how do you do number 9 on tonight's homework.
For number 9, freeze the right side and work only with the left side. So, sec^2A is equal to 1/cos^2A and then you have tan^2A which equals sin^2A/cos^2A and then there is sec^2A which again is 1/cos^2A. So since the tan part and the sec part were multplied, you multiply (sin^2A/cos^2A) by (1/cos^2A) and after multiplying that part I got sin^2A/cos^4A. Now you have to add that to 1/cos^2A part. We need a common denominator, in this case it will be cos^4A...so, once you get the commmon denominator for both the terms, you should have (cos^2A/cos^4A)+ (sin^2A/cos^4A). Cos^2A+sin^2A is a pythagorean relation equal to 1. So, you end up with 1/cos^4A, which is sec^4A.
Hope this helps!
For number 18 on pg 811, I am getting up to the part where I get 1/cos^2r + sinr/cos^2r and then getting stuck. I'm not sure if I did something I wasn't supposed to on the way, or if I'm just overthinking this. Any help would be great! Thanks!
Sorry! Scratch my question above, I just fiqured it out...sorry again.
For the 14.2 homework number 12, I used difference of two squares and used P.I. for sec2t-tan2t) to equal 1. Where am I going wrong?
I just can't solve it. I reached a dead end.How did you do it?
Can someone help me with number 24 pg. 811, please?
Use difference of two squares and stop...don't break down the next difference of two squares, because you don't have to. So, you get (sec^2t+tan^t)(sec^2t-tan^2t)...the second part is equal to 1 (from pythagorean relations) So basically yo are left with sec^2t+tan^2t. Sec^2t is equal to 1+tan^2t. So, you have 1+tan^2t+tan^2t. Which is then 1+2tan^2t.
When I freeze the right side on problem 27, I end up the point where I get -3-3cosx+ 4sin^2x/1- cos^2x...I don't know where to go from here. Can someone plz help me?
Archana- for problem #27 you are right so far, now you have to factor the top half of the fraction. Remember what she said in class today? Think algebra. So when you factor it out, you should get (1+cosx)(1-4cosx). And that is all over 1-cos^2x. Let me know if you need any more help from there! You should get the answer in the next couple of steps:)
How do you do 24 on 14.1 and 21 on 14.2?
For 24 on pg 811/812 you have to use algebra. You should be able to factor the right side and then go from there to transform it to the left side. In this problem you are actually freezing the left side, not the right!
For #24 on 14.1, you need to get a common denominator, so you multiply the first fraction by sin/sin and the second by cos/cos. The equation then reduces to 1-cos^2B/sinBcosB the top equals sin^2B from trig identity and you get sin^2/sinBcosB sins cancel leaving sinB/cosB = tan B
Hii could somebody please help me with #14 and #31 on the 14.3 homework?
Thank youu :)
Can someone please help me with number 25 and 29 on pg. 822? I totally got stuck on both, I came to a certain point, but then didn't know what to do.
Archana. #25 on p. 822, you get told tan(a-b). So you make the triangles and find out A=Root 3/3 and B=3/4. so you plug those numbers into the the equation tanA-tanB/1+tanA*tanB. You get (root3/3-3/4)/(1+root3/3*3/4) Whence you multiply it out you get (4root3-9)/1+3root3/4). So then you have to find a commor denominator which would be 12. so you get (4root3-9)/(12+3root3). And since you cannot have a root in the denominator you must multiply it through and you get (48root3-36-108+27root3)/(114-27). So when you combine like terms you get 25root3-48/39.
I think this is right but I'm not sure this is what I wrote down in class.
Does anyone know the hmwk for tonight? for some reason i can't get to the hmwk page.
Can someone please help me with number 32 on pg. 822.
The homework is pg. 822 #15, 22-32 even, 45, 47, 49.
I actually got number 32...I'm just in a big mess on number 47. Can someone help me with that one please?
You have to use certain angles like 60 and 45 degrees to get to 15 degrees. So I used 60-45 degrees for the numbers inside the parenthesis. Then you have to use the formula given on the laminated sheet, and since it is csc, you would use the sine formula, but use the reciprocal as the answer. I think.
Could someone help me with problem 47 on page 824. im not sure which side I should to freeze in the first place.
Has anyone else not been able to see the homework page? i clicked on it and i got a white screen...
Yea the homework page isn't working for some reason..
And I don't get #47 either!
If you look up just a bit Archana posted what it was for people who were having similar problems :)
Does anybody know tonights homework? The H, Alg. 2 Trig page isn't working for me! :)
As Emma said, I posted the homework for Kirin...you just need to scroll up a little.
Can someone please help me with page 828 #14?
Sry for the typing above...sorry!! Can someone please help me with number 23 pg. 828? I looked in the back of the book and they have an answer which I am unsure of where they are getting it from.
For # 14...
Freeze the left side. Then on the right side, change all the sec^2x to 1/cos^2x. Then you have (1/cos^2x)/(2-1/cos^2x). For the bottom part get a common denominator, so then you have (1/cos^2x)/(2cos^2x-1/cos^2x). 2cos^2x-1 is equal to cos 2x...look on pg 826 in the book for an explanation of how those are equal. So, now you have (1/cos^2x)/(cos2x/cos^2x). Freeze and flip to get 1/cos2x which is equal to sec2x. Hope this helps!
Yea, I had a problem with number 23 on page 828 too. If anyone can explain please let me know how to do it.
For number 20 on pg. 835, when we divide by 2 to find out if the half angle is positive or negative, are we supposed to divide the orginal angle (450<x<540) by 2 or do we divide the coterminal angle (90<x<180) by 2?
How are you supposed to do problems 16 through 20 on tonights homework, how do you find what x equals
On page 857 is the answer to number 10 360 degrees +360n? I want to check to see if I'm doing this right
Does anyone know how to do number 10? I feel like i'm missing something obvious
Nevermind, i got it and brandon i hthink the answer would be in radians because its not talking about degrees but numbers
On 14.9 why do we find co-terminals again? Also, what does the parentheses mean-included or not included?
Parentheses mean not included and brackets mean included.
for #14, I plugged in the pi/2 value as 90 degrees first. Then I used the identity sin(x-y)= sinxcosy-cosxsiny. I plugged in the values and got: sinxcos90-cosxsin90.
When you solve this, cos90=0 so sinx multiplied by cos90 is = to 0. sing90 = 1 so it would be 0-cosx which is the final answer of proving the identity, sin(x-pi/2)= -cosx
for #31, I used sec(45-30). I then converted that into 1/cos(45-30) which I further broke down into (2/sq.rt.2) X (2/sp.rt.3) + (sq.rt.2/2) X 2. This is because sec is = to 1/cos so i had to flip all of the numbers. I got an answer of 2 sp.rt.6/3 + sq.rt.2.
Can someone help me on number 21 on page 857? I don't know where to start. Thanks.
Christian, I believe you have to use GCF and factor most of them, but to make life easier, set the equation equal to zero. I think that will guide you through questions 16 till 20.
Can someone please help me with number 27 on pg. 857?
I am also confused on #27 on pg. 857!
Can someone help me with that one too?
ok for #27 on page 857 this is how you do it :
cos4x-sin2x=0 this is also equal to cos4x=sin2x. If you use the double argument property for cosine (page 847), cos of 4x is simplified to cosine sqared of 2x-sine squared of 2x.
now the equation is cosine^2(2x)-sine^2(2x)=sin2x.
now cos^2(2x) can be simplified to 1-sin^2(2x) because of the pythagoren property so now the equation is 1-sin^2(2x)-sin^2(2x)=sin2x
if you bring all that to one side and simplify the equation is
2sin^2(2x)+sin2x-1=0 which is a quadratic equation and factor it to get
(2sin2x-1)(sin2x+1) and set both equal to zero and find the angles from there on the unit cirlce :)
hope that helped!
and how do you do #3 on the previous page?
Can someone please help me with problem #18 on pg. 857?
Sahana- For problem #3, the equation is 2sin(x+47degrees)=1. First you divide both sides by 2. Then you end up with sin(x+47degrees)=1/2. Now you have to figure out which degree has a sine of 1/2. If you look on the unit circle, 30degrees and 150degrees have a sine of 1/2. So you set x+47degrees=30degrees and x+47degrees=150degrees. Then you solve the equation and get x=-17degrees and x=103degrees. But since -17degrees isnt in the domain for x then you add 360 degrees so x=343degrees.
Anna, for problem #18, these are the steps I took to solving it:
3-3sinx-2cos^2x=0 [use the pythagorean relations to change cos^2x to 1-sin^2x]
3-3sinx-2+2sin^2x=0 [combine like terms]
2sin^2x-3sinx+1=0 [factor the equation as if it was 2x^2-3x+1]
(2sinx-1)(sinx-1)=0 [set each set of parenthesis equal to 0 and solve]
30 degrees, 150 degrees
So the final answers are:
pi/6, pi/2, & 5pi/6
Anna, the problem is 3-3sinx-2cos2x=0. First turn the cos2x into 1-sin2x (it is an identity) and then distribute the negative (don't forget to change the signs. So if you simplify it, you get 2sinx-3sinx+1=0 (in standard form). Then factor to get (2sinx-1)(sinx-1)and then set the two equal to 0, thus you have sinx= 1/2 and sinx=1 so when you are in the range [-pie, pie] the possible answers are pie/6, pie/2, 5pie/6.
Sorry couldn't find the pie button.
I was doing some extra problems for tomorrow's quiz and was wondering how to do number 13 on page 857? Thanks :)
Sorry I spelled pi wrong.
Ishta - For number 13 on page 857, you are going to want to treat it like a regular algebra problem. To begin, you factor it as (2cosx - 1)(cosx -2)=0 then you just add either 1 or 2, then divide if needed, then look at the unit circle for those cosine values, make sure they are in the domain. The final answer ends up being pi/3 and 5pi/3.
For the review assignment, I just had an idea for R1 c. Instead of multiplying by 1+sinA or 1-sinA, could multiply one of them by -1 so you have a common denominator or is that wrong?
For R1 C. I multiplied the left side by (1+sinA) I did that to both the top and bottom. On the top I got (1+sinA) and the bottom I got (1-sin^2A). On the right side I multiplied it by (1-sinA), again I did that to the top and bottom. On the right top I got (1+sinA) and at the bottom I got again (1-sin^2A). Now with the common denometer you can subtract them. But since there is a minus sign in the middle you have to distribute it to the right part to get -1+sinA. Then I combined the numerators to get 2sinA. And since the bottom is 1-sin^2A, it is the same as cos^2A. Then you have 2sinA over cos^2A. Then you can separate tge 2 cos in the bottom to get 2tanAsecA
In the problem, sin2x=-√2sinx, would I square both sides first? or change the left side to 2sinxcosx first?
Randy-change the left side to 2sinxcosx then add the -rootsinx to the other side and set the whole thing equal to zero so you get 2sinxcosx+root2sinx=0 factor out the sinx so you get sinx(2cosx+root2)=0 set both factors equal to zero and you get sinx=0 2cosx+root2=0 move the root2 over and divide by two to get cosx=-root2/2. then you apply those answers to the domain and you're set.
So I was looking back on past homework to review and I was wondering if anyone knew how to solve pg 856 #35: tan1/2x+1=cosx.... I tried plugging in the formula for tan1/2x and then i squared it so I could possibly use an identity, but I couldn't get anywhere after that... any suggestions?
Mrs. Johnson teaches math at Metea Valley High School in Aurora, IL.
Ap Ab Calculus
H Algebra 2 Trig
H Pre Calculus
Regular Pre Calculus