# Quadratic Equation Edexcel F2 June 2017

#### fasih178

##### New member
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?

#### MarkFL

##### Super Moderator
Staff member
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

##### New member
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

##### Super Moderator
Staff member
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

##### New member
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

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

#### fasih178

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

Thank you.