UKC

Right, I have a question in 2 parts and have the answer for part a, and half the answer for part b (looked up the back of text book for other half of part b, but can't quite figure out how the answer is got at)

a) Express cosx cos pi/6 - sinx sin pi/6 in the form cos(A+B)
b)Hence solve cosx cos30 - sinx sin30 = 1/2 for 0<x60


So a)= cos(x+pi/6)

and b)= cos (x+30) = 1/2
so x must be 30 as cos(30+30) = 1/2
30 expressed in radians = pi/6 (given in text book as correct)

Now, they also have x being 2pi/3 (ie 120degrees) and this is the bit I don't understand. Where are they getting that from?

I'm assuming it's something to do with the -sinx bit ie -sin30 (and I hate using the quadrant thing when you have to work in negatives, just can't get my head around it) Or am I on the wrong track with that?

In reply to SonyaD:

you have got as far as:

cos(x+30) = 1/2

x+30 = 60 OR x+30 = 300 -->Remember that cos crosses 0.5 twice in the given interval (0<x60)

x = 30 OR x = 270

this explains how you get two solutions. you can check them by inserting into the original question.
120 deg is not a solution, again check by substitution in the original question. I suspect a typo - 270 deg is 3pi/2 rad

Answer should be 3pi/2 (not 2pi/3)i.e 270 degrees, since 270+30=300 and cos300=0.5. Sure you didn't misread the textbook answer?

In reply to SonyaD:

Personally I have always found the 'quadrant' approach much less useful than simply looking at a cos or sine curve. The quadrant approach could have identified a second solution but it is easier to see on a simple sin curve IMHO, that way you don't have to remember the rules for how stuff reflects about the quadrant. Certainly the quadrant approach struggles with more complex functions like cosh sinh cot etc.

There is no need to resort back to the -sin(x) and -sin(30) stuff. any solution to that part of the problem must be a solution to the later equation cos(30+x)=0.5

Tom B

In reply to Tom Barraclough: Aaaaarg!!!!! Yes!!! It's a f*cking typo!!! That is the 3rd typo on one bloody page!! I've been going back to this problem several times over the day to try and figure it out. How the hell do they expect folk to get on with understanding stuff with there are so many errors!

Any Maths teachers here? Well, be warned Heinemann Higher Maths, 2nd Edition, page 356 is dotted with errors!!!

<rant over> <and breathe>

Anyway, Tom, thank you very much! I drew myself a wee Cos graph initially, and did come up with the answer of 270degrees, but just didn't relate it to a typing error, so got sidetracked into thinking there must be some other answer, and thus got confused by the -sin30 bit, which doesn't work out at 120 degrees anyway.

Phew! Was getting stressed. Have my Unit2 NAB on Tuesday and this stuff is supposed to be simple!

In reply to SonyaD:

You have to use the quadrants. The way you do is for part b is:

cos(x+30)=1/2
X+30 = cos^-1 (1/2) (which is pi/3) (which is positive)
X+pi/6 = pi/3 ( theres twelve 30deg in 360, thus 2pi/12= pi/6)

using the quadrants All/sin (pi-ans)/tan (pi+ans)/cos (2pi-ans)

All
x= pi/3- pi/6
x= pi/6

cos
and x= 2pi - pi/3 = 5pi/3
x= 5pi/3 - pi/6 = 9pi/6
x= 3pi/2

In reply to Tom Barraclough: The thing is though, if you are working with numbers with decimal places then you kinda need to use the Quadrant approach surely, as the sin/cos graphs will be less helpful no?

In reply to benbers: Thanks. But could you possibly explain it in degrees? I prefer to work in degrees and then convert my final answer to radians, less confusing for me.

So, I'm with you as far as the ALL quadrant go, but you've lost me a bit with the COS quadrant. Now, I know that cos 60 = 1/2 but if x = 30, why arn't you doing 360 - 30? And why are you doing 360 - 60? And why are you taking away 30 from 300? Surely that is taking you over into the TAN quadrant???

In reply to SonyaD:

I see sin/cos graphs as expressing the same thing as the quarants. i.e. whether the answer is pos or neg. The quadrants approach seems more intuitive when you get around to considering Euler's equation and Agand (sp?) diagrams etc..

In reply to benbers and Tom. <lightbulb pings!> I'm with y'all now!

Cos 60 ie(30+30) = 1/2
So, I have to work with 60, and not 30!
So 360 - 60 = 300 and cos 300= 1/2 so that is correct.

So, cos 300 ie (270+30) = 1/2
thus x also = 270 (3pi/2)

Aaaaaaaah, I can relax now I've got that sussed in my head. Bloody maths! This is so sad, it's a Fri night and I'm at home doing maths, lol!

In reply to Removed User: I just find it all bloody confusing. It's going to take a miracle for me to pass my Higher Maths, lol! I need to take a break.

In reply to SonyaD:

I can happily apply the same rules that the CAST quadrant uses to the graphs and find that more intuitive. For example I came to 300 in the earlier post by looking at a graph and remembering that cos peaks at 0 and 2pi. Therefore, if a solution exists at n degrees it also exists at 2pi-n. the necessary logical constructs are more complex for a sin curve but are still easy enough. This echoes my personal preference for imagining solutions as graphical intersections.

This approach is particularity useful if you are unsure weather or not you have no, one or multiple solutions, or have questions where your trig function does not equal a constant, eg cos(x)=x. While I like this approach it won't suit all and I wouldn't recommend changing a method you are happy with if you're running up to an exam.

The main problem with CAST quadrants for me is that they only work for a limited number of trigonometric functions. I'm at my final year of university and tend to need to work with more complex functions so it's kind of redundant to me.

best of luck for the exam!

Tom

In reply to SonyaD: (don't know how to explain in degrees as for engineering we only do radians!)

Theres 4 Quadrants. the 1st one , the 'All' means whatever you get cos^-1(1/2) to be remains that number (we'll call it Z). If Z is positive , you deal with the cos quadrant which is 2pi (360) - Z.

If Z is negative you deal with the Tan quad. (pi or 180 + Z) and the sin quad. (pi or 180 - z)

You deal with that equation (2pi-Z) answer 1st then you take 30deg off/ (pi/6).

the Tan quadrant starts below but not including 270 deg.

Hope that solves it! btw, That book does contain a few errors, used it when I went to school!

Theres a great maths book called '' Mathematics for Engineers : A modern interactive Approach by Anthony Croft and Robert Davison '' that I use for engineering that contains basic maths up to degree maths in an extremely helpful and simple way. I wish now that I had it at school as its extremly simple and easy to learn.

In reply to SonyaD:

For me, I find the only way is repetition. Do as many problems as possible, then I have a firm base on which to move on to the more complex stuff.

Long time since I did any maths courses, but I always found that having things explained by two different methods and/or teachers was quite useful.

Red tea time.

In reply to SonyaD:

I type so slowly that that lot ^^ arrived after the lightbulb moment... may or may not be helpful.


I can't say I'm doing any better, answering random's trig questions on the net in preference to revising Computational Fluid Dynamics. :p

In reply to Tom Barraclough: Yeah, I do get the graph thing (it's taken a bit of time for me to grasp it though) Trigonometry was never my strong point. I enjoy Calculus more (though looking through my text book, looks like we're going to be moving on to integrating trigonometric functions, should be interesting <groan>)

Thanks all

In reply to Tom Barraclough:

PS - I'm not even going to ask what Computational Fluid Dynamics are!

In reply to SonyaD:

Groan ?

I was quite surprised to find that the aspect that I enjoyed most about the my last taught course was the maths (despite not being good at it). There was a strange beauty about the way that everything sort-of 'fitted' together and all made sense in the end.

In reply to Removed User: Yes. I complain and moan about maths, but secretly I'm rather enjoying it. I'm finding it challenging, and I like that 'fitting togetherness' you speak of. I like how things follow little rules, and if you follow a and b then you should come up with c. And I like how what you learned in one unit, fits in with doing stuff in the following unit, like using a quadratic function (that we learned about in intermediate maths) to find if there is a tangent or a line through a curve, for eg (in higher maths) It's so nerdy, it's cool, lol!