Contributed by "Atal" (GARETH WARD) who is a year 13 pupil at Hawthorn High School

 

 

This is a (relatively) easy part of the C4 syllabus, as all of the hardwork has all ready been met in C2, and expanded upon by C3. This is the question where you are told to find all values of θ between 0˚ and 360˚.

 

The two main new types of question you will now see are;

 

cos 2θ  - cos θ  +  1 = 0    

 

Which is known as a double angle, and;

 

4tan(θ+45˚) + 21tan θ = 0

 

Which is known as a compound angle. Note that whether it has a coefficient or is cos/sin etc. doesn't matter, the above are just examples. There is one more question type involving the linear comination of sin and cos, for example 3 cos θ +sin θ = 2, but that question deserves its own page.

 

Of course, we aren't going to get anywhere without first knowing our trigonometric identities and our coumpound/double angle formulae, so here they are (A and B are integers to be defined, and where you see a ±, that simply means it could be either a + or -, and whatever it is the rest will be the same, unless it is a ∓ sign in which case it is the opposite, e.g if there is a + sign then all ± sings will become a plus sign aswell, and a ∓ would become a minus sign)

 

Compound Angles

 

Sin (A ± B) = Sin A.Cos B ± Cos A.Sin B

 

Cos(A ± B) = Cos A.Cos B ∓ Sin A.Sin B

 

Tan (A ± B)=  Tan A ± Tan B 

                    1 + TanA.TanB

 

Double Angles

 

Sin 2A = 2.Sin A.Cos A

 

Cos 2A = Cos²A  - Sin²A

(and as Cos²A + Sin²A  = 1)

Cos 2A = 1 - 2Sin²A = 2Cos²A - 1

 

(Note: it was about here I started to get annoyed at the standard typeset and decided to learn how to use an equation writing program, but I'll be damned if I'm writing all of the above out again)

 

 

 

Luckily for us, the WJEC are kind enough to state all of these in the maths formula booklet (The one which we are always given but never have to use :P). The same can't be said for the C3 trig identities involving Cot, Sec and Cosec however, so make sure you are ok with them.

 

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Examples

 

 

Right, thats enough of the explanations, let's get down to some actual questions.

 

Solve the following equations for values of x from 0˚ to 360˚ inclusive.

 

a) 6 Sin x - 2 Cosec x = 1

 

b) Sin x = 6 Sin 2x

 

The first thing you should check for are any double or compound angles, and for a we have neither. This question is new, but doesn't involve any new mathematics, so don't be daunted by it.

 

First, we need to get rid of that nasty Cosec x. If you remember from C3,

 

 

and so we have:

 

 

We must multiply everything by sin x to make this nicer, giving us:

 

 

Can you see where this is going? Next we take everything to the one side giving us a bog standard quadratic equation,

 

 

Of course there are many ways of doing this, but I like to do it the long way to make sure I make no mistake. I will refrain from explaining in detail what happens now as it is C2 work. 

 

 

 

 

We now have our two values of x to solve, using the CAST diagram to find all values of x.

 

  and

 

Giving us:

 

Easy marks!

 

Now, the part of the question worthy of being called "C4" (although in this case, easier than most things on C3!).

 

b) Sin x = 6 Sin 2x

 

The first thing that you should notice about this question is that there is a double angle involved. The first step is to apply one of our formulae to deal with it. The question is, which formula? Well, we have a double angle involving sin, so we use the Sin2A formula. That is;

 

 

So, subsituting this in, we get;

 

 

Expanding the brackets;

 

 

Collect them all on one side;

 

 

You may be thinking "Oh dear, how on earth am I going to get x on its own from here?" Well, calm down, because actaully, its just a simple factorisation that after years of quadratic equations, alot of us (including myself) fail to spot!

 

Take a factor of sin x out, giving us;

 

 

And now we have our 2 values of x;

 

along with

 

and so, using CAST,

 

Hazaa, we have just completed a full C4 past paper question! but don't break out the beer yet, for a more sinister question awaits us...

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Examples Continued

 

Here we are going to meet some compound angle questions, which are a bit harder, but they are doable none the less.

 

Write Tan(x + 45°) in terms of Tan x, and then find the values of X between 0° and 360° satisfying 4Tan(x + 45°) + 21Tan x = 0.

 

Compound angles are easy to spot, and for the most part, the only thing that makes them more difficult than double angles is the increased difficulty of the algebra.

 

The first part of our question just involves us substituting in our compound angle formula at the top of this page (It is easier for you to scroll up than for me to type it out again in the equation editior!)

 

This gives us:

 

 

It just so happens that tan 45 = 1, so this can be simplified to:

 

 

It is useful to remember the angles of sin, cos and tan that come out to be nice numbers, so you may like to make a note of them elsewhere.

 

Now, for the second part of this question, 4Tan(x + 45°) + 21Tan x = 0, we need to sub the compound angle formula we just simplified back into this equation.

 

 

You'll also note that I took -21tan x to the other side, this is so that we can multiply the whole equation by (1-tan x), which gives us:

 

 

Expanding:

 

 

As you can see, this has now become a simple C2 equation, the C4 part is over. Collecting and factorising:

 

 

Our 2 possible x values are   and

 

Using CAST: 

 

And there we have it, compound angles ladies and gentlemen. However, the most horrid C4 trigonometry is still to come, visit my page on the dreaded linear combination of sin and cos if you wish to see for yourself!

 

 

 

Written by Gareth Ward, for anybody, but mainly to make sure I understand all this balony myself!

VIDEO ON DOUBLE ANGLE FORMULAE FROM MATH TV ON YOUTUBE