JEE Trigonometry 2

Geometry Level 3

Calculate ∑ r = 1 7 cos ⁡ 2 ( r π 8 ) \sum_{r=1}^{7}\cos^2\left (\dfrac{r\pi}{8} \right )

This problem is a part of My picks for JEE 2
4 6 2 5 3

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Pranjal Jain by Pranjal Jain , 23, India

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3 solutions

Chew-Seong Cheong
Jul 10, 2017

S = ∑ r = 1 7 cos ⁡ 2 r π 8 = ∑ r = 1 7 1 2 ( 1 + cos ⁡ r π 4 ) = 7 2 + 1 2 ( cos ⁡ π 4 + cos ⁡ π 2 + cos ⁡ 3 π 4 + cos ⁡ π + cos ⁡ 5 π 4 + cos ⁡ 3 π 2 + cos ⁡ 7 π 4 ) = 7 2 + 1 2 ( 1 2 + 0 − 1 2 − 1 + 1 2 + 0 − 1 2 ) = 7 2 − 1 2 = 3 \begin{aligned} S & = \sum_{r=1}^7 \cos^2 \frac {r\pi}8 \\ & = \sum_{r=1}^7 \frac 12 \left(1+\cos \frac {r\pi}4 \right) \\ & = \frac 72 + \frac 12 \left({\color{#3D99F6}\cos \frac \pi 4} + \cos \frac \pi 2 + {\color{#D61F06}\cos \frac {3\pi} 4} + \cos \pi + {\color{#3D99F6}\cos \frac {5\pi} 4} + \cos \frac {3\pi} 2 + {\color{#D61F06}\cos \frac {7\pi} 4} \right) \\ & = \frac 72 + \frac 12 \left({\color{#3D99F6}\frac 1{\sqrt 2}} + 0 \ {\color{#D61F06}- \frac 1{\sqrt 2}} - 1 + {\color{#3D99F6}\frac 1{\sqrt 2}} + 0 \ {\color{#D61F06}- \frac 1{\sqrt 2}} \right) \\ & = \frac 72 - \frac 12 = \boxed{3} \end{aligned}

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We know that cos ⁡ ( π − θ ) = − cos ⁡ θ \cos( \pi - \theta) = -\cos\theta so ( cos ⁡ 2 ( π − θ ) ) = cos ⁡ 2 θ ( \cos^{2} ( \pi - \theta)) = \cos^{2} \theta .

So the terms about the middle term i.e. cos ⁡ 2 π 4 \cos^{2} \frac{\pi}{4} are equal , and equal to 2 ⋅ ( cos ⁡ 2 π 8 + cos ⁡ 2 2 π 8 + cos ⁡ 2 3 π 8 ) = 2 ⋅ ( 0.85 + 0.5 + 0.15 ) = 3 2 \cdot ( \cos^{2} \frac{\pi}{8} + \cos^{2} \frac{2\pi}{8} + \cos^{2} \frac{3\pi}{8} ) = 2 \cdot (0.85 + 0.5 + 0.15 ) = 3

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Could be a little tougher. I solved this for general n n . Unfortunately, I asked for n = 8 n=8 . Can you think for general n n ?

Pranjal Jain - 6 years, 4 months ago

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Ok, I'll try ⌣ ¨ \ddot\smile

A Former Brilliant Member - 6 years, 4 months ago

∑ r = 1 n cos ⁡ 2 ( r π 8 ) \sum_{r=1}^{n}\cos^2\left (\dfrac{r\pi}{8} \right )

∑ r = 1 n 1 + c o s ( r π 4 ) 2 \sum_{r=1}^{n}\dfrac{1 + cos\left (\dfrac{r\pi}{4} \right )}{2}

= n 2 + 1 2 ∑ r = 1 n 2 c o s ( r π 4 ) s i n ( π 8 ) 2 s i n ( π 8 ) = \dfrac{n}{2} + \dfrac{1}{2}\sum_{r=1}^{n} \dfrac{2cos\left (\dfrac{r\pi}{4} \right )sin\left (\dfrac{\pi}{8} \right )}{2sin\left (\dfrac{\pi}{8} \right )}

π 4 = x \dfrac{\pi}{4} = x

= n 2 + 1 4 s i n ( π 8 ) ∑ r = 1 n sin ⁡ ( x ( 2 r + 1 ) 2 ) − sin ⁡ ( x ( 2 r − 1 ) 2 ) = \dfrac{n}{2} + \dfrac{1}{4sin\left (\dfrac{\pi}{8} \right )}\sum_{r=1}^{n} \sin\left( \dfrac{x(2r + 1)}{2}\right) - \sin\left( \dfrac{x(2r - 1)}{2}\right)

= n 2 + 1 4 s i n ( π 8 ) sin ⁡ ( 2 n π + π 8 ) − sin ⁡ ( π 8 ) = \dfrac{n}{2} + \dfrac{1}{4sin\left (\dfrac{\pi}{8} \right )} \sin \left( \dfrac{2n\pi + \pi}{8}\right) - \sin \left( \dfrac{\pi}{8}\right)

= n 2 + 1 4 s i n ( π 8 ) 2 cos ⁡ ( n π + π 8 ) × sin ⁡ ( n π 8 ) = \dfrac{n}{2} + \dfrac{1}{4sin\left (\dfrac{\pi}{8} \right )} 2\cos \left(\dfrac{n\pi + \pi}{8}\right) \times \sin \left(\dfrac{n\pi}{8}\right)

C = ∑ r = 1 n cos ⁡ r x C = \sum_{r=1}^n \cos rx

i S = ∑ r = 1 n i sin ⁡ r x iS = \sum_{r=1}^n i\sin rx

C + i S = ∑ r = 1 n e i r x = 1 − e i n x 1 − e i x C + iS = \sum_{r=1}^n e^{irx} = \dfrac{ 1 - e^{inx}}{1 - e^{ix}}

C + i S = ( 1 − cos ⁡ n x − i sin ⁡ n x ) ( 1 − cos ⁡ x + i sin ⁡ x ) 2 − 2 cos ⁡ x C + iS = \dfrac{(1 - \cos nx - i \sin nx)( 1 - \cos x + i \sin x)}{2 - 2 \cos x}

C = 1 2 − cos ⁡ ( n + 1 ) x 2 cos ⁡ ( n − 1 ) x 2 2 − 2 cos ⁡ x − cos ⁡ x ( n + 1 ) 2 − 2 cos ⁡ x C = \dfrac{1}{2} - \dfrac{\cos \dfrac{(n+1)x}{2} \cos \dfrac{(n -1)x}{2}}{2 - 2 \cos x} - \dfrac{\cos x(n + 1)}{2 - 2 \cos x}

∑ r = 1 n 1 + cos ⁡ r π 8 2 = n 2 + 1 2 − cos ⁡ ( n + 1 ) π 4 2 cos ⁡ ( n − 1 ) π 4 2 2 − 2 cos ⁡ x − cos ⁡ π 4 ( n + 1 ) 2 − 2 cos ⁡ π 4 \sum_{r=1}^{n} \dfrac{1 + \cos \dfrac{r\pi}{8}}{2} = \dfrac{n}{2} + \dfrac{1}{2} - \dfrac{\cos \dfrac{(n+1)\dfrac{\pi}{4}}{2} \cos \dfrac{(n -1)\dfrac{\pi}{4}}{2}}{2 - 2 \cos x} - \dfrac{\cos \dfrac{\pi}{4}(n + 1)}{2 - 2 \cos \dfrac{\pi}{4}}

U Z - 6 years, 4 months ago

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Hi Megh, if you continue from here

n 2 + 1 4 s i n ( π 8 ) 2 cos ⁡ ( n π + π 8 ) × sin ⁡ ( n π 8 ) \dfrac{n}{2} + \dfrac{1}{4sin\left (\dfrac{\pi}{8} \right )} 2\cos \left(\dfrac{n\pi + \pi}{8}\right) \times \sin \left(\dfrac{n\pi}{8}\right)

= n 2 + 1 4 ⋅ ( cos ⁡ n π 4 ) + 1 4 ⋅ ( c o t π 8 ) ( s i n n π 4 ) − 1 4 = \dfrac{n}{2} + \frac{1}{4}\cdot( \cos { \frac{n \pi}{4}} ) + \frac{1}{4} \cdot ( cot\frac{\pi}{8} )( sin \frac{n \pi}{4} ) - \frac{1}{4}

which I think will suffice .

A Former Brilliant Member - 6 years, 4 months ago

Nice! That's what I wanted.

Pranjal Jain - 6 years, 4 months ago
Arpit Sah
Jul 21, 2015

cos ⁡ 2 θ \cos^{2}\theta = c o s 2 ( π 2 − θ ) cos^{2}(\frac{\pi}{2}-\theta)

cos ⁡ 2 ( 7 π 8 ) = s i n 2 ( 3 π 8 \cos^{2}(\frac{7\pi}{8})=sin^{2}(\frac{3\pi}{8} )

cos ⁡ 2 ( 6 π 8 ) = s i n 2 ( 2 π 8 \cos^{2}(\frac{6\pi}{8})=sin^{2}(\frac{2\pi}{8} )

cos ⁡ 2 ( 5 π 8 ) = s i n 2 ( 1 π 8 \cos^{2}(\frac{5\pi}{8})=sin^{2}(\frac{1\pi}{8} )

cos ⁡ 2 θ + sin ⁡ 2 θ = 1 \cos^{2}\theta + \sin^{2}\theta = 1

cos ⁡ 2 ( π 2 ) = 0 \cos^{2}(\frac{\pi}{2}) = 0

Therefore,

sin ⁡ 2 ( 3 π 8 ) + cos ⁡ 2 ( 3 π 8 ) + sin ⁡ 2 ( 2 π 8 ) + cos ⁡ 2 ( 2 π 8 ) + sin ⁡ 2 ( 1 π 8 ) + cos ⁡ 2 ( 1 π 8 ) = 3 \sin^{2}(\frac{3\pi}{8}) + \cos^{2}(\frac{3\pi}{8}) +\sin^{2}(\frac{2\pi}{8}) +\cos^{2}(\frac{2\pi}{8}) + \sin^{2}(\frac{1\pi}{8}) + \cos^{2}(\frac{1\pi}{8}) = 3

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