Quadratic Equation Edexcel F2 June 2017

fasih178

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Q8 of FP2 IAL June 2017 paper. I was able to do parts (a) and (b), but am lost with (c). I can't understand how (c) is linked with (b).
The solution uses tan5x=0 to find the quadratic. Why?202001180350551000.jpg
 

MarkFL

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I would begin as you did with:

\(\displaystyle (\cos(\theta)+i\sin(\theta))^5=\cos(5\theta)+i\sin(5\theta)\)

\(\displaystyle (\cos(\theta)+i\sin(\theta))^5=\cos^5(\theta)+5\cos^4(\theta)i\sin(\theta)+10\cos^3i^2\sin^2(\theta)+10\cos^2(\theta)i^3\sin^3(\theta)+5\cos(\theta)i^4\sin^4(\theta)+i^5\sin^5(\theta)\)

Hence:

\(\displaystyle \cos(5\theta)+i\sin(5\theta)=(\cos^5(\theta)-10\cos^3\sin^2(\theta)+5\cos(\theta)\sin^4(\theta))+i(5\cos^4(\theta)\sin(\theta)-10\cos^2(\theta)\sin^3(\theta)+\sin^5(\theta))\)

Equating real and imaginary parts, we find:

\(\displaystyle \cos(5\theta)=\cos^5(\theta)-10\cos^3\sin^2(\theta)+5\cos(\theta)\sin^4(\theta)\)

\(\displaystyle \sin(5\theta)=5\cos^4(\theta)\sin(\theta)-10\cos^2(\theta)\sin^3(\theta)+\sin^5(\theta)\)

And so we may write:

\(\displaystyle \tan(5\theta)=\frac{\sin(5\theta)}{\cos(5\theta)}\cdot\frac{\sec^5(\theta)}{\sec^5(\theta)}=\frac{t^5-10t^3+5t}{5t^4-10t^2+1}\)

Where \(\displaystyle t=\tan(\theta)\) and \(\cos(\theta)\ne0\)

Now, we know:

\(\displaystyle \tan(\pi)=\tan(2\pi)=0\)

And so we must have (using the numerator of the above expression for \(\tan(5\theta)\)):

1.) \(\displaystyle \tan^4\left(\frac{\pi}{5}\right)-10\tan^2\left(\frac{\pi}{5}\right)+5=0\)

2.) \(\displaystyle \tan^4\left(\frac{2\pi}{5}\right)-10\tan^2\left(\frac{2\pi}{5}\right)+5=0\)

Can you now see the quadratic having the required roots?
 

fasih178

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Thank you! I understood one crucial thing: we need to substitute in pie into LHS tan(5theta) so that we get theta=pie/5, which is one of our roots. And, since tan(pie) = 0, we can set our RHS expression = 0 too. By the same intuition, tan(2pie) is zero. Thus, overall, I understood the idea behind setting tan(5theta) = 0.

But what I don't get is why did you even think about tan(pie) = 0 in the first place -- then going on to using this to solve the problem. That is, you could have instead chosen, say, tan(pie/4) = 1, but you didn't. Why not? What gave you the idea to use tan(pie) = 0?
 

MarkFL

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Thank you! I understood one crucial thing: we need to substitute in pie into LHS tan(5theta) so that we get theta=pie/5, which is one of our roots. And, since tan(pie) = 0, we can set our RHS expression = 0 too. By the same intuition, tan(2pie) is zero. Thus, overall, I understood the idea behind setting tan(5theta) = 0.

But what I don't get is why did you even think about tan(pie) = 0 in the first place -- then going on to using this to solve the problem. That is, you could have instead chosen, say, tan(pie/4) = 1, but you didn't. Why not? What gave you the idea to use tan(pie) = 0?
We previously derived the formula:

\(\displaystyle \tan(5\theta)=\frac{t^5-10t^3+5t}{5t^4-10t^2+1}\)

This gives us a way to relate \(\tan(5\theta)\) to \(\tan(\theta)\). Now consider that:

\(\displaystyle \pi=5\cdot\frac{\pi}{5}\)

\(\displaystyle 2\pi=5\cdot\frac{2\pi}{5}\)

And so if \(\theta=n\dfrac{\pi}{5}\) then \(5\theta=n\pi\).

The reason I chose values for which \(\tan(5\theta)=0\) is because we are searching for roots of a quadratic. Using the formula relating \(\tan(5\theta)\) to \(\tan(\theta)\), we see that in order for \(\tan(5\theta)=0\) we must have:

\(\displaystyle \tan^5(\theta)-10\tan^3(\theta)+5\tan(\theta)=0\)

\(\displaystyle \tan(\theta)\left(\tan^4(\theta)-10\tan^2(\theta)+5\right)=0\)

Now, when we have:

\(\displaystyle 0<\theta<\pi\)

We know \(\tan(\theta)\ne0\), and since both angles under consideration fall within that range, we may divide by \(\tan(\theta)\) to obtain the quadratic in \(\tan^2(\theta)\):

\(\displaystyle \tan^4(\theta)-10\tan^2(\theta)+5=0\)

Since \(\tan(k\pi)=0\) where \(k\in\mathbb{Z}\), we know the above quadratic is true for all:

\(\displaystyle \theta=k\frac{\pi}{5}\)

Because of the fact that on the unit circle, an angle of \(\theta\) is equivalent to \(\theta+2k\pi\), we know then that for:

\(\displaystyle \theta\in\left\{k\frac{\pi}{5},(k+1)\frac{\pi}{5}\right\}\)

\(\displaystyle \tan^2(\theta)\) must be distinct roots of the quadratic:

\(\displaystyle x^2-10x+5=0\)

Now given that the product of the two roots of a quadratic is the quotient of the constant term divided by the coefficient of the squared term, we may conclude that:

\(\displaystyle \tan^2\left(k\frac{\pi}{5}\right)\tan^2\left((k+1)\frac{\pi}{5}\right)=5\)

Since for \(k=1\) we have \(\theta\) in the first quadrant, we can therefore state:

\(\displaystyle \tan\left(\frac{\pi}{5}\right)\tan\left(\frac{2\pi}{5}\right)=\sqrt{5}\)

Does this make sense?
 

fasih178

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Oh yes, it does! It makes perfect sense now. However, are there any resources or videos I can watch to hone in on such questions, particularly those that pertain to making these connections using trigonometric functions? I know this was rather intuitive for you, but I have difficulty grasping such connections.
 

MarkFL

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It really just takes practice. The more problems you work, the easier it becomes to begin making these connections.
 

fasih178

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Indeed. The old adage ''Practice makes perfect'' applies best to Math.

Thank you.
 
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