Soviet Russian Cosine #2

Geometry Level 3

If the exact value of $\displaystyle\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}$ can be represented in the form $\displaystyle\frac{(-a)\sqrt{b}}{c}$ , and $gcd(a, b, c) = 1$ , then find $a+b+c$ .

The answer is 4.

1 solution

Mas Mus
Apr 24, 2015

Note that $~~\sin\left(\pi-\alpha\right)=\sin\alpha~~\text{and}~~\sin\left(\pi+\alpha\right)=-\sin\alpha$

$A=2\cos\frac{2\pi}{7}~\sin\frac{2\pi}{7}=\sin\frac{4\pi}{7}\\\\B=2\cos\frac{4\pi}{7}~\sin\frac{2\pi}{7}=\sin\frac{6\pi}{7}-\sin\frac{2\pi}{7}=\sin\frac{\pi}{7}-\sin\frac{2\pi}{7}\\\\C=2\cos\frac{6\pi}{7}~\sin\frac{2\pi}{7}=\sin\frac{8\pi}{7}-\sin\frac{4\pi}{7}=-\sin\frac{\pi}{7}-\sin\frac{4\pi}{7}$

Adding $A, B, C$ we have:

$2\sin\frac{2\pi}{7}\left(\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}\right)=-\sin\frac{2\pi}{7}\\\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}=\frac{\left(-1\right)\sqrt{1}}{2}$

So, $~~a+b+c=\boxed{4}$

and how we are supposed to know the value of -sin2pi/7 in fractional form .......its simply not possible....pls provide appropriate value and edit the question or bring it down...................

Rishabh Mishra - 6 years, 1 month ago

The solution does not require you to know the value of sin(2 pi/7). In the last step, both sides of the equation are divided by sin(2 pi/7).

Eric Kim - 6 years, 1 month ago

The final expression of the answer, using two 1's in place of one and all that, is truly convoluted. Not fun.

Marta Reece - 4 years ago

Soviet Russian Cosine #2

The answer is 4.

1 solution

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