Linearize the Trigonometric Powers Part 3

Geometry Level 2

cos ⁡ 4 θ = a 4 cos ⁡ ( 4 θ ) + a 2 cos ⁡ ( 2 θ ) + a 0 \cos^4 \theta = a_4 \cos (4 \theta) + a_2 \cos (2 \theta) + a_0

Above shows a trigonometric identity for constants a 0 , a 2 , a 4 a_0, a_2, a_4 .

What is the value of a 4 − a 2 + a 0 a_4 - a_2 + a_0 ?

See Part 1 , Part 2 , and Part 4 .


The answer is 0.

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Chung Kevin by Chung Kevin , 45, Australia

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2 solutions

Rohit Ner
May 2, 2015

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

Great. Like this question, can you state cos ⁡ 8 θ \cos^{8} \theta in terms of signed sum of linear powers of cosines?

Mas Mus
May 7, 2015

cos ⁡ 4 θ = ( cos ⁡ 2 θ ) 2 = ( cos ⁡ 2 θ + 1 2 ) 2 = 1 4 ( cos ⁡ 2 2 θ + 2 cos ⁡ 2 θ + 1 ) = 1 4 ( cos ⁡ 4 θ + 1 2 ) + 2 4 cos ⁡ 2 θ + 1 4 = 1 8 cos ⁡ 4 θ + 4 8 cos ⁡ 2 θ + 3 8 = a 4 cos ⁡ 4 θ + a 2 cos ⁡ 2 θ + a 0 \cos^4\theta=\left(\cos^2\theta\right)^2\\=\left(\dfrac{\cos2\theta+1}{2}\right)^2=\dfrac{1}{4}\left(\cos^2{2\theta}+2\cos2\theta+1\right)\\=\dfrac{1}{4}\left(\dfrac{\cos4\theta+1}{2}\right)+\dfrac{2}{4}\cos2\theta+\dfrac{1}{4}\\=\dfrac{1}{8}\cos4\theta+\dfrac{4}{8}\cos2\theta+\dfrac{3}{8}=a_{4}\cos4\theta+a_{2}\cos2\theta+a_{0}

Hence, a 4 − a 2 + a 0 = 0 ~\large{a_{4}-a_{2}+a_{0}=0}

0 Helpful 0 Interesting 0 Brilliant 1 Confused

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