A Level Maths: Pure 3 formulas

Every chapter of A Level Maths: Pure 3 on one page: the 63 formulas, definitions and facts to remember, in syllabus order. Use it for a last look before a test, then check yourself.

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Algebra

If k > 0, |f(x)| = k means f(x) = k or f(x) = −k; and |x − a| < b means a − b < x < a + b
For |f(x)| = |g(x)| or |f(x)| > |g(x)|, square both sides to remove the modulus signs
Remainder theorem: when p(x) is divided by (x − a) the remainder is p(a); when divided by (ax − b) it is p(b/a)
Factor theorem: (x − a) is a factor of p(x) exactly when p(a) = 0
Partial fractions: (x − a)(x − b) gives A/(x − a) + B/(x − b); a repeated factor (x − a)² gives A/(x − a) + B/(x − a)²; a factor (x² + c) gives (Bx + C)/(x² + c)
(1 + x)ⁿ = 1 + nx + [n(n − 1)/2!]x² + [n(n − 1)(n − 2)/3!]x³ + ..., valid for |x| < 1
(a + bx)ⁿ = aⁿ(1 + bx/a)ⁿ, valid for |bx/a| < 1

Logarithmic and exponential functions

log_a x = y means a^y = x; so log_a 1 = 0 and log_a a = 1
log x + log y = log(xy); log x − log y = log(x/y); log(xⁿ) = n log x
ln(e^x) = x and e^{ln x} = x (for x > 0)
a^x = b gives x = ln b ÷ ln a
In an inequality, dividing by ln a reverses the sign when 0 < a < 1, because ln a is then negative
y = kxⁿ gives ln y = n ln x + ln k: plot ln y against ln x; gradient = n, intercept = ln k
y = k(a^x) gives ln y = (ln a)x + ln k: plot ln y against x; gradient = ln a, intercept = ln k

Trigonometry

sec θ = 1/cos θ, cosec θ = 1/sin θ, cot θ = 1/tan θ = cos θ/sin θ
sec²θ = 1 + tan²θ and cosec²θ = 1 + cot²θ
sin(A ± B) = sin A cos B ± cos A sin B
cos(A ± B) = cos A cos B ∓ sin A sin B (the sign in the middle is the opposite one)
tan(A ± B) = (tan A ± tan B)/(1 ∓ tan A tan B)
sin 2A = 2 sin A cos A; cos 2A = cos²A − sin²A = 2cos²A − 1 = 1 − 2sin²A; tan 2A = 2 tan A/(1 − tan²A)
For a, b > 0: a sin θ + b cos θ = R sin(θ + α), where R = √(a² + b²) and tan α = b/a; the greatest value is R and the least is −R

Differentiation

d/dx(e^{ax + b}) = ae^{ax + b}; d/dx[ln f(x)] = f′(x)/f(x), so d/dx(ln x) = 1/x
d/dx(sin ax) = a cos ax; d/dx(cos ax) = −a sin ax; d/dx(tan ax) = a sec² ax
d/dx(tan⁻¹ x) = 1/(1 + x²)
Chain rule: dy/dx = dy/du × du/dx
Product rule: if y = uv, dy/dx = u(dv/dx) + v(du/dx)
Quotient rule: if y = u/v, dy/dx = [v(du/dx) − u(dv/dx)]/v²
Parametric: dy/dx = (dy/dt) ÷ (dx/dt). Implicit: d/dx(yⁿ) = ny^{n − 1}(dy/dx) and d/dx(xy) = y + x(dy/dx)

Integration

∫ e^{ax + b} dx = (1/a)e^{ax + b} + c; ∫ 1/(ax + b) dx = (1/a) ln|ax + b| + c
∫ sin(ax + b) dx = −(1/a) cos(ax + b) + c; ∫ cos(ax + b) dx = (1/a) sin(ax + b) + c
∫ sec²(ax + b) dx = (1/a) tan(ax + b) + c
∫ 1/(x² + a²) dx = (1/a) tan⁻¹(x/a) + c
sin² x = ½(1 − cos 2x); cos² x = ½(1 + cos 2x); tan² x = sec² x − 1
∫ f′(x)/f(x) dx = ln|f(x)| + c
By parts: ∫ u(dv/dx) dx = uv − ∫ v(du/dx) dx; choose u = ln x when ln x appears, otherwise u = the power of x

Numerical solution of equations

Change of sign: if f is continuous on the interval and f(a), f(b) have opposite signs, a root lies between a and b
The method fails if the curve has a break (an asymptote) in the interval, and it can miss roots when there is an even number of them in the interval
To show a root is 1.53 correct to 2 decimal places, show a change of sign between 1.525 and 1.535
If x_n+1 = F(x_n) converges to α, then α = F(α); rearrange this to find the equation that α satisfies
Keep at least 4 decimal places in each iterate when the answer is wanted to 2 decimal places, and stop when two iterates agree to the accuracy needed
Different rearrangements of the same equation can converge to different roots, or not converge at all
Use radians for any iteration with sin, cos or tan

Vectors

Magnitude: |xi + yj + zk| = √(x² + y² + z²); unit vector = vector ÷ its magnitude
Vector AB = b − a; the midpoint of AB has position vector ½(a + b)
Line: r = a + tb, where a is a point on the line and b is its direction; the line through A and B has direction b − a
Lines are parallel if one direction vector is a multiple of the other; to test for intersection, equate two components, solve for s and t, then check the third component; if the check fails and the lines are not parallel, they are skew
Scalar product: a·b = a₁b₁ + a₂b₂ + a₃b₃ = |a||b| cos θ
a·b = 0 means the vectors are perpendicular; for the angle between two lines use only their direction vectors
Foot of the perpendicular N from P to a line: write N as a general point on the line, then solve PN·b = 0 for t; the distance is |PN|

Differential equations

"The rate of increase of x is proportional to x" means dx/dt = kx; a rate of decrease needs a minus sign: dx/dt = −kx
"Inversely proportional to x" means dx/dt = k/x
Separate: dy/dx = f(x)g(y) becomes ∫ [1/g(y)] dy = ∫ f(x) dx
Write one constant of integration, on one side only, as soon as you integrate
Put in the given condition to find c before you rearrange to make y the subject
dy/dx = ky has the solution y = Ae^kx, where A is the value of y when x = 0
As t becomes large, e^−kt tends to 0 (for k > 0); use this to find the long-term value

Complex numbers

For z = x + iy: |z| = √(x² + y²), z* = x − iy, z + z* = 2x and zz* = x² + y²
arg z: find tan⁻¹|y/x|, then use a sketch to put the angle in the correct quadrant, with −π < arg z ≤ π
To divide, multiply the top and bottom by the conjugate of the bottom
Polar form: multiply the moduli and add the arguments; to divide, divide the moduli and subtract the arguments
Square roots of a + bi: put (x + iy)² = a + bi, then x² − y² = a and 2xy = b
Loci: |z − a| = r is a circle, centre a, radius r; |z − a| = |z − b| is the perpendicular bisector of the line joining a and b; arg(z − a) = θ is a half-line from a at angle θ
Real coefficients: if p + qi is a root, so is p − qi; together they give the factor z² − 2pz + (p² + q²)