De Moivre's Theorem and Trig Identities

 Hi Guys and Gals Hope you have a great New Years eve party if you are having one. Hopefully 2021 will see the end of Covid and life can get back to normal. IMHO the restrictions have been unnecessarily punitive and the shut down of the pubs, theatres and concert halls has been totally unnecessary anyway I'm not here to gripe about the restrictions laid down by Mrs Oliver Cromwell (aka Nicola Sturgeon) up in Scotland during the lockdown. I have been amusing myself to see if I can do some Cambridge Maths past papers set for their Natural science students and available here 

https://www.maths.cam.ac.uk/undergradnst/pastpapers/2019

In doing the 2019 1st paper I came across this question which at first sight looks nigh impossible the question was 

Use De Moivre's theorem to write out the expression 

$$ 8 cos 6\theta + 15sin 4\theta sin 2\theta $$ in terms of $ cos \theta$ and $ sin \theta $ 

However I discovered for myself a simple method to write out the expansion of $cos (n\theta)$ or $sin (n\theta) $ Almost instantly from Pascal's triangle 

First let us recall De Moivre's theorem it states that 

$$exp(in\theta) = [exp(i\theta)]^n $$ using 

$$ exp(in\theta) = cos(n\theta) + i sin(n\theta) $$ 

it is seen that 

$$cos(n\theta) = Re [cos(\theta)  + i sin(\theta)]^n $$ and 

$$sin(n\theta) = Im [cos(\theta) + i sin(\theta)]^n $$ 

Ok that looks really messy how do I remember the coefficients in a binomial expansion and how do I pick out the Real and Imaginary parts of each coefficient and all in about 5-10 mins which is all I have time to answer the question

Well a short cut is as follows Recall that the binomial coefficients in a binomial expansion are given by the rows of Pascal's triangle . Namely 

                                n = 0                                 1

                                n  = 1                              1     1

                                n  = 2                         1     2    1

                                n  =  3                     1     3     3    1

                                n  =  4                   1     4     6     4    1

                                n =  5                1     5    10   10   5     1

                               n   = 6            1    6    15     20    15  6      1

where each number in a given row is given by adding the two numbers above it and the coefficient corresponds to the following expansion 

$$ (a + b)^n = C_1  a^n + C_2  a^{n-1} b + C_3 a^{n-2} b^2 + ...........C_{n-1}.a b^{n-1} + C_n b^n$$

The coefficients are just the corresponding element in the nth row of Pascal's triangle. So for example 

$$(a + b)^5 = a^5 + 5 a^4 b + 10 a^3 b^2 + 10 a^2 b^3 + 5 a^1 b^4 + b^5 $$

Now to apply this to De Moivre's theorem we would have 

$$[(cos(\theta) + i sin(\theta)]^n = C_1 cos^n (\theta) + C_2 cos ^{n-1}(\theta) (i sin(\theta)) + $$

                                                   $$   C_3 cos^{n-2}(\theta) (i sin(\theta))^2 + ..... $$

So the odd coefficients in the expansion will correspond to the real part and the even coefficients will correspond to the imaginary part of the expansion, Furthermore as $i^2 = -1$ and $i^4 = +1$ the coefficients will alternate in sign 

So this gives us a nice quick method of writing out $cos(n\theta)$ and $sin(n\theta)$ in terms of $cos(\theta)$ and $sin(\theta)$

For a given n work out the corresponding row in Pascal's triangle then the expansions of $cos(n \theta)$ and $sin(n \theta)$ are as follows 

$$ cos(n\theta) = C_1 cos^n(\theta) - C_3 cos^{n-2}(\theta) sin^2(\theta) + $$

$$ C_5 cos^{n-4}(\theta) sin^4(\theta)+ ...$$

$$ sin(n\theta) = C_2 cos^{n-1}(\theta) sin(\theta) - C_4 cos^{n-3}sin^3(\theta)+ .... $$ 

Hence given a horrendous expression involving $cos(n\theta)$ and $sin(n\theta)$ the expansions of these functions in individual powers of $cos (\theta)$ or $sin(\theta)$ can be written down by sight 

So for our problem we have 

$$sin(2\theta) = 2 sin(\theta) cos(\theta) $$ as the second row of pascals triangle is 1 2 1 and the even coefficient is 2 

$$sin(4\theta) = 4 cos^3(\theta) sin(\theta) - 4 cos(\theta) sin^3(\theta) $$ as the fourth row of Pascal's triangle is 1 4 6 4 1 and the even terms are 4 and 4 

Finally 

$$cos(6\theta) = cos^6(\theta) - 15 cos^4(\theta) sin^2(\theta) + 15 cos^2(\theta) sin^4(\theta) - sin^6(\theta) $$ 

as the 6th row of Pascal's triangle is 1 6 15 20 15  6 1 and the odd numbered coefficients are 1 15 15 and 1 

So 

$$8 cos(6\theta) =8[ cos^6(\theta) - 15 cos^4(\theta) sin^2(\theta) + 15 cos^2(\theta) sin^4(\theta) -$$ $$sin^6(\theta) ]$$ 

and 

$$15sin4\theta sin 2\theta=15 [4 cos^3(\theta)sin(\theta)-4cos(\theta)sin^3(\theta) ] $$

$$ \times 2 sin(\theta)cos(\theta)$$

This simplifies to 

$$ 15 \times  8 [cos^4(\theta)sin^2(\theta) - cos^2(\theta) sin^4(\theta)] $$ 

and it is seen that the mixed terms in $cos(\theta)$ and $sin(\theta)$ cancel leaving 

$$ 8 cos 6\theta + 15sin 4\theta sin 2\theta = 8(cos^6(\theta) -sin^6(\theta)) $$

So next time you are faced with a seemingly horrendous expression involving $sin(n\theta)$ or $cos(n\theta)$ just remember Pascal's triangle and De Moivre's theorem.

Hope every one reading this has a great new year I will write another post over the weekend summarising my plans for 2021 and hopefully achieving them :)