De Moivre's Theorem: Find Expansion of cos 5θ

[jammed]

Homework Statement

Using De Moivre's Theorem, find the expansion of cos 5θ in terms of cos θ.
Hence find the exact value of
cos(pi/10) x cos(3pi/10)

Homework Equations

Well i used the equation (cosθ + isinθ)^5 and then equated the real parts to get cos5θ in terms of cosθ.

The Attempt at a Solution

I expanded the equation above using binomial expansion and then equated the real parts to get cos5θ in terms of cosθ

c= cosθ, s = sinθ
(cosθ+isinθ)^5 = c^5 + 5c^4is-10c^3s^2-10c^2is^3+5cs^4- 1is^5
After equating the real parts I got
cos5θ = 16c^5 - 20c^3 + 5c (1)
For the second part

I used (1) to make cosθ the subject and then substituted θ = pi/10 and θ = 3pi/10 respectively but I am unable to to get the exact value and most importantly without calculator.

 

[Mentallic]

Have you been taught how to do this in class? Because there is probably a more elegant way than I'm about to show you.

Given that $$cos(5\theta) = 16cos^5(\theta) - 20cos^3(\theta) + 5cos(\theta)$$

and we need to find the value of $$cos(\frac{\pi}{10})cos(\frac{3\pi}{10})$$

Substituting $$\theta=\frac{\pi}{10}$$ into the equation gives us

$$cos(\frac{\pi}{2})=16cos^5(\frac{\pi}{10})-20cos^3(\frac{\pi}{10})+5cos(\frac{\pi}{10})$$

Now the left side, $$cos(\frac{\pi}{2})=0$$

So now we need to solve for $$16cos^5(\frac{\pi}{10})-20cos^3(\frac{\pi}{10})+5cos(\frac{\pi}{10})=0$$

Since $$cos(\frac{\pi}{10})\neq 0$$ then we can divide through by that, leaving us a quadratic in $$cos^2(\frac{\pi}{10})$$

After you do this, solve for $$cos(\frac{\pi}{10})$$ and notice that you'll have a plus or minus from the quadratic, and obviously it can't be equal to both, it must be just one of them. Well, if you don't have a calculator handy then do the same process for $$cos(\frac{3\pi}{10})$$, you'll find you get the same answer. Since $$cos(\frac{\pi}{10})\neq cos(\frac{3\pi}{10})$$ then obviously in the plus or minus, one of them is the answer to the first and the other is the answer to the second. You can quickly give a short explanation as to why the + would be for the $\pi/10$ since $$cos(\frac{\pi}{10})>cos(\frac{3\pi}{10})$$

Now just multiply both values together to find $$cos(\frac{\pi}{10})cos(\frac{3\pi}{10})$$ as needed.

 

[jammed]

No I wasnt taught in school. It just came in the exam. After I posted this thread I tried to substitute
x = cos$$\theta$$ in the equation

Then used the product of the roots 16x^4 - 20x^2 + 5 to find the result. What do you say about this way of solving

 

[Mentallic]

Oh yes of course that's much better

You might want to let $$x=cos^2\theta$$ such that you end up with a quadratic so then you only have two roots. With the quartic you have, you would need to argue that since there are 4 roots in the quartic equation, two of them are equal roots.

 

[jammed]

Thanx for your idea!