Quintuple of cosine

Geometry Level 3

$\large \cos{5\theta}=a\cos^5{\theta}+b\cos^3{\theta}+c\cos{\theta}$

Given that the above is a trigonometric identity for constants $a,b$ and $c$ , find the value of $a^2+b^2+c^2$ .

The answer is 681.

1 solution

Akhil Bansal
Oct 29, 2015

Formula's used in the question,

$1) \cos(A + B) = \cos(A)\cos(B) - \sin(A)\sin(B)$
$2) \cos(3A) = 4\cos^3(A)- 3\cos (A)$
$3) \sin(3A) = 3\sin(A) - 4\sin^3(A)$
$4) \cos(2A) = 2\cos^2(A) - 1$
$5) \sin(2A) = 2\sin(A)\cos(A)$
$6) \sin^2(A) = 1 - \cos^2(A)$

You must need to know above formula's to simplify given expression.If you're not able to do so,try again using above formula's.

You could also equate real parts of the binomial expansion of $(\cos \theta + i\sin \theta)^5$ to $(\cos 5 \theta + i\sin 5 \theta)$ (through De moivre's Theorem).

Ryan Macha - 5 years, 7 months ago

Or just use a Chebyshev Table to get

$T_5 (x) = 16x^5 - 20x^3 + 5x$

and replace $x$ with $\cos \theta$ but where's the fun in that? :P

Sharky Kesa - 5 years, 7 months ago

There's a typo in the second formula, you wrote a sine instead of the cosine.

A Former Brilliant Member - 5 years, 7 months ago

Thanks,.. Edited!!

Akhil Bansal - 5 years, 7 months ago

No problem, man. It happens, especially when you have to worry about the latex code while writing as well. :D

A Former Brilliant Member - 5 years, 7 months ago

Quintuple of cosine

The answer is 681.

1 solution

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