Linearize the Trigonometric Powers Part 4

Geometry Level 3

cos ⁡ 5 θ = a 5 cos ⁡ ( 5 θ ) + a 3 cos ⁡ ( 3 θ ) + a 1 cos ⁡ ( θ ) \cos^5 \theta = a_5 \cos (5 \theta) + a_3 \cos (3 \theta) + a_1 \cos ( \theta)

Above shows a trigonometric identity for constants a 1 , a 3 , a 5 a_1, a_3, a_5 .

What is the value of 3 a 3 − a 1 a 5 \large \frac {3a_3 - a_1}{a_5} ?

See Part 1 , Part 2 , and Part 3 .


The answer is 5.

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Chung Kevin by Chung Kevin , 45, Australia

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3 solutions

Jake Lai
May 22, 2015

I don't know why this never occurred to me before. Here's a way to find a way to express cos ⁡ n θ \cos^{n} \theta as a linear combination of cos ⁡ ( k θ ) \cos(k\theta) for general n n :

Recall that

cos ⁡ θ = e i θ + e − i θ 2 \cos \theta = \frac{e^{i\theta}+e^{-i\theta}}{2}

Let y = e i θ y = e^{i\theta} . Taking both sides to the power of n n we get

cos ⁡ n θ = ( y + 1 y 2 ) n = 1 2 n ∑ k = 0 n ( n k ) y k y n − k + 1 2 n − 1 ∑ k = 0 n ( n k ) y 2 k − n 2 \cos^{n} \theta = \left( \frac{y+\frac{1}{y}}{2} \right)^{n} = \frac{1}{2^{n}} \sum_{k=0}^{n} \binom{n}{k} \frac{y^{k}}{y^{n-k}} + \frac{1}{2^{n-1}} \sum_{k=0}^{n} \binom{n}{k} \frac{y^{2k-n}}{2}

= 1 2 n − 1 ∑ k = 0 ⌈ n / 2 ⌉ ( n k ) y 2 k − n + y 2 ( n − k ) − n 2 = 1 2 n − 1 ∑ k = 0 ⌈ n / 2 ⌉ ( n k ) cos ⁡ ( ( n − 2 k ) θ ) = \frac{1}{2^{n-1}} \sum_{k=0}^{\lceil n/2 \rceil} \binom{n}{k} \frac{y^{2k-n} + y^{2(n-k)-n}}{2} = \boxed{\displaystyle \frac{1}{2^{n-1}} \sum_{k=0}^{\lceil n/2 \rceil} \binom{n}{k} \cos((n-2k)\theta)}

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Moderator note:

Great observation!

The conversion of trigonometric expressions into Euler form often allows us to easily relate cos ⁡ k θ \cos ^k \theta with cos ⁡ ( n θ ) \cos (n \theta) .

Very nicely done. It's amazing you came to this realization on your own!

Chung Kevin - 6 years ago

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Thank you! I don't think it's anything special, but the closed form was surprisingly neat.

Jake Lai - 5 years, 11 months ago
Rohit Ner
May 2, 2015

cos ⁡ 5 θ = cos ⁡ 3 θ . cos ⁡ 2 θ = ( 3 cos ⁡ θ + cos ⁡ 3 θ 4 ) ( 1 + cos ⁡ 2 θ 2 ) = 1 8 ( 3 cos ⁡ θ + 3 cos ⁡ θ cos ⁡ 2 θ + cos ⁡ 3 θ + cos ⁡ 3 θ cos ⁡ 2 θ ) = 1 8 ( 3 cos ⁡ θ + 3 2 ( cos ⁡ 3 θ + cos ⁡ θ ) + cos ⁡ 3 θ + 1 2 ( cos ⁡ 5 θ + cos ⁡ θ ) ) = 1 16 ( cos ⁡ 5 θ + 5 cos ⁡ 3 θ + 10 cos ⁡ θ ) cos ⁡ 5 θ = 1 16 cos ⁡ 5 θ + 5 16 cos ⁡ 3 θ + 10 16 cos ⁡ θ \cos ^{ 5 }{ \theta } =\cos ^{ 3 }{ \theta } .\cos ^{ 2 }{ \theta } \\ \\ \quad \quad \quad \quad \quad =\left( \frac { 3\cos { \theta } +\cos { 3\theta } }{ 4 } \right) \left( \frac { 1+\cos { 2\theta } }{ 2 } \right) \\ \\ \quad \quad \quad \quad \quad =\frac { 1 }{ 8 } \left( 3\cos { \theta } +3\cos { \theta } \cos { 2\theta +\cos { 3\theta } +\cos { 3\theta } \cos { 2\theta } } \right) \\ \\ \quad \quad \quad \quad \quad =\frac { 1 }{ 8 } \left( 3\cos { \theta } +\frac { 3 }{ 2 } \left( \cos { 3\theta } +\cos { \theta } \right) +\cos { 3\theta } +\quad \frac { 1 }{ 2 } \left( \cos { 5\theta } +\cos { \theta } \right) \right) \\ \\ \quad \quad \quad \quad \quad =\frac { 1 }{ 16 } \left( \cos { 5\theta } +5\cos { 3\theta } +10\cos { \theta } \right) \\ \\ \cos ^{ 5 }{ \theta } =\frac { 1 }{ 16 } \cos { 5\theta } +\frac { 5 }{ 16 } \cos { 3\theta } +\frac { 10 }{ 16 } \cos { \theta }

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Chew-Seong Cheong
May 22, 2015

From the following two identities:

{ cos ⁡ 5 θ = 16 cos ⁡ 5 θ − 20 cos ⁡ 3 θ + 5 cos ⁡ θ cos ⁡ 3 θ = 4 cos ⁡ 3 θ − 3 cos ⁡ θ \begin{cases} \cos{5\theta} = 16\cos^5{\theta} - 20\cos^3{\theta} + 5 \cos{\theta} \\ \cos{3\theta} = 4\cos^3{\theta} - 3 \cos{\theta} \end{cases}

cos ⁡ 5 θ = 16 cos ⁡ 5 θ − 20 cos ⁡ 3 θ + 5 cos ⁡ θ = 16 cos ⁡ 5 θ − 5 ( 4 cos ⁡ 3 θ − 3 cos ⁡ θ ) − 15 cos ⁡ θ + 5 cos ⁡ θ = 16 cos ⁡ 5 θ − 5 cos ⁡ 3 θ − 10 cos ⁡ θ ⇒ 16 cos ⁡ 5 θ = cos ⁡ 5 θ + 5 cos ⁡ 3 θ + 10 cos ⁡ θ cos ⁡ 5 θ = 1 16 cos ⁡ 5 θ + 5 16 cos ⁡ 3 θ + 10 16 cos ⁡ θ \begin{aligned} \cos{5\theta} &= 16\cos^5{\theta} - 20\cos^3{\theta} + 5 \cos{\theta} \\ & = 16\cos^5{\theta} - 5(4\cos^3{\theta} - 3\cos{\theta}) - 15\cos{\theta} + 5\cos{\theta} \\ & = 16\cos^5{\theta} - 5\cos{3\theta} - 10\cos{\theta} \\ \Rightarrow 16\cos^5{\theta} & = \cos{5\theta} + 5\cos{3\theta} + 10\cos{\theta} \\ \cos^5{\theta} & = \frac{1}{16}\cos{5\theta} + \frac{5}{16}\cos{3\theta} + \frac{10}{16}\cos{\theta} \end{aligned}

⇒ 3 a 3 − a 1 a 5 = 3 × 5 16 − 10 16 1 16 = 5 \Rightarrow \dfrac{3a_3-a_1}{a_5} = \dfrac {3\times \frac{5}{16}-\frac{10}{16}}{\frac{1}{16}} = \boxed{5}

2 Helpful 1 Interesting 0 Brilliant 0 Confused

Moderator note:

You were able to tease out the identity by comparing coefficients.

Can this approach be generalized?

Thanks for your challenge. I see Jake Lai has answer the challenge. Here is how it is done with induction.

\(\begin{array} {} \cos{\theta} = \cos{\theta} & \\ \cos{2\theta} = 2\cos^2{\theta} - 1 & \Rightarrow \cos^2{\theta} = \frac{1}{2} \left( \cos{2\theta} + 1 \right) \\ \cos{3\theta} = 4\cos^3{\theta} - 3\cos{\theta} & \Rightarrow \cos^3{\theta} = \frac{1}{4} \left( \cos{3\theta} + 3\cos{\theta} \right) \\ \cos{4\theta} = 8\cos^4{\theta} - 8\cos^2{\theta} + 1 & \Rightarrow \cos^4{\theta} = \frac{1}{8} \left( \cos{4\theta} + 4\cos{2\theta} + 6 \right) \\ \cos{5\theta} = 16 \cos^5 {\theta} - 20\cos^3{\theta} + 5\cos{\theta} & \Rightarrow \cos^5{\theta} = \frac{1}{16} \left( \cos{5\theta} + 5\cos{3\theta} + 10\cos{\theta} \right) \end{array} \)

⇒ cos ⁡ n θ = 1 2 n − 1 ∑ k = 0 ⌈ n 2 ⌉ ( n k ) cos ⁡ ( [ n − 2 k ] θ ) \Rightarrow \cos^n {\theta} = \dfrac{1}{2^{n-1}} \displaystyle \sum_{k=0}^{\left \lceil \frac{n}{2} \right \rceil} {\begin{pmatrix} n \\ k \end{pmatrix} \cos{([n-2k]\theta)}}

Chew-Seong Cheong - 6 years ago

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This is not induction.

Pi Han Goh - 6 years ago

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I think he means that we "induct" to find cos ⁡ n θ \cos n \theta in terms of cos ⁡ n θ \cos^n \theta and other powers, and then substitute back in to obtain the multiple angle form.

Though, I don't see how the final step actually occurs.

Chung Kevin - 6 years ago

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@Chung Kevin – The Pascal triangle could help determine the coefficients of (cos(theta) )^6

Vijay Simha - 2 years, 5 months ago

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