Multiple angles

Geometry Level 4

Let θ = π 11 \theta = \dfrac \pi{11} . Find

sec ⁡ ( θ ) sec ⁡ ( 2 θ ) sec ⁡ ( 3 θ ) sec ⁡ ( 4 θ ) sec ⁡ ( 5 θ ) \large \sec (\theta) \sec (2\theta) \sec (3\theta) \sec (4\theta) \sec (5\theta)


The answer is 32.

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Naren Bhandari by Naren Bhandari , 21, India

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4 solutions

Abhijit Dixit
Feb 26, 2017

let

k = cos ⁡ θ × cos ⁡ 2 θ × cos ⁡ 3 θ × cos ⁡ 4 θ × cos ⁡ 5 θ k=\cos\theta\times\cos2\theta\times\cos3\theta\times\cos4\theta\times\cos5\theta

multiplying both side by

2 5 × sin ⁡ θ × sin ⁡ 2 θ × sin ⁡ 3 θ × sin ⁡ 4 θ × sin ⁡ 5 θ 2^5\times\sin\theta\times\sin2\theta\times\sin3\theta\times\sin4\theta\times\sin5\theta

We have

2 5 × k × sin ⁡ θ × sin ⁡ 2 θ × sin ⁡ 3 θ × sin ⁡ 4 θ × sin ⁡ 5 θ = sin ⁡ 2 θ × sin ⁡ 4 θ × sin ⁡ 6 θ × sin ⁡ 8 θ × sin ⁡ 10 θ 2^5\times k\times\sin\theta\times\sin2\theta\times\sin3\theta\times\sin4\theta\times\sin5\theta = \sin2\theta\times\sin4\theta\times\sin6\theta\times\sin8\theta\times\sin10\theta

RHS can be written as

= sin ⁡ 2 θ × sin ⁡ 4 θ × sin ⁡ ( 11 θ − 5 θ ) × sin ⁡ ( 11 θ − 3 θ ) × sin ⁡ ( 11 θ − θ ) = \sin2\theta\times\sin4\theta\times\sin(11\theta-5\theta)\times\sin(11\theta-3\theta)\times\sin(11\theta -\theta)

putting θ = π 11 \theta=\frac{\pi}{11}

= sin ⁡ θ × sin ⁡ 2 θ × sin ⁡ 3 θ × sin ⁡ 4 θ × sin ⁡ 5 θ = \sin\theta\times\sin2\theta\times\sin3\theta\times\sin4\theta\times\sin5\theta

thus

2 5 × k × sin ⁡ θ × sin ⁡ 2 θ × sin ⁡ 3 θ × sin ⁡ 4 θ × sin ⁡ 5 θ = sin ⁡ θ × sin ⁡ 2 θ × sin ⁡ 3 θ × sin ⁡ 4 θ × sin ⁡ 5 θ 2^5\times k\times\sin\theta\times\sin2\theta\times\sin3\theta\times\sin4\theta\times\sin5\theta = \sin\theta\times\sin2\theta\times\sin3\theta\times\sin4\theta\times\sin5\theta

k = 1 32 k=\frac{1}{32}

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Chew-Seong Cheong
Feb 27, 2017

P = sec ⁡ π 11 sec ⁡ 2 π 11 sec ⁡ 3 π 11 sec ⁡ 4 π 11 sec ⁡ 5 π 11 = 1 cos ⁡ π 11 cos ⁡ 2 π 11 cos ⁡ 3 π 11 cos ⁡ 4 π 11 cos ⁡ 5 π 11 Note that cos ⁡ ( π − x ) = − cos ⁡ x = 1 cos ⁡ π 11 cos ⁡ 2 π 11 cos ⁡ 4 π 11 ⋅ − 1 cos ⁡ 3 π 11 cos ⁡ 6 π 11 = sin ⁡ π 11 sin ⁡ π 11 cos ⁡ π 11 cos ⁡ 2 π 11 cos ⁡ 4 π 11 ⋅ − sin ⁡ 3 π 11 sin ⁡ 3 π 11 cos ⁡ 3 π 11 cos ⁡ 6 π 11 = 2 sin ⁡ π 11 sin ⁡ 2 π 11 cos ⁡ 2 π 11 cos ⁡ 4 π 11 ⋅ − 2 sin ⁡ 3 π 11 sin ⁡ 6 π 11 cos ⁡ 6 π 11 = 4 sin ⁡ π 11 sin ⁡ 4 π 11 cos ⁡ 4 π 11 ⋅ − 4 sin ⁡ 3 π 11 sin ⁡ 12 π 11 = 8 sin ⁡ π 11 sin ⁡ 8 π 11 ⋅ − 4 sin ⁡ 3 π 11 sin ⁡ 12 π 11 Note that sin ⁡ ( π − x ) = sin ⁡ x = 8 sin ⁡ π 11 sin ⁡ 3 π 11 ⋅ − 4 sin ⁡ 3 π 11 − sin ⁡ π 11 Note that sin ⁡ ( π + x ) = − sin ⁡ x = 32 \begin{aligned} P & = \sec \frac \pi{11} \sec \frac {2 \pi}{11} \sec \frac {3 \pi}{11} \sec \frac {4 \pi}{11} \sec \frac {5 \pi}{11} \\ & = \frac 1{\cos \frac \pi{11} \cos \frac {2 \pi}{11} \cos \frac {3 \pi}{11} \cos \frac {4 \pi}{11} \color{#3D99F6}\cos \frac {5 \pi}{11}} & \small \color{#3D99F6} \text{Note that }\cos (\pi - x) = - \cos x \\ & = \frac 1{\cos \frac \pi{11} \cos \frac {2 \pi}{11}\cos \frac {4 \pi}{11}} \cdot \frac{-1} {\cos \frac {3\pi}{11} \color{#3D99F6}\cos \frac {6 \pi}{11}} \\ & = \frac {\color{#3D99F6}\sin \frac \pi{11}}{{\color{#3D99F6}\sin \frac \pi{11}}\cos \frac \pi{11} \cos \frac {2 \pi}{11} \cos \frac {4 \pi}{11}} \cdot \frac {- \color{#D61F06}\sin \frac {3 \pi}{11}}{{\color{#D61F06}\sin \frac {3\pi}{11}}\cos \frac {3\pi}{11} \cos \frac {6 \pi}{11}} \\ & = \frac {2\sin \frac \pi{11}}{\sin \frac {2\pi}{11} \cos \frac {2 \pi}{11} \cos \frac {4 \pi}{11}} \cdot \frac {- 2\sin \frac {3 \pi}{11}}{\sin \frac {6\pi}{11} \cos \frac {6 \pi}{11}} \\ & = \frac {4\sin \frac \pi{11}}{\sin \frac {4\pi}{11} \cos \frac {4 \pi}{11}} \cdot \frac {- 4\sin \frac {3 \pi}{11}}{\sin \frac {12 \pi}{11}} \\ & = \frac {8\sin \frac \pi{11}}{\color{#3D99F6}\sin \frac {8\pi}{11}} \cdot \frac {- 4\sin \frac {3 \pi}{11}}{\color{#D61F06}\sin \frac {12 \pi}{11}} & \small \color{#3D99F6} \text{Note that }\sin (\pi - x) = \sin x \\ & = \frac {8\sin \frac \pi{11}}{\color{#3D99F6}\sin \frac {3\pi}{11}} \cdot \frac {- 4\sin \frac {3 \pi}{11}}{\color{#D61F06}- \sin \frac \pi{11}} & \small \color{#D61F06} \text{Note that }\sin (\pi + x) = - \sin x \\ & = \boxed{32} \end{aligned}

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W e h a v e i d e n t i t y Π k = 1 n − 1 C o s ( k π n ) = S i n ( n π 2 ) 2 n − 1 . L e t n = 11. ∴ Π k = 1 11 C o s ( k π 11 ) = Π k = 1 10 C o s ( k π 11 ) ∗ C o s ( 11 π 11 ) = S i n ( 11 π 2 ) 2 10 ∗ C o s ( 11 π 11 ) = S i n ( π 2 ) 2 10 ∗ ( − 1 ) N o t e t h a t C o s ( π − α ) = − C o s α . ⟹ C o s ( k π 11 ) = − C o s ( π − k π 11 ) ∴ Π k = 1 5 ∗ C o s ( k π 11 ) = − S i n ( 11 π 2 ) 2 10 ∗ ( − 1 ) = S i n ( 1 2 ∗ π ) 2 10 = 1 2 5 . ∴ given expression= reciprocal of the above = 32. \Large~\displaystyle We~ have~identity~~{\Huge\Pi}_{k=1}^{n-1} Cos(\frac{k\pi} n) = \dfrac{Sin(\frac{n\pi} 2)}{2^{n-1}}.\\ Let~n=11.~~\therefore~ {\Huge\Pi}_{k=1}^{11}Cos(\frac{k\pi} {11})= {\Huge\Pi}_{k=1}^{10} Cos(\frac{k\pi} {11})* Cos(\frac{11\pi} {11}) \\ = \dfrac{Sin(\frac{11\pi} 2)}{2^{10}}* Cos(\frac{11\pi} {11}) = \dfrac{Sin(\frac{\pi} 2)}{2^{10}}*(-1)\\ Note~that~Cos(\pi- \alpha)= - Cos\alpha.~\implies~ Cos(\frac{k\pi}{11}) =- Cos(\pi -\frac{k\pi} {11})\\ \therefore~ {\Huge\Pi}_{k=1}^5* Cos(\frac{k\pi} {11})=\sqrt{ - \dfrac{Sin(\frac{11\pi} 2)}{2^{10}}*(-1)} \\ \ \ \ \ =\sqrt{\dfrac{Sin(\frac 1 2 *\pi)}{2^{10}}}=\dfrac 1 {2^5}.\\ \therefore~\text{given expression= reciprocal of the above =} \Large \color{#D61F06}{32}.

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Naren Bhandari
Feb 27, 2017

S = s e c θ × s e c 2 θ × s e c 3 θ × s e c 4 θ × s e c 5 θ S = sec\theta\times sec2\theta\times sec3\theta \times sec4\theta\times sec5\theta

= 1 c o s θ × c o s 2 θ t i m e s c o s 3 θ × c o s 4 θ × c o s 5 θ =\frac{1}{cos\theta\times cos2\theta\ times cos3\theta \times cos4\theta\times cos5\theta}

Multiplying numerator and denominator by 2 s i n θ 2sin\theta

= 2 s i n θ ( 2 s i n θ c o s θ ) × c o s 2 θ × c o s 3 θ × c o s 4 θ × c o s 5 θ =\frac{2sin\theta}{ (2sin\theta cos\theta)\times cos2\theta\times cos3\theta \times cos4\theta\times cos5\theta}

= 2 s i n θ s i n 2 θ × c o s 2 θ t i m e s c o s 3 θ × c o s 4 θ × c o s 5 θ =\frac{2sin\theta}{sin2\theta\times cos2\theta\ times cos3\theta \times cos4\theta\times cos5\theta}

= 4 s i n θ ( 2 s i n 2 θ × c o s 2 θ ) t i m e s c o s 3 θ × c o s 4 θ × c o s 5 θ =\frac{4sin\theta}{(2sin2\theta\times cos2\theta)\ times cos3\theta \times cos4\theta\times cos5\theta}

= 8 s i n θ ( 2 s i n 4 θ × c o s 4 θ ) t i m e s c o s 3 θ × c o s 5 θ =\frac{8sin\theta}{(2sin4\theta\times cos4\theta)\ times cos3\theta\times cos5\theta}

Note that

c o s 3 θ = c o s ( 11 θ − 3 θ ) = c o s ( π − 3 θ ) = − c o s 8 θ cos3\theta = cos(11\theta - 3\theta) = cos(\pi - 3\theta) = - cos8\theta

= 8 s i n θ ( 2 s i n 8 θ t i m e s ( − c o s 8 θ ) × c o s 5 θ =\frac{8sin\theta}{(2sin8\theta\ times (- cos8\theta)\times cos5\theta}

= 16 s i n θ ( − 2 s i n 8 θ t i m e s c o s 8 θ ) × c o s 5 θ =\frac{16sin\theta}{(-2sin8\theta\ times cos8\theta)\times cos5\theta} = 16 s i n θ − s i n 16 θ × c o s 5 θ =\frac{16sin\theta}{-sin16\theta\times cos5\theta}

Note that

s i n 5 θ = s i n ( 11 θ + 5 θ ) = − s i n 5 θ sin5\theta = sin(11\theta + 5\theta )= - sin5\theta = 32 s i n θ + s i n 5 θ × c o s 5 θ =\frac{32sin\theta}{+sin5\theta\times cos5\theta}

= 32 s i n θ s i n 10 θ =\frac{32sin\theta}{sin10\theta}

= 32 s i n θ s i n ( 11 θ − θ ) =\frac{32sin\theta}{sin(11\theta- \theta )}

= 32 s i n θ s i n θ =\frac{32sin\theta}{sin\theta}

= 32 = \boxed{32}

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