Thread: Cos of multiple angles to the powers of cos - I need help understanding these steps!

1. Cos of multiple angles to the powers of cos - I need help understanding these steps!

I was doing these questions, and did them all up to question (e), but then I got stuck. I looked at the mark scheme but still don't understand it.

Please could someone explain the steps to me?
Here's the mark scheme:

So I understand everything up to the quadratic equations - it's the manipulating from there that I don't understand.
How does 20 become 4.5, when 32 becomes 4.8? What did they multiply by?
I don't understand the $\displaystyle cos\frac{\pi}{10}$ line onwards. 'since max value of cosine => angle closest to zero' - is this because pi by 10 is the closest value to 0 that is a solution?
Why not simplify the $\displaystyle \sqrt{400-4(16)(5)$?
How do you know that's the value of $\displaystyle cos\frac{7\pi}{10}$?

I realise that this is a long post and that I have a lot of questions, but help with this question will be greatly appreciated.

2. Re: Cos of multiple angles to the powers of cos - I need help understanding these ste

I imagine part of your confusion stems from the fact that what they did is wrong. The correct simplification is

$\cos\left(\dfrac \pi {10}\right)=\sqrt{\dfrac{20+\sqrt{400 - 4(16)(5)}}{32}}=$

$\sqrt{\dfrac{20+4\sqrt{25- 4(5)}}{32}}=$

$\sqrt{\dfrac{5+\sqrt{25- 4(5)}}{8}}=\sqrt{\dfrac{5+\sqrt{5}}{8}}$

They did come up with the right answer though. I wouldn't get to bothered by how they did it as long as you can work the algebra yourself and get the same answer.

3. Re: Cos of multiple angles to the powers of cos - I need help understanding these ste

Originally Posted by romsek
I imagine part of your confusion stems from the fact that what they did is wrong. The correct simplification is

$\cos\left(\dfrac \pi {10}\right)=\sqrt{\dfrac{20+\sqrt{400 - 4(16)(5)}}{32}}=$

$\sqrt{\dfrac{20+4\sqrt{25- 4(5)}}{32}}=$

$\sqrt{\dfrac{5+\sqrt{25- 4(5)}}{8}}=\sqrt{\dfrac{5+\sqrt{5}}{8}}$

They did come up with the right answer though. I wouldn't get to bothered by how they did it as long as you can work the algebra yourself and get the same answer.
Ah, that makes sense now, thanks.
Where do the signs of the quadratic come in for the pi/10 and 7pi/10 values?

4. Re: Cos of multiple angles to the powers of cos - I need help understanding these ste

They were looking at solutions of the equation $\cos(5\theta) = 0$. You apparently showed in part (c) that $\cos(5\theta) = 16\cos^5 \theta-20\cos^3 \theta+5\cos \theta$. So, if the LHS equals zero, then the RHS equals zero. So, setting the RHS to zero, you can factor out $\cos \theta$. Then:

$0 = 16\cos^5 \theta-20\cos^3 \theta+5\cos \theta = \cos\theta(16(\cos^2 \theta)^2 - 20(\cos^2 \theta) + 5)$

If you let $x = \cos^2 \theta$, then:

$0 = x^{1/2}(16x^2-20x+5)$

So, either $x^{1/2} = 0$ or $16x^2-20x+5 = 0$.

The latter is a quadratic equation. Plugging it into the quadratic formula, you have $x = \dfrac{20 \pm \sqrt{400-4(16)(5)}}{32}$.

Substituting back for $x = \cos^2 \theta$, you have: $\cos^2 \theta = \dfrac{5 \pm \sqrt{5}}{8}$. Hence, $\cos \theta \in \left\{-\sqrt{\dfrac{5+\sqrt{5}}{8}},\sqrt{\dfrac{5+\sqrt{ 5}}{8}},-\sqrt{\dfrac{5-\sqrt{5}}{8}},\sqrt{\dfrac{5-\sqrt{5}}{8}}\right\}$. So, you just need to figure out which it is. Since $\cos \theta$ is decreasing on the interval $[0,\pi]$, it must be that $\cos 0 = 1 > \cos \dfrac{\pi}{10} > \cos \dfrac{\pi}{6} = \dfrac{\sqrt{3}}{2} \approx 0.866$. So, out of the possible values, you can discount three of them, making the only possible value $\sqrt{\dfrac{5+\sqrt{5}}{8}}$. For $\cos \dfrac{7\pi}{10}$, you know $\cos \dfrac{\pi}{2} = 0 > \cos \dfrac{7\pi}{10} > \cos \dfrac{5\pi}{6} \approx -0.866$, so again, there is only one possible value among the four that works.