Linearize the Trigonometric Powers Part 2

Geometry Level 2

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

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

What is the value of a 1 ÷ a 3 a_1 \div a_3 ?

See Part 1 , Part 3 , and Part 4 .


The answer is 3.

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1 solution

Jake Lai
May 6, 2015

From the compound angle formula and double-angle formula, we may obtain the identity 4 cos 3 θ 3 cos θ = cos 3 θ 4\cos^{3} \theta-3\cos \theta = \cos 3\theta . Rearrangment gives

cos 3 θ = 1 4 cos 3 θ + 3 4 cos θ \cos^{3} \theta = \frac{1}{4} \cos 3\theta+\frac{3}{4} \cos \theta

As such, a 1 = 3 a 3 = 3 4 a_{1} = 3a_{3} = \frac{3}{4} , and a 1 ÷ a 3 = 3 a_{1} \div a_{3} = \boxed{3} .

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