3

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where

$cos5\theta=16cos^5\theta-20cos^3\theta+5cos\theta$ This is the question I don't understand, and I don't undersatnd the markscheme either.

This question is the fifth part of a few, but here are the previous questions.enter image description here

The mark scheme for e is:

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I've done all the parts before this correctly, but I don't get this final part of the question. I'd really appreciate it if someone could explain all of it to me.

Jim
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1 Answers1

2

Let $\theta=\frac{\pi}{10}$. Then $\cos 5\theta=0$. It follows that $$16\cos^5\theta-20\cos^3\theta+5\cos\theta=0.$$ Since $\cos\theta\ne 0$, we have $$16\cos^4\theta-20\cos^2\theta+5=0.$$ This is a quadratic equation in $\cos^2\theta$. Solve, using the Quadratic Formula. There is a choice of root for $\cos^2\theta$, but it should not be difficult to select the right one.

André Nicolas
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  • Sorry but I don't understand what they've done in the mark scheme - how have they cancelled everything down? This is from a non-calculator paper so how did they get 4.8? I understand the max value of cosine -> closest to zero bit, but not how the signs are both negative for $cos\frac{7\pi}{10}$... Is it because $\frac{7\pi}{10}$ is in the 2nd quadrant so it's negative? – Jim May 12 '14 at 12:42
  • After solving the quadratic for $cos^2\theta$, we need to write down $\cos \theta$, and yes, in the case of $7\pi/10$ we choose negative because it is in second quadrant. For solving the quadratic, I think the arithmetic can be done in one's head, by the Quadratic Formula we want $\frac{20+\pm\sqrt{200^2-(4)(16)(5)}}{32}$. This is $(20\pm\sqrt{80})/32$, which they simplify to $(5\pm\sqrt{5})/8$. There is no $4.8$, they are observing that $32=(4)(8)$. – André Nicolas May 12 '14 at 14:55