By using de Moivre's theorem to express sin 5θ and cos 5θ in terms of sin θ and cos θ, show that $$ \tan 5\theta = \frac{5\tan \theta - 10\tan^3 \theta + \tan^5 \theta}{1 - 10\tan^2 \theta + 5\tan^4 \theta} $$ where t = tan θ. Show that the roots of the equation x^4 - 10x^2 + 5 = 0 are tan{nπ/5} for n = 1, 2, 3, 4. By considering the product of the roots of this equation, find the exact value of tan{π/5} tan{2π/5}.

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By using de Moivre's theorem to express sin 5θ and cos 5θ in terms of sin θ and cos θ, show that $$ \tan 5\theta = \frac{5\tan \theta - 10\tan^3 \theta + \tan^5 \theta}{1 - 10\tan^2 \theta + 5\tan^4 \theta} $$ where t = tan θ - CIE - A-Level Further Maths - Question 10 - 2010 - Paper 1

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$By-using-de-Moivre's-theorem-to-express-sin-5θ-and-cos-5θ-in-terms-of-sin-θ-and-cos-θ,-show-that--$$-\tan-5\theta-=-\frac{5\tan-\theta---10\tan^3-\theta-+-\tan^5-\theta}{1---10\tan^2-\theta-+-5\tan^4-\theta}-$$--where-t-=-tan-θ-CIE-A-Level Further Maths-Question 10-2010-Paper 1.png$

By using de Moivre's theorem to express sin 5θ and cos 5θ in terms of sin θ and cos θ, show that $$ \tan 5\theta = \frac{5\tan \theta - 10\tan^3 \theta + \tan^5 \th... show full transcript

Worked Solution & Example Answer:By using de Moivre's theorem to express sin 5θ and cos 5θ in terms of sin θ and cos θ, show that $$ \tan 5\theta = \frac{5\tan \theta - 10\tan^3 \theta + \tan^5 \theta}{1 - 10\tan^2 \theta + 5\tan^4 \theta} $$ where t = tan θ - CIE - A-Level Further Maths - Question 10 - 2010 - Paper 1

Step 1

By using de Moivre's theorem to express sin 5θ and cos 5θ

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Answer

We can express sin and cos in terms of exponential functions using de Moivre's theorem:

(c + i s)^n = r^n (\cos n\theta + i \sin n\theta)

for ( n = 5 ). This gives:

\cos 5\theta = c^5 - 10c^3s^2 + 5cs^4.

And similarly,

\sin 5\theta = 5c^4s - 10c^2s^3 + s^5.

Thus,

\tan 5\theta = \frac{\sin 5\theta}{\cos 5\theta} = \frac{5c^4s - 10c^2s^3 + s^5}{c^5 - 10c^3s^2 + 5cs^4}.

Substituting ( c = \cos\theta ) and ( s = \sin\theta ), we can simplify further to arrive at:

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

where ( t = \tan\theta ).

Step 2

Show that the roots of the equation x^4 - 10x^2 + 5 = 0 are tan{nπ/5} for n = 1, 2, 3, 4

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Answer

Let ( y = x^2 ). The equation transforms to:

y^2 - 10y + 5 = 0.

Using the quadratic formula,

y = \frac{10 \pm \sqrt{(10)^2 - 4 \cdot 1 \cdot 5}}{2 \, 1} = \frac{10 \pm \sqrt{100 - 20}}{2} = \frac{10 \pm \sqrt{80}}{2} = 5 \pm 2\sqrt{5}.

This gives us:

x^2 = 5 + 2\sqrt{5} \quad \text{and} \quad x^2 = 5 - 2\sqrt{5}.

Thus the solutions for ( x = \tan{n\pi/5} ) for ( n = 1, 2, 3, 4 ).

Step 3

By considering the product of the roots of this equation

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Answer

Using Vieta's formulas, the product of the roots of the equation ( x^4 - 10x^2 + 5 = 0 ) is given as:

x_1 x_2 x_3 x_4 = (-1)^4 \cdot \frac{5}{1} = 5.

Knowing the values of the roots as ( \tan{n\pi/5} ), we can express:

an{\pi/5} \tan{2\pi/5} \tan{3\pi/5} \tan{4\pi/5} = 5.

Then, utilizing the property that ( \tan{\pi - x} = -\tan{x} ), we can find:

an{3\pi/5} = -\tan{2\pi/5} \quad \text{and} \quad \tan{4\pi/5} = -\tan{\pi/5}.

Therefore, the exact value of ( \tan{\pi/5} \tan{2\pi/5} = \sqrt{5}. )

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By using de Moivre's theorem to express sin 5θ and cos 5θ in terms of sin θ and cos θ, show that $$ \tan 5\theta = \frac{5\tan \theta - 10\tan^3 \theta + \tan^5 \theta}{1 - 10\tan^2 \theta + 5\tan^4 \theta} $$ where t = tan θ - CIE - A-Level Further Maths - Question 10 - 2010 - Paper 1

By using de Moivre's theorem to express sin 5θ and cos 5θ

Show that the roots of the equation x^4 - 10x^2 + 5 = 0 are tan{nπ/5} for n = 1, 2, 3, 4

By considering the product of the roots of this equation

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