Cosines And Pies!

Geometry Level 3

cos ⁡ 2 π 15 cos ⁡ 4 π 15 cos ⁡ 8 π 15 cos ⁡ 14 π 15 = 2 n \large \cos\dfrac{2{\pi}}{15}\cos\dfrac{4{\pi}}{15}\cos\dfrac{8{\pi}}{15}\cos\dfrac{14{\pi}}{15}\ = 2^n

Find n n .


The answer is -4.

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Rakshit Joshi by Rakshit Joshi , 26, India

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1 solution

Ikkyu San
Apr 25, 2016

Relevant wiki: Proving Trigonometric Identities - Intermediate

cos ⁡ ( 2 π 15 ) cos ⁡ ( 4 π 15 ) cos ⁡ ( 8 π 15 ) cos ⁡ ( 14 π 15 ) \cos\left(\dfrac{2\pi}{15}\right)\cos\left(\dfrac{4\pi}{15}\right)\cos\left(\dfrac{8\pi}{15}\right)\cos\left(\dfrac{14\pi}{15}\right)

= − cos ⁡ ( π 15 ) cos ⁡ ( 2 π 15 ) cos ⁡ ( 4 π 15 ) cos ⁡ ( 8 π 15 ) =-\cos\left(\dfrac{\pi}{15}\right)\cos\left(\dfrac{2\pi}{15}\right)\cos\left(\dfrac{4\pi}{15}\right)\cos\left(\dfrac{8\pi}{15}\right)

= − 2 sin ⁡ ( π 15 ) cos ⁡ ( π 15 ) cos ⁡ ( 2 π 15 ) cos ⁡ ( 4 π 15 ) cos ⁡ ( 8 π 15 ) 2 sin ⁡ ( π 15 ) =-\dfrac{2\sin\left(\dfrac{\pi}{15}\right)\cos\left(\dfrac{\pi}{15}\right)\cos\left(\dfrac{2\pi}{15}\right)\cos\left(\dfrac{4\pi}{15}\right)\cos\left(\dfrac{8\pi}{15}\right)}{2\sin\left(\dfrac{\pi}{15}\right)}

= − 2 sin ⁡ ( 2 π 15 ) cos ⁡ ( 2 π 15 ) cos ⁡ ( 4 π 15 ) cos ⁡ ( 8 π 15 ) 4 sin ⁡ ( π 15 ) =-\dfrac{2\sin\left(\dfrac{2\pi}{15}\right)\cos\left(\dfrac{2\pi}{15}\right)\cos\left(\dfrac{4\pi}{15}\right)\cos\left(\dfrac{8\pi}{15}\right)}{4\sin\left(\dfrac{\pi}{15}\right)}

= − 2 sin ⁡ ( 4 π 15 ) cos ⁡ ( 4 π 15 ) cos ⁡ ( 8 π 15 ) 8 sin ⁡ ( π 15 ) =-\dfrac{2\sin\left(\dfrac{4\pi}{15}\right)\cos\left(\dfrac{4\pi}{15}\right)\cos\left(\dfrac{8\pi}{15}\right)}{8\sin\left(\dfrac{\pi}{15}\right)}

= − 2 sin ⁡ ( 8 π 15 ) cos ⁡ ( 8 π 15 ) 16 sin ⁡ ( π 15 ) =-\dfrac{2\sin\left(\dfrac{8\pi}{15}\right)\cos\left(\dfrac{8\pi}{15}\right)}{16\sin\left(\dfrac{\pi}{15}\right)}

= − sin ⁡ ( 16 π 15 ) 16 sin ⁡ ( π 15 ) =-\dfrac{\sin\left(\dfrac{16\pi}{15}\right)}{16\sin\left(\dfrac{\pi}{15}\right)}

= sin ⁡ ( π 15 ) 16 sin ⁡ ( π 15 ) =\dfrac{\sin\left(\dfrac{\pi}{15}\right)}{16\sin\left(\dfrac{\pi}{15}\right)}

= 1 16 = 2 − 4 =\dfrac1{16}=2^{-4}

Thus, n = − 4 n=\boxed{-4}

Notes:

sin ⁡ ( 2 θ ) = 2 sin ⁡ θ cos ⁡ θ \sin(2\theta)=2\sin\theta\cos\theta

sin ⁡ ( π + θ ) = − sin ⁡ θ \sin(\pi+\theta)=-\sin\theta

cos ⁡ ( π − θ ) = − cos ⁡ θ \cos(\pi-\theta)=-\cos\theta

6 Helpful 0 Interesting 0 Brilliant 0 Confused

Exactly!! and what to say about your representation.(+1)

Rakshit Joshi - 5 years, 1 month ago

Did the same way.

Niranjan Khanderia - 5 years, 1 month ago

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