Maths: Further Pure 2 (4756)

    Polar co-ordinates

      P1 Introduction:

      • One co-ordinate system is cartesian, (x,y)
      • The polar co-ordinate system contains the following:
        - A fixed point O, called the pole
        - A line coming from this point, usually going to the right. This is the initial line
      • Then, positions are defined with (r,θ), where:
        - r is the distance between the pole and point
        - θ is the angle between the initial line and an imaginary line between the pole and point, going anti-clockwise
      • The pole has the co-ordinates (0,undefined)
      • If r is negative, then the point will be the opposite side to where θ actually indicates, i.e. (r,θ)(r,θ+π)
      • If r>0, draw a solid line between the pole and point
      • If r<0, draw a dashed line between the pole and point

      Converting between polar and cartesian co-ordinates:

      • To convert from polar co-ordinates to cartesian, use arctan and Pythagoras' theorem
      • You can check with a Casio scientific calculator:
      • Cartesian polar:
        - Enter Pol(x,y) where x and y are the cartesian co-ordinates
        - Pol is SHIFT and +
        - Get a comma with SHIFT and the right bracket
      • Polar cartesian:
        - Enter Rec(r,θ) where r is the distance and θ is the angle
        - Rec is SHIFT and
        - You do not need to specify the angle units (deg/rad), just set them in the calculator options

      Converting between polar and cartesian equations:

      • Use the following relationships:
        - x2+y2=r2
        - x=rcosθ
        - y=rsinθ
      • For example, to convert r=8sinθ to cartesian:
        - Multiply both sides by r: r2=8rsinθ
        - Use the third relationship listed above: r2=8y
        - Use the first relationship above: x2+y2=8y
        - This is a cartesian equation. To simplify further, move the 8y to the LHS and complete the square:
        x2+ [(y4)216] =0

      P2 Sketching polar curves:

      • First, create a table of values - for several values of θ at equal intervals, calculate r
      • Now use these to sketch the curve, using the rules in P2
      • Many graphical calculators can sketch polar curves. For example, in CASIO calculators, change from Y= mode to r=, and then enter the RHS using the θ button

      P3 Finding the area enclosed by a polar curve:

      • Use the formula area=12r2 dθ where r is the polar equation (e.g. r=2cosθ)


      • The graph of r=θ3 forms a spiral as shown below:
      • We will find the area between the curve and y-axis shown in the graph. This is in the range [0,π2]
      • First, substitute the polar equation into the area formula: A=120π2(θ3)2 dθ
      • This simplifies to A=120π219θ2 dθ
      • Now we can integrate to A=12[127θ3]0π2
      • So A=12((π÷3)3270)=π30.5×8×27=π3108


        c1 Inverse trig functions:

        • sin, cos and tan functions all have many-to-one mapping
        • Therefore, their inverses would be one-to-many
        • A function cannot be one-to-many, so the range is restricted in the inverse trig functions to:
          - π2arcsinπ2 (domain is 1x1)
          - 0arccosπ (domain is 1x1)
          - π2<arctan<π2 (domain is xR)

        Inverse reciprocal trig functions in a calculator:

        • To enter arcsecx in a calculator, enter cos1(1x)
        • For arccosecx, enter sin1(1x)
        • For arccotx, enter π2tan1x

        c2 Differentiating inverse trig functions:

        • Use one of the following results (in the formula booklet)
          - ddx(arcsinx)= 11x2
          - ddx(arccosx)= 11x2
          - ddx(arctanx)= 11+x2
        • Also, you should remember the identity ddx(arccotx)=ddx(arctanx)

        Example: find d/dx(arcsinx):

        • This is the same as differentiating y=arcsinx
        • Find x in terms of y: siny=x
        • Now differentiate this implicitly (C3): dydxcosy=1
        • Solve for dydx: dydx=1cosy
        • Now you need to replace the cos with a sin. The steps are listed below:
          - Pick an identity sin2θ+cos2θ=1
          - We want this in terms of cos: cos2θ=1sin2θ
          - We also need to remove the powers: cosθ=±1sin2θ
        • Now substitute: dydx=1±1sin2y
        • Because of the restricted range of the arcsin function, the ± can be removed
        • A similar method can be used to differentiate arccos

        Example: find d/dx(arctanx):

        • This is the same as differentiating y=arctanx
        • So we need to implicitly differentiate tany=x
        • This gives us sec2y×dydx=1 (remember that ddx(tana)=sec2a)
        • Now make this in terms of dydx: dydx=1sec2y
        • We said earlier that y=arctanx, so this is the same as 1sec2(arctanx)
        • Now use the identity sec2a=tan2a+1:
        • We can write this as 1(tan(arctanx))2+1
        • Because tan(arctanx)=x, this simplifies to dydx=1x2+1

        c3 Integration:

        • From the above differentiation results, the following can be derived for integration; they are given in the formula booklet:
          - 1a2x2dx=arcsin(xa)+c
          - 1a2+x2dx=1aarctan(xa)+c
        • They will only work when x2 has no coefficient. Therefore, you must remove coefficients by rearranging to get a fraction outside of the integral. Example:
          Find I where I=1163x2dx.
          1) Factorise the bracket to get x2 on its own: I=13(163x2)dx
          2) Now move the 3 outside of the square root sign: I=13163x2dx
          3) Now move the 3 outside of the integral: I=131163x2dx
          4) Now use the relevant result: I=13arcsinx34+c
        • The alternative method would be to use the substitution u=x3

        c4 Using identities to integrate functions:

        • The below identities can easily be derived from the formula booklet (see C4)
        • If you rearrange these to make sin2x and cos2x the subject, you get the following identities, which are particularly useful (since it's very easy to integrate sin2x and cos2x):

        Example: find sin2x dx:

        • Start with sin2x=12(1cos2x) from above
        • Substitute this into the integral: I=12(1cos2x)dx
        • Move the 12 to the left of the and integrate to I=12(x12sin2x)+c
        • Expand the brackets to I=x214sin2x+c

        Example: find sin3x dx:

        • Rearrange to get a quadratic: I=(sinx×sin2x) dx
        • Rearrange sin2x+cos2x=1 to sin2x=1cos2x
        • Combine the two: I=sinx(1cos2x) dx
        • Let u=cosx. Use integration by substitution to get I=sinx(1u2)sinxdu =1+u2 du=u21 du
        • Integrate this to I=13u3x=13cos3xx+c

        Example: find sin4x dx:

        • Rearrange to get the integral in terms of sin2x: I=(sin2x)2 dx
        • Rearrange cos2x=1sin2x to sin2x=12(1cos2x)
        • Combine the two: I=(12(1cos2x))2 dx
        • Expand the brackets: I=14(12cos2x+cos22x)
        • The third term needs to be simplified further to be integrated:
          - Rearrange 2cos2a1=cos2a to cos2a=12(1+cos2a)
          - We have cos22x and not cos2x, so let a=2x: cos22x=12(1+cos4x)
          - This simplifies to 12+12cos4x
        • Now substitute this in: I=141 2cos2x+12+ 12cos4x dx =1432 2cos2x+12cos4x dx
        • Now integrate to I=14(32x sin2x+18sin4x)+c
        • Simplify to I=38x14sin2x +132sin4x+c

        Example: find tanx dx:

        • I=tanx dx=sinxcosx dx
        • Let u=cosx. dx=1sinx du
        • Use integration by substitution: I=sinxsinx×u du
        • Simplify: I=1u du
        • Use the result 1x dx=ln(|x|)+c:
        • This can be simplified further, since ln(a)=ln(a1). So I=ln(|secx|)+c


          s1 Maclaurin series:

          • A Maclaurin series is an approximation of a function as a sum of infinite terms
          • They are centred around x=0 (unlike Taylor series) so the approximation will be the most accurate near to x=0
          • The more terms you use, the more accurate the approximation will become
          • The formula (given in the formulae booklet) is f(x)=f(0)+xf(0) + x22!f(0) + ... + xrr!f(r)(0) + ...
          • To find an expansion to a specified degree of accuracy (e.g. 5 decimal places), keep adding terms until adding another does not affect your answer to the degree of accuracy required

          Example expansion:

          • We'll be finding the expansion of f(x)=cosx
          • The first term is f(0) which is 1
          • The second term is xf(0). We don't know x so we'll leave that as it is. This is multiplied by the derivative of f(x) at x=0 which is sin0=0. So the second term is x×0=0
          • The third term is x22!f(0). f(0) is cos0=1 so the third term is x22
          • With the same method, this continues. So the fourth term is x33!×sin0=0, and fifth is x44!×cos0= x44!
          • This trend continues. So our expansion up to the fourth power would be 1+0  x22!+ 0  x44! (which can be further simplified)

          s2 Convergence:

          • With infinite terms, some series such as the one above for f(x)=cosx will be accurate for all values of x
          • But for some series, the approximation will only be valid for a small domain, and outside of this, adding more terms will decrease the accuracy
          • For example, f(x)= 11x and the first 8 terms of its Maclaurin series are shown below. You can see that it is only valid for a limited domain

            With more terms, the values for x-coordinates outside of the valid domain would be even less accurate. For example, the Maclaurin series with only one term is just y=1. At x=4, this is much more accurate than with more terms

          s3 Maclaurin series for standard functions:

          • You need to be famililar with the following Maclaurin expansions from the Infinite series section of the formula booklet
            ex=1+x+x22!+...+ xrr!+..., all x
            ln(1+x)=xx22+ x33...+(1)r+1xrr+..., 1<x1
            sinx=xx33!+x55! ...+(1)rx2r+1(2r+1)!+..., all x
            cosx=1x22! +x44!...+ (1)rx2r(2r)!+..., all x
          • You also need to be familiar with the (1+x)n binomial expansion from the C4 module (in the Binomial expansions section of the formula booklet). This is also a Maclaurin series; you can derive it by using the chain rule and the Maclaurin series general formula in s1
            (1+x)n=1+nx+ n(n1)2!x2+...+ n(n1)...(nr+1)12...rxr+..., |x|<1

          Complex numbers

            j1 Polar form:

            • Polar form or modulus-argument form is a way of writing complex numbers with the notation z=r(cosθ+jsinθ) where r=|z| (the modulus) and θ=argz (the argument, usually written in radians), as seen in FP1
            • Rectangular form (z=a+bj) and polar form can be converted on fx-991 series calculators, see here for more details
            • Ensure that r>0 and π<θπ
            • To get the argument within this range, keep adding/subtracting 2π

            j2 Multiplying and dividing numbers in polar form:

            • To multiply two complex numbers in polar form, multiply their moduli and add their arguments
            • To divide two complex numbers in polar form, divide their moduli and subtract their arguments

            j3 de Moivre's theorem:

            • For any integer n, (cosθ+jsinθ)n=cosnθ+jsinnθ
            • When the modulus of a complex number is not 1, de Moivre's theorem is often useful if you treat it separately:
              [r(cosθ+jsinθ)]n =rn(cosθ+jsinθ)n =rn(cosnθ+jsinnθ)

            j4 Applying de Moivre's theorem:

              Example: express cos3θ in terms of cosθ only:

            • 1) Write this as a complex number (cos3θ is real so j=0): cos3θ=(cosθ+jsinθ)3
            • 2) Expand this with binomial expansion (see the C1 module). (I use c to mean cosθ etc):
              (cosθ+jsinθ)3 =1×c3×(js)0  +  3×c2×(js)1  +  3×c1×(js)2  +  1×c0×(js)3
            • 3) Now set j=0 and simplify, to c33cs2
            • 4) Because sin2θ+cos2θ1, the s can be eliminated, to c33c(1c2)= c33c+3c3= 4c33c
            • 5) So cos3θ=4cos3θ3cosθ

              Similarly, if expressing a sin function in terms of sinθ only, you would instead eliminate the real part in step 3

              For a tan function, remember that tanθ= sinθcosθ

            Useful deductions:

            • Let z=cosθ+jsinθ
            • The following deductions can be made from the main theorem:
            • zn=cosnθjsinnθ
            • cosnθ= zn+zn2
            • sinnθ= znzn2j

            Example: express cos5θ in terms of multiple angles:

            • Let z=cosθ+jsinθ
            • With the first identity above, we can say that 2cosθ=z+z1
            • If we raise this to the 5th power, we get 25cos5θ=(z+z1)5
            • Now expand and simplify the right side to z5+5z3+10z +10z1+5z3+z5
            • By factorising into groups, RHS =(z5+z5)+5(z3+z3) +10(z+z1)
            • Using the definition that z=cosθ+jsinθ, this simplifies to 2cos5θ+10cos3θ +20cosθ
            • The left hand side is still 25cos5θ, so we need to divide both sides by 25 to get the final answer, cos5θ= 2cos5θ+10cos3θ+20cosθ32. This can still be simplified - divide the numerator and denominator by 2

            Summing series:

            • Aim to simplify the series to a form where reverse binomial expansion or a sum of series formula can be applied
            • To get to this stage, it is often useful to find a way to use de Moivre's theorem
            • For example, when simplifying a cos series, try introducing a sin series called S and try adding jS to the original sequence. de Moivre's theorem can now be used

            Summing series example:

            • Here is an exam question, exactly as it was written in the paper I took it from:

              The infinite series C and S are defined as follows.

              By considering C+jS, show that

              and find a similar expression for S

            • First, you need to add the two series, multiplying each term in the S series by j:
              C+jS=(cosθ+jsinθ) +13(cos3θ+jsin3θ) +19(cos5θ+jsin5θ) +
            • Now de Moivre's theorem can be used:
              C+jS=(cosθ+jsinθ)1 +13(cosθ+jsinθ)3 +19(cosθ+jsinθ)5 +
            • Convert to exponential form (this is covered in j5):
              C+jS=ejθ+13e3jθ +19e5jθ
            • You can tell that this is a geometric series where
              a=ejθ and
            • From the formula booklet, S= a1r, so S=C+jS= ejθ113e2jθ
            • It's generally best to get rid of fractions at this stage:
              C+jS= 3ejθ3e2jθ
            • Now multiply the top and bottom of the fraction by the conjugate of the denominator:
              C+jS= 3ejθ3e2jθ × 3e2jθ3e2jθ = 9ejθ3ejθ93e2jθ3e2jθ+1
            • Convert back into cosθ+jsinθ form:
              C+jS= 9(cosθ+jsinθ)3(cosθjsinθ)103(cos2θjsin2θ)3(cos2θ+jsin2θ)
            • Simplify to C+jS= 6cosθ+12jsinθ106cos2θ
            • C will be the real part of the numerator, with the entire denominator, so
              C= 6cosθ106cos2θ
            • S will be whatever you need to add to this to get your simplified C+jS, so
              S= 12sinθ106cos2θ
            • (you can simplify the above fractions by dividing numerator and denominators by 2)

            j5 Exponential form:

            • If z=r(cosθ+jsinθ), then:
            • de Moivre's theorem is simply (ejθ)n=ejnθ
            • The following can now be derived from the relationships in the previous section:
              cosθ= ejθ+ejθ2,
              sinθ= ejθejθ2j

            j6 nth roots of a complex number:

            • All non-zero complex numbers have n nth roots. For example, z3=5+5j has three roots
            • On an Argand diagram, these form the vertices of a n-gon (e.g. the roots of z4=2+2j would form a square)

            j7 Finding nth roots:

            • The roots of rejθ are nr eθ+2kπn where k is the root number (starting at k=0 for the first root, and finishing at k=n1 for the last)

            Roots of unity:

            • A root of unity is a complex number which gives 1 when raised to the power of a positive integer
            • For example, the polynomial z51 has 5 roots of unity (values of z which multiply together to make 1)
            • The argument of 1 is 0, so you should use the formula above with θ=0
            • The first root of unity is therefore always 1
            • The symbol ω is used to represent the root of unity with the smallest positive argument
            • ω2 is used to represent the next root, ω3 for the next and so on

            j8 Sum of roots of unity:

            • 1+ω+ω2+ω3++ωn1= a(1rn)1r = 1(1ωn)1ω = 01ω =0 (since ωn=ω0=1)
            • Therefore, the sum of roots of unity is always 0. You need to memorise the derivation above
            • You can also prove this geometrically - the shape formed when you connect the roots is always centred about (0,0)

            j9 Multiplication by a complex number:

            • In an Argand diagram, if you multiply by rejθ, there will be an enlargement of scale factor r and a rotation of θ about the origin
            • For example, if you multiply by j (which can be written as 1ej×π2), there will be a rotation of π2 about the origin (and no enlargement, since r=1)


              m1 3×3 determinants with the cofactor method:

              • If M is the matrix (abcdefghi), detM = a|efhi| b|dgfi|+c|degh|
              • The second component of this sum is subtracted because b is negative on the sign matrix (+++++)
              • The 2×2 determinants here are called minors; the minor of a is the determinant of the 2×2 matrix when the row and column of a is crossed out
              • A minor with the correct sign from the sign matrix above is called a cofactor
              • You don't have to use the sum of the top row element cofactors to calculate the determinant, you can use any row or column in the matrix. Therefore, it's easier to pick a row/column with some zeroes

              3×3 determinants with the Sarrus method:

              • Create two new rows under the matrix, and copy into them the first two rows of the matrix, e.g.: (abcdefghiabcdef)
              • Multiply the 3 elements in each of the 6 diagonals
              • The determinant is the sum of the ↙ subtracted from the sum of the ↘ products; ↘
              • So the determinant would be (aei+dhc+gbf)(ceg+fha+ibd)

              Finding the inverse of a (non-singular) 3×3 matrix:

              • Find the cofactor of each element and create a 3×3 matrix of cofactors
              • Then transpose it (the first column becomes the first row, the second column becomes the second row, etc) to get the adjugate, adjM
              • Finally, multiply this by 1detM to get M1

              m2, m6 Eigenvectors:

              • If s is a non-zero vector and Ms=λs, then s is called an eigenvector and λ is called an eigenvalue
              • An eigenvector is a vector which is mapped to a multiple of itself by M
              • Each eigenvector can be multiplied by any integer and it will still work. Therefore, the elements are simplified as much as possible. For example, the eigenvectors (24) and (12) are the same
              • The trace of a matrix is the sum of its eigenvalues. It's also the sum of the top-left to bottom-right diagonal on the matrix. You don't need to know this, but it's useful for checking that you haven't made an error

              Finding eigenvalues and eigenvectors:

              • Start with Ms=λs from above
                MsλIs=0 (multiplying by the identity matrix)
                (MλI)s=0 (factorising)
              • This gives the equations det(MλI)=0 and (MλI)s=0
              • Use the first to find the eigenvalues; find the determinant of the matrix with λ subtracted from all values on the top-left to top-right diagonal. Set this equal to 0 and solve for λ to get a polynomial
              • The polynomial you get is called the characteristic equation. It will have the same order as the matrix (e.g. it'll be a cubic if the matrix is 3×3). Solve it to find the eigenvalues
              • Substitute each eigenvalue into the second equation [(MλI)s=0] to find the eigenvectors for each eigenvalue

              m3 Diagonal form:

              • The matrix P is created by writing the eigenvectors next to each other. For example, with the eigenvectors (11) and (21), P=(1211)
              • The matrix D is created by diagonally writing the eigenvalues onto a matrix of zeroes. For example, with the eigenvalues 2 and 5, D=(2005)
              • If you construct P and D correctly, M=PDP1

                Note: the symbols S and Λ are used in the textbook instead of P and D respectively. I used these because they're what the exams use

              m4 Finding powers of matrices:

              • The formula Mn=PDnP1 can be used to raise a matrix M to a power n
              • Calculating Dn is very simple - because each eigenvalue in the matrix only shares its row and column with zeroes, you can simply raise each eigenvalue by the power n

              m5 Simultaneous equations:

              • If the determinant of a matrix is not 0, you can solve three variable simultaneous equations with a 3×3 matrix as in FP1
              • If it is 0, there are either no solutions or infinite solutions
              • To find out which of the above is the case, you need to eliminate one of the variables. Add or substract a multiple of an equation to/from another equation to do this. If the answer is the same as the remaining equation, then the equations are consistent and there are infinite solutions
              • To solve a set of consisten simultaneous equations, introduce a new variable λ. Let it equal one of your variables and then find the solutions in terms of λ

              m7 The Cayley–Hamilton Theorem:

              • The Cayley–Hamilton Theorem says that a matrix satisfies its own characteristic equation
              • For example, let M=(1214). This has the characteristic equation λ25λ+2=0
                Let's say we want to calculate M4. First, substitute M into the characteristic equation, replacing λ. Multiply the integer by I. This gives us M25M+2I=0
                We now need to rearrange this in terms of M2, leaving us M2=5M2I
                This relationship can be used to calculate M4:

                M4=105(1214) 46(1001)
                      =(105210105420) (460046)

              Hyperbolic functions

                a4 The hyperbolic functions:

                • The three main hyperbolic trig functions are sinh, cosh and tanh
                • There's no 'correct' pronunciation, but they are usually pronounced 'shine', 'cosh' (rhyming with 'posh') and 'than' (with the 'th' being said as it is in 'theta')
                • These also have inverses; arsinh, arcosh and artanh (note: it's arsinh, not arcsinh; the same applies for arcosh and artanh)
                • And there are cosech, sech and coth and their inverses. These are the reciprocals of their related standard hyperbolic trig function

                Identities involving hyperbolic trig functions:

                • You can use the standard trigonometric identities. However, you must flip the sign if you are squaring sinh
                • For example, if you apply the identity cos2θ+sin2θ1 to sinhθ and coshθ, you must flip the sign in front of sin2θ since you are squaring sin. So this identity would become cosh2θsinh2θ1


                • Has an infinite domain and range
                • Is an odd function
                • sinhx=12(exex)


                • Has an infinite domain. Range is limited: y1
                • Is an even function
                • coshx=12(ex+ex)


                • Has an infinite domain. Range is 1<y<1
                • Is an odd function
                • Use the identity tanhx= sinhxcoshx
                • Therefore, tanhx= exexex+ex

                a5 Differentiating hyperbolic functions:

                • ddx(sinhx)=coshx
                • ddx(coshx)=sinhx
                • Sometimes you need to convert to 12(ex±ex) form, simplify, then convert back to hyperbolic functions

                Integrating hyperbolic functions:

                • The above applies for integrating

                a6 Inverse hyperbolic functions:

                • These functions are the same as the above, except reflected in the line y=x
                • y=arsinhx:
                • y=arcoshx:
                • y=artanhx:
                • To prevent it being a one-to-many function, the domain of y=arcoshx is restricted to x1
                • Similarly, y=artanhx is restricted to 1<x<1
                • The functions arcosechx, arsechx and arcothx are equivalent to arsinh(1x), arcosh(1x) and artanh(1x) respectively

                a7 Logarithmic form of inverse hyperbolic functions:

                • The formula book contains the following, which can be used to differentiate and integrate inverse hyperbolic functions:
                • arsinhx=ln(x+x2+1)
                • arcoshx=ln(x+x21), x1
                • artanhx=12ln(1+x1x), |x|<1

                a8 Integrating functions of the form (x2 ± a2 )1/2:

                • For these, you just need to use integration by substitution with the results in the formula booklet. These are:
                  1a2+x2 dx=arsinh(xa)+c
                  1x2a2 dx=arcosh(xa)+c