to get the problem started I use the theorems of:
1-tan(^2)x = sec(^2)x
1+cot(^2)x = csc(^2)x
I thought maybe then it could be converted to sine and cosine, but I am really confused by this problem!
If someone knows who to solve it can you please help us?

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^-1e-v 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.

Instead, can anyone help me with number 38 from 7.6 day 3??

Maddie, what problem in particular?

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(a-b) 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.

Why wouldn't you SUBTRACT pi from the 11pi/6 to find the other value? From what I did in the homework, that is what I am getting; to add pi to the values.

If someone can please explain this to me, that would be great.

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 16-19 from 7.5)
You have to take the csc of square root of 2. since csc is the same thing as 1/sin, you have to do 1 over the sqaure root of 2. after getting rid of the radical in the denominator you get sqaure root of 2 over 2. Now look on the unit circle. where u find root 2 over 2 as sign is the answer which is pi/4.

for nineteen you do it similarly. since cot = 1/tan. you do 1/ - (root 3/3). after simplifying you end up with negative root 3. now go to your unit circle. where u find tan as a negative root three then this is ur answer, which is 2pi/3

Bethany F

2/13/2011 06:30:41 am

I dont understand #37 or #41 or #36 from section 7.6.

Sejzelle Erastus-Obilo

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.

For number 37 from 7.6 first you should draw a triangle for the sec^-1 (5/3)
That would be a 3-4-5 triangle in the first quadrant.
Then find the equation for cot(x+y) on the laminated sheet, which is
(cotxcoty-1) all over (coty + cotx)
Y is equal to pi/6 so you will just have to solve for cot (cos/sin)
X is the angle from the 3-4-5 triangle, so going by SOHCAHTOA it is A/O

Plug it in and you should get

(3/4 times square root of 3) - 1
all over
(square root of three) plus (3/4)

Which, when multiplied out to get rid of radicals in the denominator gets you the answer in the back of the book

The Doctor (Also Known As James Hunter)

2/13/2011 01:36:30 pm

For number 41 from 7.6
Draw the triangle that gives you sin^-1 3/5
Which again is a 3-4-5 triangle in the first quadrant
then, by calling the angle at the origin x, you have a cos(2x) case
So you just make it
cos^2x - sin^2x
Which is
(4/5)^2 - (3/5)^2
Equal to 7/25

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!)

One triangle (the one for tan^-1 4/3) is a 3-4-5 triangle in the first quad

The second triangle (for cot^-2 5/12) is a 5-12-13 triangle in the first quad

Personally, i named the angle at the origin in the 3-4-5 triangle x and the angle at the origin of the 5-12-13 triangle y

Plugging back in these variables you get

sec(X+Y) which is the same as 1/cos(X+Y)
So you should then solve for cos(X+Y) which is equal to cosXcosY-sinXsinY
which is (3/5)(5/13) - (4/5)(12/13)
That gets you (-33/65) but remember that we are solving for sec!
So flip! and the answer is

(-65/33)

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((1-x^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
understand what to do when there are just x's i thought you had to have two separate angles and such... where do i begin??:( and also 3.b if i pluged it into the calculator and made sure it was in radians will i get the right
answer?

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.

#23 -

tan(θ/2) = (sin θ) / (1 + cos θ)

Multiply (1 + cos θ) with the above solution, then cancel out (1 + cos θ) from the fraction. You finally get to sin θ!

#33 (assuming you made a typo) -

Use half-angle formulas!

sin (θ/2) = sqrt[(1 - cos θ) / 2]
Substitution...
sin (330/2) = sqrt[(1 - cos 330) / 2)]

*cos 330 = sqrt(3)/2

Find a common denominator for 1 and the above identity, then divide.

sin (330/2) = sqrt[(2 - sqrt(3)) / 4]

Simplify

sin (330/2) = sqrt[(2 - sqrt(3)] / 2 !

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.

Sohaila: I did that one by accident, but the answer that I got was pi/2 and 3pi/2. I got this by utilizing the angle addition formula for cos(S+B), where in this case A and B are 2pi...so I solved it out and simplified the 1-sin^2theta to cos^2theta...and I got the final equation as cos^2theta=0...where theta has to be pi/2 and 3pi/2. That's what I thought...i have no idea if that's right though since it's not on the answers sheet or book...

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

For #19:

ok im making theta = to x, just so its less confusing.

1+sinx = 2cos^2x
1+sinx = 2(1-sin^2x)
1+sinx = 2-2sin^2
2sin^2x+sinx-1 = 0

(2sinx-1)(sinx+1) = 0

2sinx-1 = 0
2sinx=1
sinx= 1/2, which would be pi/6 and 5pi/6

sinx+1 = 0
sinx=-1, which would be 3pi/2

Kyle Wong link

2/15/2011 03:15:58 pm

How would you simplify sin(4x) or sin (6x)?

Chandan Yashraj

2/16/2011 08:51:41 am

Thanks Sohaila.

Kyle: What I do is use the angle addition for the those kind of functions...so it would be sin(2x+2x), or sin(3x+3x), respectively for each of the functions, then go further.

But we didn't have those on the homewor because they were said to be omitted, so I guess just know it, but you don't have to use it for the test...I think.

Jacob Hayes link

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:

sin2x(cos2xcos2x-sin2xsin2x) + cos2x(2sin2ccos2x)

Then simplify all the double angles and it would be incomprehensable if written here because it would end up just being too long... sorry but i hope i helped to give the process

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 half-angle formula based off of the pi/4...because half of that is pi/8. is this the right way?

Jacob, I agree with you on that...that is the best way Kyle for sin(6x).

Alex: I will get back to you in ten minutes on number 48...I'm not there but if I get it I will reply to you.

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.

I don't know if the part that both of them are in the same quadrant is confusing you...but it does not matter...DRAW 2 SEPARATE TRIANGLES to make it less confusing, and solve accordingly.

Let me know if this helped you or not...is there something particular about the problem you were stuck on?

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.
(For example: The tangent of pi/4 is 1. Taking the inverse tangent of this gives you pi/4)

Because 4pi/7 is out of the range of the inverse tangent graph (the range is from -pi/2 to pi/2) you'll need to subtract pi. This will give you the answer, -3pi/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

Well, you know that tan(4pi/7) is in the 2nd quadrant (less that 7pi/7 but more than 3.5pi/7). However, this is not the right quadrant for tangent. So, you find the principal solution by measuring it from the x-axis. You do 7pi/7 - 4pi/7 which gives you 3pi/7. But, you are going backwards from the x-axis, so you make it negative. So, the answers is (-3pi/7).

Chandan Yashraj

2/16/2011 09:50:01 am

Oh ok I understand now...

Ok so is the domain of the tangent inverse graph from -pi/2 to pi/2, and for sin and cos, it's from 0 to pi...???

Nithya Sridhar

2/16/2011 10:04:35 am

no, chandan
the domain for sin inverse and tangent inverse is: pi/2 to -pi/2
the domain for cosine inverse is: 0 to pi

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.
someone help me please?

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.
you know that sec(theta)= 1/cos(theta)
so you can rewrite the expression as:
cos(theta)[1-sin(theta)]
this is because you freeze and flip.
so after that you can multiply this by the "conjugate" so you would do:
cos(theta)[1-sin(theta)] multliplied by:[1+sin(theta)
after distributing, you end up with:
cos(theta)[1-sinsquared(theta)]all divided by [1+ sin(theta)]

now you'll notice that theres an identity! [1-sinsquared(theta)] is equal to [cossquared(theta)]

then you should end up with cos(theta) multiplied by cossquared(theta). so now you have it solved and it equals:

cos3theta/1+sintheta

(sorry if that explanation was weird...ha hope that helped)

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?

Jacob Hayes link

2/16/2011 12:49:10 pm

Would anyone mind explaining how to do #19 in the review?

Christian Noblett

2/16/2011 01:15:18 pm

For # 19 you manipulate the left side so:

cos(a+b)\cosasinb=cotb-tana
first you start by using the sum and differance relations for the numerator to get:
cosacosb-sinasinb\cosasinb=cotb-tana
next you can split the left side up to get:
cosacosb\cosasinb-sinasinb\cosasinb= cotb-tana
lastly you can cancel cosa in the first part on the left and sinb in the second and you will get:
cosb\sinb-sina\cosa
which you can simplify to cotb-tana

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?
(for 19)

Justin Temple

2/16/2011 02:08:19 pm

Julianne-
You have to split it up in the second step in order to cancel the cosa in the first fraction and sinb in the second fraction:

cosacosb sinasinb
-------- - -------- = cotb - tana
cosasinb cosasinb

then you are left with:

cosb sina
---- - ---- = cotb - tana
sinb cosa

I hope that helped!

Shivani D

2/16/2011 02:15:47 pm

For 71:
the problem is:
cos(sin(-1)(3/5)-cos(-1)(1/2))
The identity cos(A + B) can be applied here. To do this you need to find the sin and cosine of A and B.
Sine A:3/5
Cosine A:4/5
Sine B:root3/2
Cosine B:1/2
Then you substitute these values in for the identity cos(A +B)=cosacosb-sinasinb= (4/5)(1/2)-(3/5)(root3/2)=2/5 - 3root2/10
Therefore you have as your final answer:
4/10 - 3root3/10
= 4-3root3/10

Shivani D

2/16/2011 02:20:26 pm

Alec Kestler

2/16/2011 02:22:48 pm

Kyle:
For number 71 i changed the equation to cos(a-b). Substituting a for the inverse sine of 3/5 and b for the inverse cosine of 1/2.
Since a = the inverse sine of 3/5 then sin(a) would equal 3/5. Since b = the inverse cosine of 1/3 then sin(b) would equal 1/2.
Next, you would make the triangle for both a and b because you need to plug in all of the answers for (cos(a)cos(b)-sin(a)sin(b)) (from the equation cos(a-b). So the triangle for a would be a 3-4-5 triangle, the 3 being the opposite and the 4 the adjacent. And the triangle for b would be a 1-(root 3)-2 triangle, the root3 being the opposite and the 1 the adjacent.
Then you would find cos(a) from your a triangle, which is 4/5. And then sin(b) from your b triangle, which is (root 3)/2.
Then you would plug in what you have found into the equation cos(a)cos(b)-sin(a)cos(b) which would look like (4/5)(1/2)-(3/5)(root 3/ 2). When you multiply it out you get (4/10)-(3root3/10). And your final answer would be ((4-3root3)/10)

Shivani D

2/16/2011 02:23:23 pm

Sorry! My message got deleted:)
for 81:
the problem is sin (2x) + 1=0
You subtract 1 from both sides
sin(2x)=-1
The possible values for 2x is 3pi/2
2x=3pi/2
you divide both sides by 2
x=3pi/4
since, when you divided your answer by two, you also divided your interval, making it go from [0,2pi] to [0,pi]], you add pi to 3pi/4, and your two answers are 3pi/4 and 7pi/4.

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,
Haha no problem!

Connor Allison

2/16/2011 02:48:13 pm

For #15 in the review, i don't
know how to start the problem.
Or rather, which side to
manipulate.

Connor Allison

2/16/2011 02:51:52 pm

Oh wait it's already been answered
sorry.
Also 97, i know it should probably
be easy but i don't know if you
can divide out the 2 on the left
or the 5 on the right and when
I do i don't know where to go from
there

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.

H Precalculus Ch. 7

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