The equation of the circle centre (a,b) radius r is (x-a)^2 + (y-b)^2 = r^2.
A chord is a line that joins two points on the circumference of a circle.
The perpendicular from the centre of a circle to a chord bisects the chord.
The angle in a semicircle is a right angle.
A tangent is a line that meets a circle at one point only.
The angle between a tangent and a radius is 90°.
Chapter 5 summary - The binomial expansion
You can use Pascal’s Triangle to multiply out a bracket.
You can use combinations and factional notation to help you expand binomial expressions. For larger indices it is quicker than using Pascal’s Triangle.
where x and y are the coordinates of P and r is the radius of the circle.
In the first quadrant, where \theta is acute, All trigonometric functions are positive.
In the second quadrant, where \theta is obtuse, only Sine is positive.
In the third quadrant, where \theta is reflex, 180^\circ < \theta < 270^\circ, only Tangent is positive.
In the fourth quadrant, where \theta is reflex, 270^\circ < \theta < 360^\circ, only Cosine is positive.
The trigonometric ratios of angles equally inclined to the horizontal are related:
For an increasing function {\rm f}(x) in the interval (a, b), {\rm f}'(x)>0 in the interval a \leq x \leq b.
For a decreasing function {\rm f}(x) in the interval (a, b), {\rm f}'(x)<0 in the interval a \leq x \leq b.
The points where {\rm f}(x) stops increasing and begins to decrease are called maximum points.
The points where {\rm f}(x) stops decreasing and begins to increase are called minimum points.
A point of inflexion is a point where the gradient is at a maximum or minimum value in the neighbourhood of the point.
A stationary point is a point of zero gradient. It may be a maximum, a minimum or a point of inflexion.
To find the coordinates of a stationary point find \displaystyle {{\rm d}y\over {\rm d}x}, i.e. {\rm f}'(x), and solve the equation {\rm f}'(x)=0 to find the value, or values, of x and then substitute into y={\rm f}(x) to find the corresponding values of y.
The stationary value of a function is the value of y at the stationary point. You can sometimes use this to find the range of a function.
You may determine the nature of a stationary point by using the second derivative.
If \displaystyle {{\rm d}y\over {\rm d}x}=0 and \displaystyle {{\rm d}^2y\over {\rm d}x^2}>0, the point is a minimum point.
If \displaystyle {{\rm d}y\over {\rm d}x}=0 and \displaystyle {{\rm d}^2y\over {\rm d}x^2}<0, the point is a maximum point.
If \displaystyle {{\rm d}y\over {\rm d}x}=0 and \displaystyle {{\rm d}^2y\over {\rm d}x^2}=0, the point is either a maximum or a minumum point or a point of inflexion.
Hint: In this case you need to use the tabular method and consider the gradient on either side of the stationary point.
If \displaystyle {{\rm d}y\over {\rm d}x}=0 and \displaystyle {{\rm d}^2y\over {\rm d}x^2}=0, but \displaystyle {{\rm d}^3y\over {\rm d}x^3}\not= 0, then the point is a point of inflexion.
In problems where you need to find the maximum or minimum value of a variable y, first establish a formula for y in terms of x, then differentiate and put the derived function equal to zero to find x and then y.
Chapter 10 summary - Trigonometrical identities and simple equations
\displaystyle \tan \theta = {\sin \theta\over \cos \theta} (providing \cos \theta \not= 0, when \tan \theta is not defined)
\sin^2 \theta + \cos^2 \theta = 1
A first solution of the equation \sin x = k is your calculator value, \alpha = \sin^{-1} k. A second solution is (180^\circ - \alpha), or (\pi - \alpha) if you are working in radians. Other solutions are found by adding or subtracting multiples of 360° or 2\pi radians.
A first solution of the equation \cos x = k is your calculator value, \alpha = \cos^{-1} k. A second solution is (360^\circ - \alpha), or (2\pi - \alpha) if you are working in radians. Other solutions are found by adding or subtracting multiples of 360° or 2\pi radians.
A first solution of the equation \tan x = k is your calculator value, \alpha = \tan^{-1} k. A second solution is (180^\circ + \alpha), or (\pi + \alpha) if you are working in radians. Other solutions are found by adding or subtracting multiples of 360° or 2\pi radians.
Chapter 11 summary - Integration
The definite integral \displaystyle \int_a^b {\rm f}'(x){\rm d}x = {\rm f}(b) - {\rm f}(a).
The area beneath the curve qith equation y={\rm f}(x) and between the lines x=a and x=b is
Area \displaystyle = \int_a^b {\rm f}(x){\rm d}x
The area between a line (equation y_1) and a curve (equation y_2) is given by