C3 Algebraic fractions
… Adding or subtracting by making the denominators the same.
… Dividing algebraic polynomials using the bus stop method.
… Dividing algebraic polynomials using the coefficients method and the remainder theorem. If the numerator is an order of 1 greater than the denominator (the powers), then:
F(x) = (quotient x divisor) + Remainder
F(x) = (Quotient x divisor) + D
… If the numerator has a order of 4 (the highest power of x) vs. the denominator order of 2, then:
F(x) = (Quotient x divisor) + Dx + E
This is because the the remainder will contain an ‘extra’ power of x term.
For example:
(2x^4 + 3x^3 – 2x^2 + 4 – 6) / (x^2 + x – 2)
We can say:
2x^4 + 3x^3 – 2x^2 + 4 – 6 = (ax^2 + bx + c) (x^2 + x – 2) + dx + e
Where dx + e is the remainder.
First find the remainder by factoring the divisor into (x – 1)(x + 2), which can be made to give 0 if:
x = 1: giving f(1) = 1, so
1 = d(1) + e
or x = -2: giving f(-2) = -14, so
-14 = d(-2) + e
Solving these two simultaneous equations we get d = 5 and e = -4.
So now:
2x^4 + 3x^3 – 2x^2 + 4 – 6 = (ax^2 + bx + c) (x^2 + x – 2) + 5x – 4
And we can equate coefficients in the normal way to find a, b and c.
It’s actually much easier to divide using a bus stop!
Differentiation Techniques
Product Rule
… Using the product rule to differentiate functions that multiply, e.g. (2y)(lny)
… If y = uv, then
dy/dx = u’v + v’u
… When the product rule is made to go in ‘reverse’, we get a way to integrate multiplying functions called ‘integration by parts’ … We’ll cover this soon!
The Quotient Rule
… Using the quotient rule to differentiate dividing functions
If y = u/v, then
dy/dx = (u’v – v’u)/v²
… Take care that you have u, v, u’ and v’ in the right places when using the quotient rule (the minus sign means that everything has to be in the right order!)
Chain Rule
… Using the chain rule to differentiate ‘compound’ functions (potatoes within potatoes) such as (x³ + 7)². The “u =” method is a good step by step method to use for more difficult cases, making u = ‘inner potato’.
… The reverse chain rule can be used for integrating a ‘function within a function’, for example:
∫(5 – 3x)²dx
Start by guessing a reasonable answer:
= (5 – 3x)³
Then check it by differentiating (we should get back to the expression in the original integral):
-3(5 – 3x)² … Ah, but we don’t want the -3, so modify by multiplying by a correcting factor as required:
= (-1/3)(5 – 3x)³ + c
So this process is: “Guess – Check (differentiate) – Modify”
Differentiation from first principles
… This uses the general result that:
dy/dx = lim(δx→0) δy/δx = lim(δx→0) { f(x+δx) – f(x) ) / δx }
For any function, substitute the value of x and x+δx into the function. Then put these values into the above formula and simplify, remembering that δx→0 (i.e. disappears!)
… To prove the differentiation from first principles formula, start by drawing a curve with two points:
(x , f(x) )
(x+h , f(x+h) )
Then find the change in y between the two points and write down an expression for the gradient:
δy/δx = (change in y / change in x) = ( f(x+h) – f(x) ) / h
Once you have simplified, then use the limit notation:
dy/dx = lim as h→0 | δy/δx
… and set h = 0.
… Differentiating sin or cos from first principles.
Remember to use the compound angle formulae and the small angle approximations:
h→0 then
cos(h) → 1 – h²/2
and
sin(h) → h
Differentiating sin, cos and tan
… We saw graphically how when we differentiate sin(x) we get cos(x). When we differentiate cos(x) we get -sin(x). This forms a continuous cycle for sin and cos:
sin(x) → cos(x) → -sin(x) → -cos(x) → back to sin(x)
Note that going backwards in this cycle is integration:
back to sin(x) ← cos(x) ← -sin(x) ← -cos(x) ← sin(x)
… The differential of tan(x) gives sec²(x), where sec(x) = 1/cos(x)
Differentiating exponentials
… The trick for differentiating y = a^x lies in knowing how to differentiate e^f(x):
First rewrite y = a^x as a power of Euler’s number:
y = e^(lna^x)
y = e^(xlna)
Now differentiate:
dy/dx = (lna)e^(xlna) = (lna)e^(lna^x) = (lna)a^x
or by taking natural logs of both sides:
y = a^x
ln(y) = ln(a^x)
ln(y) = xln(a)
x = [1/ln(a)] ln(y)
dx/dy = [1/ln(a)] 1/y
dy/dx = yln(a)
but y = a^x, so
dy/dx = (a^x)ln(a)
Alternatively, you can take natural logs of both sides and implicit differentiation:
y = a^x
ln(y) = ln(a^x)
ln(y) = xln(a)
Then implicitly differentiate each side:
(1/y)dy/dx = ln(a)
dy/dx = yln(a)
dy/dx = y = (a^x)ln(a)
… Note that whenever you see e^lnB then the e and ln ‘cancel’ each other out and you will be left with B.
This is also the case for ln(e^B) = B
How to differentiate inverse trig functions
… for example y = arcsin(x)
We first need to rewrite this to get rid of the ‘inverse bit’:
x = sin(y)
Now differentiate with respect to y
dx/dy = cos(y)
and ‘flip’:
dy/dx = 1/cos(y)
Finally, square both sides of x = sin(y), and use an identity:
x² = sin²(y)
x² = 1 – cos²(y)
cos²(y) = 1 – x²
cos(y) = √(1 – x²)
So
dy/dx = 1/√(1 – x²)
The same method can be used for arccos and arctan.
Functions
… Functions have inputs and outputs
- What can go into a function is called the Domain, e.g. x ∈ {1, 2, 3, 4}
- What may POSSIBLY come out of a function is called the Codomain, e.g. f(x) ∈ {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
- What ACTUALLY comes out of a function is called the Range, e.g. f(x) ∈ {1, 2, 3, 4}
… The Range is a subset of the Codomain.
Sometimes we don’t know the exact range, because the function may be complicated or not fully known. However, we may know the set it lies in (such as integers or reals), so we define the codomain as a ‘good start’!
… f:N → N
This says that the function “f” has a domain of “N” (the natural numbers), and a codomain of “N” also.
Check out https://www.mathsisfun.com/sets/domain-range-codomain.html for an very good explanation of these terms.
… A function is ‘defined’ (allowed) if it is ‘many to one’, or ‘one to one’… but NOT ‘one to many’.
… If you have a function which could result in two different outputs (such as a square root), first restrict the range and then rearrange to find out what the domain must be. For example, if
f(x) = √x
then restrict the range:
f(x) ≥ 0
and so x ≥ 0 (to avoid having a square root of a negative number)
… Set notation for describing solutions of a function. For example:
1) If a solution is given by x < a, OR x > b (i.e. two separate intervals), then we can write:
{x ∈ R : x < a} ∪ {x ∈ R : x > b}
2)) If the solution is given by x > a AND x < b (i.e. a single interval), then: {x ∈ R : x > a} ∩ {x ∈ R : x < b}
or just
{x ∈ R : a < x < b}
… Note that a function cannot give an ‘undefined’ output, such as infinity. For example, if:
f(x) = 1/(x-2)
then as 1/0 is not allowed (it results in infinity!), we can restrict the domain:
{x ∈ R : x ≠ 2}
… The set of natural numbers according to MEI is {1, 2, 3…}.
(Sometimes 0 (zero) is also included in other boards!)
… The ‘natural domain’ is the largest possible domain for a particular function.
… Symmetric functions in the line y = x are ‘self inverse’. This means that the inverse function is exactly the same as the original.
… An inverse function can only be found if the original function is a ‘one-to-one’.
… Finding the inverse function by making x the subject of the function and then swapping x and y around.
You can visualise the inverse function by sketching a graph of the original function and then flipping the page so that you are viewing the graph from the other side of the paper with the x and y axes swapped over!
… Reflection in the line y = x will give the inverse function.
… The domain of the inverse function is the RANGE of the original function, but in terms of x.
… The range of the inverse function is the DOMAIN of the original function, but in terms of f(x).
… If A is a point on a function f(x) and B is the reflection of point A on the line y=x, then:
The gradient of the tangent at A = reciprocal of gradient of tangent at B on the inverse of f(x).
… To find the range of a function:
- Draw a graph of the function using your calculator. This is a great way to see the whole picture quickly to start with.
- Test the extreme values of the domain (or at least close to them).
- Find f(0), if allowed
- If you have a quadratic of cubic, find dy/dx and use this to find the max or min stationary points.
- Note that for log(x) and ln(x), x must always be greater than zero
If in doubt, sketch it out!
… For sketching graphs of functions, you can ask:
– What happens to y if x = 0?
– What happens to y as x tends to positive infinity?
– What happens to y as x tends to negative infinity?
– What happens to y if x approaches an ‘asymptotic’ value [check the denominator for any factors such as (x – 2) etc]?
… Even functions: f(x) = f(-x). These are symmetric about the y axis, for example y = cos(x)
… Odd functions: f(x) = -f(-x). These usually go through the origin and have rotational symmetry of 2, for example y = sin(x)
… Completing the square for quadratic functions can help to find the function’s range (as it identifies the minimum or maximum y coordinate).
… Composite functions
gf(x), i.e. feed the output of function f(x) into function g(x).
g² means gg(x)
(fh)² means fhfh(x) !
… To find the domain of a composite function, for example: gf(x) where:
f(x) = 2x + 3 (x < 0)
g(x) = x² – 2x + 1 (x > 1)
1) What can we put into f?
x < 0
2) What can we put into g?
f(x) > 1 (because g only accepts x > 1 and we’re putting the output of f into g)
2x + 3 > 1
2x > -2
x > -1
3) Are there any other x values which will not work with gf(x)? Here we need to find the simplified expression gf(x) = 4(x + 1)²
x ∈ R (because there are no reciprocals or square roots which can lead to undefined outputs)
So putting all this together means that the domain of gf(x) is: -1 < x < 0
… Working backwards to write expressions in x as functions in terms of f(x) and g(x). For example:
x → √(x+4) , given that f(x)=√x and g(x)=x+4
This mapping is then given by fg
… Finding a function by working backwards from a composite function. For example:
if f(x) = 16x – e^(2x) and fg(x) = 16/x – e^(2/x),
then what is the function g(x)?
… Writing an algebraic expression in terms of two or more given functions.
Example 2:
“if y = x², what is a sequence of transformations which maps this curve onto the curve y = (2x – 3)² ?”
When you have two transformations in the x direction, do the TRANSLATION first…
So this is a translation +3 in the x direction followed by a stretch scale factor 1/2 parallel to the x axis.
Example 3:
… “Describe a single geometrical transformation that will transform the graph of y = f(x) into the graph of y = 3f(2q – x).”
This can be written as:
y = 3f( -x + 2q )
As we have two transformations in the x direction, do the translation first:
So these transformations are:
A translation of -2q in the x direction
A reflection in the y-axis (which is the same as a stretch scale factor -1 parallel to the x axis)
A stretch, scale factor 3 parallel to the y-axis.
The modulus function
… Modulus means always take the positive value of whatever is inside the modulus brackets. For example, |-5| = 5
… y = |f(x)|, for example y = |2x – 3|. To sketch this function, first sketch the line y = 2x – 3 and then reflect the negative y value part in the x axis. The modulus function is then only the positive y sections.
… Think of a modulus function as actually two functions. For example, y = |2x – 3| can be thought of as:
y = 2x – 3
and
y = -(2x – 3)
but ignoring the negative y sections for both equations.
… To plot the function y = 4 – |2x + 1|
Split up into the positive modulus part and the negative modulus part:
y = 4 – (2x + 1)
y = -2x + 3
and
y = 4 – (-2x -1)
y = 2x + 5
Then sketch these lines, noting which parts of the line segments you will need to ignore (putting in experimental values of x into the original modulus function can help identify the correct parts of the line).
or
… To plot the function y = 4 – |2x + 1|
First sketch the line y = 2x + 1.
Reflect the negative y section in the x-axis. Label this part y = -2x – 1. Now we have the graph y = |2x + 1| which should look like a ‘V’.
Transform this into y = -|2x + 1| by reflecting in the x-axis (it now looks like a ‘Λ’)
Transform again by translating the graph +4 in the y-direction. We now have the graph of y = 4 – |2x + 1|
Make sure that you have labelled the equations of the two straight line parts which make up the modulus function. In this case, we would have the lines: y = -2x + 3 and y = 2x + 5
… Solving an equation, such as:
x = 4 – |2x + 1|
This is an intersection between the line y = x and the line y = 4 – |2x + 1|.
Use the equations which make up the modulus function to find the intercepts. For example:
x = -2x + 3
x = 1
and
x = 2x + 5
x = -5
… You can sometimes simplify a modulus equation by squaring both sides because squaring always gives you a positive result. For example:
|x + 3| = 5a
(x + 3)(x + 3) = 25a² … You may then be able to solve this quadratic to find values for x.
… y = f(|x|). Here we are always making sure the ingredients are positive as they go into the function. We are ‘in the brackets’. So this is a reflection of ONLY the positive x part of the function in the y-axis.
For example, the function y = 2|x| – 3 is made of two parts:
y = 2x – 3 in the positive x part of the graph,
and
y = 2(-x) – 3 which is reflection in the y axis of the positive x part of the graph .
… Sketching y = 3 |x – 5|
The 3 can be put inside the modulus as it has no effect on the sign of the function:
y = |3x – 15|
This function can then be considered as two separate lines:
y = 3x – 15 for positive y
and
y = -3x + 15 also for positive y.
Exponential functions
… Exponential graphs, e.g. y = a(b)^x always intercept the y axis at (0,a) and have an asymptote along the negative x axis.
… You can use your knowledge of function transformations to work out what an exponential function looks like. For example, y = 4 + a(b)^(-x) would be the graph of y = a(b)^x, but first reflected in the y axis and then translated + 4 in the y direction.
… For sketching graphs of functions, you can ask:
“What happens to y if x = 0?”
“What happens to y as x tends to positive infinity?”
“What happens to y as x tends to negative infinity?”
“What happens to y if x approaches an ‘asymptotic’ value [check the denominator for any factors such as (x – 2) etc]?”
Logarithms
… Logs tell us what power we need to raise a number to, to equal another number. E.g. What power do you need to raise 2 to, to give 16? Log(base2) of 16 would tell us the answer.
… So logs are actually powers
… “log” usually refers to log(base10).
… Watch out for logs which look positive, but are in fact negatives in disguise! For example, log(0.2). This is particularly important when solving inequalities because if you multiply or divide both sides by a negative number then you must swap the inequality sign around.
… Using the ‘twirly whirly’ rule (or ‘unravelling’) to understand logs.
… Adding or subtracting same base logs: logA + logB = log(AB) or logA – logB = log(A/B)
… The power rule for logs, e.g. log(C^p) = p(logC) … where ^ means ‘to the power of’
… You can always ‘log’ both sides of an equation, or remove a log from both sides.
For example, if
LogAB = LogCD,
then
AB = CD
Logs are generally used to bring powers ‘down to the normal world’. For example, to solve 3^x = 10 we would log both sides, and use the power rule which would then enable us to solve the equation by making x the subject.
… You may need to ‘convert’ a term in an equation into a log, so that you can combine it with the other logs and simplify. For example if we were using logs of base 10:
log(x) = log(4) + 2
log(x) = log(4) + 2log10 <—- because log(base10)10 = 1
log(x) = log(4) + log(10^2)
log(x) = log(4 × 10^2)
log(x) = log(400)
x = 400
… or with Euler’s number, e:
ln(x) = ln(4) + 2
ln(x) = ln(4) + 2lne
ln(x) = ln(4) + ln(e²)
ln(x) = ln(4e²)
x = 4e²
… This can also apply to powers as you may need to convert an ‘odd one out’ power to get common bases:
4^x + 2^x = 8
(2^2)^x + 2^x = 8
(2^x)^2 + 2^x = 8 … Now substitute using y = 2^x to form a quadratic. Remember to go back to 2^x for your solutions.
… 3^(x+1) can be written as (3^x)(3^1) = (3^x)(3)
… Log(0) or log(negative number) are NOT possible, why not?
… Log (1/a) = log(a^-1) = -log(a)
… y = log(x) (base 10) is the inverse function of y = 10^x. These graphs will be reflections of each other in the line y = x.
Euler’s number (e) and the natural logarithm
… y = ln(x) is the inverse function of y = e^x
… e and ln tend to ‘cancel each other out’. E.g. e^(ln5) = 5
This is why you’ll usually need to use the natural logarithm when solving e^x type equations.
… d/dx(lnx) = 1/x
… d/dx (e^x) = e^x.
This is a special property of Euler’s number which means that in the graph of y = e^x, the gradient at a point P is the same value as the y ordinate at that point.
For other similar exponential graphs such as:
y = 2e^(3x)
The gradient at a point P will be directly proportional to the y-ordinate at that point.
… “Function-derivative” type integrals:
∫ f'(x)/f(x) dx = ln(f(x)) + c.
So if the result of differentiating the denominator of a fraction gives the numerator, then the differential of the fraction is a ‘ln’ function. For example:
∫ 4/(2 + x) dx = 4∫1/(2+ x) = 4ln(2 + x) + c
also
∫ 1/x dx = lnx + c
Note that using this standard result:
∫2/2x dx = ln(2x) + c
∫4/4x dx = ln(4x) + C
How is this possible as the two integrals should be exactly the same!?
The reason is that the constant of integration in each case is DIFFERENT and so accounts for the extra ‘bit’ that comes from the ln:
∫1/x dx = ln(x) + A
∫2/2x dx = ln(2x) + B = ln(x) + ln2 + B
∫4/4x dx = ln(4x) + C = ln(x) + ln4 + C
… When integrating any function, it’s always best to see if you can simplify it first – e.g. multiply it out or divide it up into smaller chunks!
… d/dx(ln2x) is a sneaky one to watch out for! First break it up into smaller chunks: ln2 + lnx … Then differentiate… = 1/x.
Numerical methods
Fixed Point Iteration
… This uses iteration to find the roots (solutions) of an equation f(x) = 0 by rearranging an equation to the form x = g(x).
… The solutions of f(x) = 0 will be the same as for x = g(x), which is where the new function g(x) intersects the line y=x. (Note that these x values are the same in both cases, but the corresponding y values will be different).
… Staircase and cobweb diagrams to illustrate the iteration and their convergence or divergence.
If g(x) is increasing (positive gradient) and above the line y = x, then the staircase will step to the right.
If g(x) is increasing and below y = x, then the staircase will step to the left.
If g(x) is decreasing (negative gradient) then you get a ‘cobweb’ diagram and it will usually converge.
In any of the above cases, if |g'(x)| > 1 , then the iteration often fails to converge at a nearby root (it might converge at another root).
… To show that f(x) = 0 has a root (a solution) between for example x = 1 and x = 2, substitute each x value into the function and find the y values. If there has been a ‘change of sign’, then it is likely that the function has intersected the x axis between those points.
This method may fail if the function has a repeated root, OR if the function is non-continuous, for example y = 1/x. Here, even though f(-1) = negative and f(1) = positive, the curve does NOT intersect the x axis.
Newton-Raphson method
… If x1 is a first approximation to the root of f(x) = 0, then a better approximation, x2, is given by:
x2 = x1 – f(x1) / f'(x1)
As for the other numerical methods of finding roots, the Newton-Raphson method can also fail if the curve is not continuous, or if the gradient f'(x) is very close to zero or very large.
Here’s a nice animation which shows how the Newton-Raphson method works: https://en.m.wikipedia.org/wiki/File:NewtonIteration_Ani.gif
… If you are asked to show that your better approximation (for example x = 0.678) is accurate to 3 decimal places, then try f(0.6775) and f(0.6785) and you should see a sign change.
Exponential Growth and Decay
… Exponential growth for situations involving natural phenomena often involve Euler’s number ‘e’ and can be modelled as A = Be^(kt), where B = initial quantity (when t = 0) and k is the growth constant (or decay constant if negative).
… However, if a particular percentage rate multiplier is mentioned (for example, “a bacteria population increases at a rate of 10% each day”) then it’s best to model the growth or decay using y = a(b)^t.
In this case, b = 1.10, t = number of days and a = starting population. Note that there is no constant multiplying the ‘t’ here.
… Note that you can use EITHER equation to model an exponential situation, but one may be easier to use than the other 🙂
… Exponential graphs, e.g.
y = a(b)^kt
or
y = ae^(kt)
always intercept the y axis at (0,a) and have an asymptote along the negative t axis.
… The variable k will be positive for exponential growth, and negative for exponential decay.
… First start by defining the variables from the question, for example, n = number of people, t = time. Then write down the basic exponential formula, in this case:
n = a(b)^t if you are given a specific % growth or decay rate, or A = Be^(kt) if the question only mentions ‘exponential growth’.
… Secondly, use the data conditions given in the question to find the value of the constants (try to set one of the conditions with t = 0 as this simplifies the process).
… Thirdly, having found the constants, write down the exponential formula that you have found to model the situation. You can then use this to answer any further questions.
… You’ll often need to take logs or natural logs of both sides to make t the subject.
Trigonometry
… Inverse trig functions, for example arcsin(0.5), are asking ‘what angle will give you a right angled triangle which has an Opposite/Hypotenuse ratio of 0.5/1?’.
On a quadrant graph, this means the hypotenuse arm would be length 1 and the opposite would be 0.5, which occurs in the first and second quadrants.
… Once you have identified the ratio from the inverse trig function (use SOH CAH TOA) then draw the triangle on a quadrant graph. You should then be able to work out the angle from the triangle geometry, assuming the triangle is equilateral or 45 degrees. Remember to find the angle which moves anti-clockwise from the positive X axis.
… The ‘CAST’ quadrant mnemonic just tells you where the output (triangle sides ratio) of sin, cos or tan is positive. This can be used to find the other solutions in a given angle range.
Always start from the positive x-axis when finding alternative angles and remember that clockwise is negative, anticlockwise is positive.
However… If you can, I recommend using the graphs of sin, cos to work out the other solutions when solving trig equations. There is usually no need to draw the tan graph because the alternative angle solutions repeat every +/- 180° (or +/- π radians).
… Sketching inverse trig functions, for example y = arcsin(x), and then checking the allowed range so that the function is always “one to one”.
… Note that, for example, y = arccos(x) is the reflection in the line y = x of the function y = cos(x), as the two functions are inverses. The same is true for arcsin and arctan.
Try sketching the graph of y = cos(x) and then flipping the paper over to view the graph from behind so that the x axis becomes the vertical axis and the y axis the horizontal axis!
… To become more familiar with the graphs of arcsin, arccos and arctan functions, try making yourself some Q&A cards so you can test yourself and become more familiar with their shapes, ranges and domains.
… To solve equations such as:
arccos(x) = arcsin(x)
Say that θ = arccos(x), so in this case, θ = arcsin(x) as well.
So…
cosθ = x and sinθ = x
cos²θ = x² sin²θ = x²
Now eliminate θ by adding the two equations:
cos²θ + sin²θ = 2x²
1 = 2x²
x = 1/√2
… It can be useful to turn equations around, e.g.
y = arcsin(x)
becomes
x = sin(y)
Then solve for x in the usual way.
–
… Sketching the reciprocal trig functions: secant, cosecant and cotangent. Whenever the cos, sin or tan curves equal zero, this will be an asymptote on the reciprocal functions – so it’s a good plan to first (lightly) sketch sin, cos or tan.
… The identities for cot, sec and cosec can be derived from sin²A + cos²A= 1:
1 + cot²A = cosec²A
tan²A + 1 = sec²A
It’s a good plan to write these out on your exam paper at the beginning of the exam as they are not in the formula book :).
… Trigonometry proofs. Much of this is recognition through practice. I highly recommend making Q&A flashcards for every ‘breed’ of trig proof question you come across. Test yourself on these cards and they will become your memorised ‘toolkit’ for these types of question. It’s also useful to answer these exam questions last as it is possible to end up down a blind alley!
… Identities have the ≡ symbol. This means that you can only consider either the LHS or the RHS and see if you can simplify it to get to the other side.
There is no need to add an “=” sign as you are dealing with an expression.
… Try to keep the denominator of an algebraic fraction factorised.
… Factorise expressions where possible, e.g. in the numerator.
… When tackling a proof question, start by observing the major differences between the LHS and RHS. You’ll need to pick one of the sides to work on to match the other.
For example, does the LHS have a double angle whereas the RHS only has single angles? This would be a clue that you would need to use the double angle formulae.
… Take care when you are cancelling terms from the numerator and denominator. Ask yourself, “What am I dividing the top and bottom by?” Note that you’ll have to divide every SEPARATE term by this.
… Note that trig questions are often divided into two parts: a proof part, and then a “hence solve” part. It is usually possible to do the second part if you get stuck on the first.
… “Deconstructing” trig functions containing cosec, sec and cot can be useful if the other side of the proof has cos or sin terms in it.
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Expressing 2-term trig functions as a single trig term (in ‘harmonic form’).
… For example, express cosX + 3sinX in the form of Rcos(X – B).
Use the formula book to select the appropriate compound angle formula. I recommend writing the identity beneath the expression you are working on as it’s easier to see the matching parts.
… Once you have formed the two equations by ‘matching up’ the identity with the expression, e.g.
4 = RsinA
3 = RcosA
- Then divide one by the other to get tanA = 4/3 to solve for A
- square both equations and add them together to solve for R.
… The maximum value of, for example, 5cos(x – 30) is 5 because cos varies between 1 and -1. This maximum value would occur when cos(x – 30) = 1 … Solve for values of x in the allowed range in the normal way 🙂
… Watch out for equations where you’re tempted to cancel out trig terms. For example:
cosA = 2sinAcosA
First see if you can factorise the terms by bringing them all on one side. That way, you won’t lose possible solutions.
… Also watch out for square roots and quadratics when solving trig equations as they give you two answers, and therefore two groups of solutions.
… For trig questions that do not have any indication of radians or degrees: if in doubt, use radians.
… We can use the compound angle formulae to work out other trig ratios. For example:
Given that
sinA = -3/5, and 180 < A < 270
cosB = -12/13, and A is obtuse
Find cos(A-B)
From the compound angle formula, we know that:
cos(A-B) = cosAcosB + sinAsinB
We can find cosA by squaring both sides of sinA = -3/5 and using the identity cos²A + sin²A = 1
As A is between 180 and 270, this is where cosA is negative (so reject the positive answer from the square root)
We can find sinB in the same way, however, as A is obtuse, sinB will be positive (use the CAST diagram or graph of sin).
Volumes of revolution:
… The volume of revolution about the x axis can be thought of as summing up of lots of discs of radius y and thickness dx.
… Using V = π∫y²dx for revolutions around the x-axis and V = π∫x²dy for revolving around the y-axis
… Rearrange the original function to make x or y the subject, then substitute into the integral.
… Do draw a sketch of the curve showing intersections with the coordinate axes. This will help you to visualise the revolution.
… For finding ‘compound’ volumes of revolution, it is very useful to use the formula for the volume of a cone or cylinder to reduce the amount of working required. This will be the larger volume from which you then subtract the integrated volume of the revolved curve.
… You can also get a rough idea of the volume to check your answer by approximating the curve to a cone and using V = ⅓πr²h
… Don’t forget to multiply by the pi (π)!
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