Hey there! Sign in to join this conversationNew here? Join for free
    • Thread Starter
    Offline

    0
    ReputationRep:
    Hi,

    Whilst working through the standard prepatory questions for science at Cambridge, there seems to be quite a difference in the content of the Advanced Higher Maths course (which I have done) and the A level course with which these questions are based.

    Although the majority of this is fine, A level seems to go further into a number of things particularly in trigonometry, and AH vice versa. Any help would be most appreciated:

    1) Is there a standard way (ie. rote method) of sketching graphs such as  y = e^{-x}sin x or  y = tan x . x^{3} ?

    Advanced Higher was very into sketching long winded, asymptotic graphs but not those such as these.

    2) Sketching graphs given by two parametric equations: ie.

     x = tan t  y = sec t

    What is the best way to proceed here?

    3) Finding angles between lines specified, for example as:

     x = y = z &  x = y = 2z + 1

    Can this be done using a vector approach or do we simply compare the gradients of the line with respect to the z axis?

    4) Is there a less painful way of expanding fuctions with the binomial theorem such as:

     \dfrac{(1 + 2x)^{1/2}}{(2 + x)^{1/2}}

    without using the binomial theorem on each part, evaluating the components of pascals triangle individually and doing some nasty division?

    5) Tangent half angles? What quite does this method achieve and how exactly are  cos \theta ,  sin \theta derived in terms of t? Is it simply by jiggling around trigonometric double angle formulae?

    6) Expressing equations such as:

     \sqrt 3 sin \theta + cos \theta in the form  A sin (\theta + C)

    Whilst there is an obvious intuitive method, is there a more concrete forumla/rote way of doing this when the numbers get horrid?

    7) This is rather embarassing, but what is the best way to attack something along the lines of:

     cos \theta + cos 3\theta = sin \theta + sin 3\theta

    8) Last small point is where tangent half angles fit into integration. Have an idea but not quite sure how to set the whole the out and the right way about it.

    Apologies for the long list, but despite searching around on the internet fairly extensively there are no concrete answers to any of the questions. It's all rather embarassing as AH (and presumably further maths) spend a lot of time with other ideas (complex numbers, matrices, vectors) etc. and I seemed to have missed some of this relatively basic stuff out.

    Thanks
    Offline

    12
    ReputationRep:
    Q.2. looks like you can put it in a friendlier cartesian form after squaring and using the identity  \tan^2 \theta + 1 = \sec^2 \theta .

    Q.7. looks like the sum-to-product formulae could be useful.

    edit:
    Spoiler:
    Show
    Q.6. If we have A \cos \theta + B \sin \theta and we want to express it in the form, say,  R\sin (\theta + C) , where A and B are known constants, and R and C are constants to be found, a fairly standard way to go is to expand the compound angle trig term, in this case  R\sin (\theta + C) and equate coefficients of  \sin \theta and  \cos \theta , and then you end up with simultaneous equations in R and C.

    For example: Let us write  \sqrt{3}\sin \theta + \cos \theta in the form  R \sin (\theta + C) , and so:

     \sqrt{3} \sin \theta + \cos \theta \equiv R\sin \theta \cos C + R \cos \theta \sin C

    Evaluating coeffcients give us two equations:

     \sqrt{3} = R \cos C

     1 = R\sin C

    So we just need to solve for R and C. We can easily solve for C by dividing one equation by the other as R cancels, for example the bottom one over the top gives:

     \dfrac{1}{\sqrt{3}} = \dfrac{R\sin C}{R\cos C} = \tan C and therefore  C = \pi /6

    We can solve for R by squaring these two equations and adding and using a standard trig identity to remove C:

    3 + 1 = R^2 \cos^2 C + R^2 \sin^2 C = R^2(\cos^2 C + \sin^2 C) = R^2 and therefore  R = 2 .

    And so,  \sqrt{3} \sin \theta + \cos \theta = 2\sin \left( \theta + \frac{\pi}{6} \right)

    Perhaps try something similar with  2 \sin \theta + \sqrt{3} \cos \theta and write it in the form  R\cos (\theta + C) .
    Offline

    12
    ReputationRep:
    around's guide to sketching graphs:

    work out any roots
    work out any asymptotes
    work out and classify turning points
    work out what happens as x gets very big in either direction

    then join the dots.

    question 2: see if you can spot any nice values of t and work out x and y values. do this for a few different t values and see if you can join the dots. if all else fails see the graph sketching guide.

    question 3: see if you can put that into vector form, ie:

    \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} a \\ b \\ c \end{pmatrix} + \lambda \begin{pmatrix} p \\ q \\ r \end{pmatrix}, and then using the dot product. It might help equating everything into a dummy variable.

    question 5: work out tan \theta in terms of half-theta, and use a trig identity to get sec, and then everything else can be derived easily.
    Offline

    2
    ReputationRep:
    (Original post by abstraction98)
    6) Expressing equations such as:

     \sqrt 3 sin \theta + cos \theta in the form  A sin (\theta + C)

    Whilst there is an obvious intuitive method, is there a more concrete forumla/rote way of doing this when the numbers get horrid?

    7) This is rather embarassing, but what is the best way to attack something along the lines of:

     cos \theta + cos 3\theta = sin \theta + sin 3\theta
    6) See attached pdf for some insight (there's some sign error if I remember correctly, but you'll get the gist of it).

    7) You could attack it with any valid method really, expansions of cos(3x) and sin(3x) can easily be derived by considering the polar form of cis(3x).

    edit: Are you sure the statement is true? Doesn't seem true to me...
    Spoiler:
    Show
    sin(pi/4)+sin(3pi/4)=0.5rt2+0.5rt2=rt2 and cos(pi/4)+cos(3pi/4)=0.5rt2+ -0.5rt2=0
    Attached Images
  1. File Type: pdf hyeree.pdf (32.0 KB, 59 views)
    Offline

    1
    ReputationRep:
    (Original post by GHOSH-5)
    Spoiler:
    Show
    ...
    and therefore .
    ...
    Sorry if this is irrelevant, but how do you know it's 2 and not -2?

    EDIT: Ah, don't worry, I realise you oculd just check it with \sqrt3=R\cos\frac\pi6
    • Study Helper
    Online

    13
    Study Helper
    (Original post by abstraction98)
    5) Tangent half angles? What quite does this method achieve and how exactly are  cos \theta ,  sin \theta derived in terms of t? Is it simply by jiggling around trigonometric double angle formulae?
    Can be used for solving a\cos\theta+b\sin\theta=c and is a useful change of variable in some integrations.

    One derivation for \sin\theta is:

    \sin\theta= \frac{2\sin\frac{\theta}{2}\cos\  frac{\theta}{2}}{\cos^2\frac{\th  eta}{2}+\sin^2\frac{\theta}{2}}=    \frac{2\tan\frac{\theta}{2}}{1+\  tan^2\frac{\theta}{2}}=\frac{2t}  {1+t^2}

    On the second step above you divide through by cos^2.

    Cos is in a similar fashion.
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by ghostwalker)
    Can be used for solving a\cos\theta+b\sin\theta=c and is a useful change of variable in some integrations.

    One derivation for \sin\theta is:

    \sin\theta= \frac{2\sin\frac{\theta}{2}\cos\  frac{\theta}{2}}{\cos^2\frac{\th  eta}{2}+\sin^2\frac{\theta}{2}}=    \frac{2\tan\frac{\theta}{2}}{1+\  tan^2\frac{\theta}{2}}=\frac{2t}  {1+t^2}

    On the second step above you divide through by cos^2.

    Cos is in a similar fashion.
    Clever, thanks.

    1, 2 now make complete sense. I am used to using Arounds method on specific type of functions - I was an idiot not to use it on these.

    3, for some reason I am lost at what to do. It's probably deceptively simple. I'm so used to finding angle between multiple planes etc. I'll get there eventually.

    By the the lack of response, I take it 4 is as painstaking as it looks.

    The derivations in 5 are a great help thanks, although I am still at a loss to see what this achieves and why it is useful.

    6 makes sense, although from a quick scan will need to have a look at where it comes from.

    7, I can whittle it down to  cot \theta = - tan 2 \theta but not sure how to proceed from here.

    8, I should probably be more specific. The specific question is  \int^{\pi / 2}_0  \frac{1}{3 + 5 cos \theta} \ d\theta and suggests, use  t = tan (\frac{1}{2} \theta)
    Offline

    12
    ReputationRep:
    Try substituting t = tan \theta/2 (keeping in mind the expression for cos in terms of t). What it basically does is convert certain kinds of trigonometric integrals onto algebraic integrals which should be easier to do.

    I don't really know what you mean by 4. Are you just trying to work out what f(x) is for various values of x? Raising something to the half is the same as square rooting, dunno if this is what you're after.

    For 3, if you write each line in vector form, then take the dot product of the directions, you should be able to work out the angle between them.
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by around)
    Try substituting t = tan \theta/2 (keeping in mind the expression for cos in terms of t). What it basically does is convert certain kinds of trigonometric integrals onto algebraic integrals which should be easier to do.

    I don't really know what you mean by 4. Are you just trying to work out what f(x) is for various values of x? Raising something to the half is the same as square rooting, dunno if this is what you're after.

    For 3, if you write each line in vector form, then take the dot product of the directions, you should be able to work out the angle between them.
    I will try that, wrt tangent half angles.

    Sorry, my apologies. I am trying to expand with the binomial theorem. I know what to do, it just takes forever, looks ugly and is a bore. Wondering if there is a better method?

    Yeh, I'm just being an idiot on three.
    • Study Helper
    Online

    13
    Study Helper
    (Original post by abstraction98)
    4) Is there a less painful way of expanding fuctions with the binomial theorem such as:

     \dfrac{(1 + 2x)^{1/2}}{(2 + x)^{1/2}}

    without using the binomial theorem on each part, evaluating the components of pascals triangle individually and doing some nasty division?
    You shouldn't require any "nasty" divisions, if you rewrite it as:

    \displaystyle \frac{1}{\sqrt{2}}(1+2x)^\frac{1  }{2}(1+\frac{x}{2})^{-\frac{1}{2}}

    Then it's just a long winded multiplications, depending on how many terms you require.
    Offline

    13
    ReputationRep:
    For 7)

    A good technique for dealing with different multiples of theta in equations is the 'Sum to Product Formula'. There are four, I've noted the 2 you need to use.

     cos u + cos v = 2 cos\left(\frac{u+v}{2}\right) cos\left(\frac{u-v}{2}\right)

     sin u + sin v = 2 sin\left(\frac{u+v}{2}\right) cos\left(\frac{u-v}{2}\right)

    So cos 3x + cos x =.....
    sin3x + sinx = ......

    Turns out to be me much easier to deal with, and gets to a point where you can start picking solutions much quicker.

    The formula above, are something you probably should be aware that they exist, I usually just derive them when I need them.


    For anecdotal and interest sake.

    Derivation


    cos (A + B) + cos (A - B) = 2cosAcosB

    cos (A + B) - cos(A - B) = 2sinA sin B

    (You can check these yourself)

    Now we do a substitution.

    let A = (u + v)/2 and B = (u - v)/2

    (A + B) = u and (A - B) = v

    We do this, because we need to be able to pick values of theta on the LHS at our own free will, which we can't do if its left as (A + B) and (A - B)

    Applying that substitution to the equations

    cos (A + B) + cos (A - B) = 2cosAcosB

    cos u + cos v = 2 cos (u + v)/2 cos (u - v)/2

    cos (A + B) - cos(A - B) = 2sinA sin B

    cos u - cos v = 2 sin (u + v)/2 cos (u - v)/2

    So now we can pick theta values for u and v, like x and 3x, or 5x,7x.... For which we can then given an expression which is a product and can be dealt with abit easier.



    Damn didn't see GHOSH-5's.
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    Would you rather give up salt or pepper?
    Useful resources

    Make your revision easier

    Maths

    Maths Forum posting guidelines

    Not sure where to post? Read the updated guidelines here

    Equations

    How to use LaTex

    Writing equations the easy way

    Student revising

    Study habits of A* students

    Top tips from students who have already aced their exams

    Study Planner

    Create your own Study Planner

    Never miss a deadline again

    Polling station sign

    Thinking about a maths degree?

    Chat with other maths applicants

    Can you help? Study help unanswered threads

    Groups associated with this forum:

    View associated groups
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

    Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

    Write a reply...
    Reply
    Hide
    Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.