A Cosine Expression

$\begin{aligned} \cos 5\theta &= \cos (4\theta + \theta)\\ &= \cos 4\theta \cos \theta - \sin 4\theta \sin \theta \\ &= (2\cos^2 2\theta - 1) \cos \theta - (2\cos2\theta \sin 2\theta) \sin \theta \\ &= (2[2 \cos^2 \theta - 1]^2 -1) \cos \theta - (2 [2\cos^2 \theta - 1] [2 \sin\theta \cos\theta])\sin\theta\\ &= (2[2x^2 - 1]^2 - 1)x - (2[2x^2 - 1][2xy])y \\ &= (2[4x^4 - 4x^2 + 1] - 1)x - 4xy^2(2x^2 - 1) \\ &= (8x^4 - 8x^2 + 2 - 1)x - 8x^3y^2 + 4xy^2 \\ &= (8x^4 - 8x^2 + 1)x - 8x^3(1-x^2) + 4x(1-x^2) \\ &= 8x^5 - 8x^3 + x - 8x^3 + 8x^5 + 4x - 4x^3 \\ &= 16x^5 - 20x^3 + 5x \\ &= 16 \cos^5 \theta - 20 \cos^3 \theta + 5\cos\theta \end{aligned}$

I plugged in values for pi/3, pi/4, and pi/6, and then solved the three simultaneous equations.

It is the application of de moivre's theorem

$\cos5\theta = \cos5\theta - 10\cos^{3}\theta \sin^{2}\theta + 5\cos\theta \sin^{4}\theta$

$\cos5\theta = \cos5\theta - 10\cos^{3}\theta (1 - \cos^{2}\theta) + 5\cos\theta (1 - \cos^{2}\theta)^{2}$

$\cos5\theta = \cos5\theta - 10\cos^{3}\theta + 10\cos^{5} \theta + 5\cos\theta (1 + \cos^{4}\theta - 2\cos^{2}\theta$

$\cos5\theta = 11\cos5\theta - 10\cos^{3}\theta + 5\cos\theta + 5\cos^{5}\theta - 10\cos^{3}\theta$

$\cos5\theta = 16\cos5\theta - 20\cos^{3}\theta + 5\cos\theta$

so a+ b + c = 41

You can solve this problem using trigonometric identities , just you need this propeties below:

cos(a+b) = cos(a).cos(b)-sin(a).sin(c)

cos(2x) = 2cos^{2}(x)-1

sin(2x) = 2.sin(x).cos(x)

Then, you can modify cos(5x) as cos(2x + 3x) or cos(x + 4x) or cos(6x - x) and other manipulating of trigonometry. But, I think simple that you choose cos(2x + 3x) . Because, cos(2x + 3x) = cos(2x).cos(3x)-sin(2x).sin(3x) (required cos(3x) and sin(3x) => find it using trigonometric identities too). The results, must be cos(5x) = 16cos^{5}(x)-20cos^{3}(x)+5cos(x) .

But, there is another way to solve this problem too. I just realized and read about De Moivre's Formula , you just required the Binomial Theorem . Check this out. It's easy to apply the formula, be patient.

Use DeMoivre's Formula. The equation become: $cos(5x) + i sin(5x) = (cos(x) + i sin(x))^5$ $= \binom{5}{0} cos^5(x) + \binom {5}{1} cos^4(x) i sin(x) + \binom{5}{2} cos^3(x) i^2 sin^2(x) + \\ \binom{5}{3} cos^2(x) i^3 sin^3(x) + \binom{5}{4} cos(x) i^4 sin^4(x) + \binom{5}{5} i^5 sin^5(x)$ $= cos^5(x) + 5i cos^4(x) sin(x) - 10 cos^3(x) sin^2(x) - 10i cos^2(x) sin^3(x) +\\ 5 cos(x) sin^4(x) + i sin^5(x)$ Regard the real part,we get: $cos(5x) = cos^5(x) - 10 cos^3(x) sin^2(x) + 5 cos(x) sin^4(x)$ let $sin^2(x) = 1 - cos^2(x)$ Finally,the result is: $cos 5θ = 16 cos^5 θ -20 cos^3 θ + 5 cos θ$

cos (5x)= cos (2x +3x)

        = cos 2x cos 3x - sin 2x sin 3x

        = (2 cos^2 x-1)(4 cos^3 x-3 cos x) - (2 sin x cos x)(3 sin x - 4 sin^3 x)

        = 8 cos^5 x -6 cos^3 x -4 cos^3 x +3 cos x -6sin^2 x cos x +8 sin^4 x cos x

        = 8 cos^5 x -10 cos^3 x +3 cos x -6 cos^3 x +12 cos^3 x -6 cos x +8 cos^5 x -16 cos^3 x +8 cos x

        = 16 cos^5 x -20 cos^3 x +5 cos x

a+b+c=16+20+5

      =41

cos(5x) = {(cosx + isinx)^{5} + (cosx - isinx)^{5}}/2 ............(Applying De Moivre's Formula )

Let cosx = c, sinx = s.

Therefore,

cos(5x) = {(c^{5} + i 5 c^{4}s - 10 c^{3}s^{2} - i 10 c^{2}s^{3} + 5 cs^{4} + i s^{5}) + (c^{5} - i 5 c^{4}s - 10 c^{3}s^{2} + i 10 c^{2}s^{3} + 5 cs^{4} - i s^{5})}/2

       = c^{5} - 10 c^{3}s^{2} + 5 cs^{4}

       = c^{5} - 10 c^{3}(1 - c^{2}) + 5 c(1 - c^{2})^{2}

       = 11 c^{5} - 10 c^{3} + 5 c^{5} - 10 c^{3} + 5 c

       = 16 c^{5} - 20 c^{3} + 5 c

Clearly, a = 16, b = 20, c = 5 and a + b + c = 41 (Ans)

De Moivre's Theorem : $(cos(\theta) + i\sin(\theta))^n = \cos n(\theta) + i\sin n(\theta)$ .

$\begin{aligned} \cos 5\theta + i \sin 5\theta &= (cos\theta + i\sin\theta)^5\\ &= \cos^5\theta + 5 i \cos^4\theta \sin\theta + 10 i^2 \cos^3\theta \sin^2\theta\\ &\quad + 10 i^3 \cos^2\theta \sin^3\theta + 5 i^4 \cos\theta \sin^4\theta + i^5 \sin^5\theta\\ &= (\cos^5 \theta - 10 \cos^3 \theta \sin^2 \theta + 5 \cos \theta \sin^4 \theta)\\ &\quad + i (5 \cos^4 \theta \sin \theta - 10 \cos^2 \theta \sin^3 \theta + \sin^5 \theta)\\ Using : \sin^2 \theta = 1 - \cos^2 \theta.\\ \cos 5\theta + i \sin 5\theta &= (16 cos^5 \theta - 20 cos^3 \theta + 5 cos \theta)\\ &\quad + i (5 \cos^4 \theta \sin \theta - 10 \cos^2 \theta \sin^3 \theta + \sin^5 \theta)\\ Take-the-real-parts:\\ \cos 5\theta + i \sin 5\theta &= (16 cos^5 \theta - 20 cos^3 \theta + 5 cos \theta)\\ \\ a + b + c = 16 + 20 + 5 = 41\\ \end{aligned}$

Put theta = 0 to get a - b +c = 1 , put theta = pi/3 to get a/32 - b/8 +c/2 = 1/2 , put theta = pi/4 to get a/4 - b/2 + c = -1, solve to get b =20, now, a + b+ c = a - b +c +2b = 1 +2*20 =41.

I used Chebyshev polynomials

θ=π,−1=−a+b−c

θ=π/3,1/2=a/25−b/23+c

θ=π/4,21/2=a2−5/2−b2−3/2+c2−1/2

a=16,b20,c=5,a+b+c=41

Staff test solution. Hi everyone.

i used cos(3theta) and cos(2theta) and changed sine to cos but my method was a lengthy one it took me at least 5 mins.

(cos(x) + iSin(x))^n = cos(nx) + iSin(nx) here n=5 expanding this and comparing the real and imaginary terms we get cos(5x) = 16cos(x)^5 - 20cos(x)^3 + 5cos(x)

Let $\cos \theta$ = x , $\sin \theta$ = y . After expanding $\cos 5\theta$ we'll get 4(x^5) - 4(x^3) + x - 12(x^3)(y^2) + 4x(y^2). Since (y^2) = 1 - (x^2) , we can further expand it and finally we'll get 16 $\cos^5 \theta$ - 20 $\cos^3 \theta$ + 5 $\cos \theta$ . By comparison, a = 16, b = 20, c = 5. Therefore, a + b + c = 16 + 20 + 5 = 41.