A problem by Manas Kumar

Algebra Level pending

1 + 1 2 cos ⁡ 2 θ + 1 ⋅ 3 2 ⋅ 4 cos ⁡ 4 θ + 1 ⋅ 3 ⋅ 5 2 ⋅ 4 ⋅ 6 cos ⁡ 6 θ + ⋯ = ( a sin ⁡ θ ) b cos ⁡ ( π c − θ d ) 1+\frac 12\cos 2 \theta +\frac {1\cdot3}{2\cdot4} \cos 4\theta+ \frac {1\cdot3\cdot5}{2\cdot4\cdot6} \cos 6\theta+ \cdots =(a \sin \theta)^{b}\cos \left (\frac {\pi}{c}-\frac {\theta}{d}\right)

This equation holds true when a a , b b , c c and d d are real numbers. Find a + b + c + d a+b+c+d


The answer is 7.5.

Monu Kumar by Monu Kumar , 19, India

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1 solution

Monu Kumar
Mar 28, 2020

C = 1 + 1 2 cos ⁡ 2 θ + 1 ⋅ 3 2 ⋅ 4 cos ⁡ 4 θ + 1 ⋅ 3 ⋅ 5 2 ⋅ 4 ⋅ 6 cos ⁡ 6 θ + ⋯ C=1+\frac 12\cos 2 \theta +\frac {1\cdot3}{2\cdot4} \cos 4\theta+ \frac {1\cdot3\cdot5}{2\cdot4\cdot6} \cos 6\theta+ \cdots

S = 1 2 sin ⁡ 2 θ + 1 ⋅ 3 2 ⋅ 4 sin ⁡ 4 θ + 1 ⋅ 3 ⋅ 5 2 ⋅ 4 ⋅ 6 sin ⁡ 6 θ + ⋯ S=\frac 12\sin 2 \theta +\frac {1\cdot3}{2\cdot4} \sin 4\theta+ \frac {1\cdot3\cdot5}{2\cdot4\cdot6} \sin 6\theta+ \cdots

C + i S = 1 + 1 2 ( cos ⁡ 2 θ + i sin ⁡ 2 θ ) + 1 ⋅ 3 2 ⋅ 4 ( cos ⁡ 4 θ + i sin ⁡ 4 θ ) + 1 ⋅ 3 ⋅ 5 2 ⋅ 4 ⋅ 6 ( cos ⁡ 6 θ + i sin ⁡ 6 θ ) + ⋯ C+iS=1+\frac 12\left(\cos 2 \theta+ i\sin 2 \theta\right) +\frac {1\cdot3}{2\cdot4} \left(\cos 4\theta+ i\sin 4 \theta\right)+ \frac {1\cdot3\cdot5}{2\cdot4\cdot6} \left(\cos 6\theta+ i\sin 6 \theta\right)+ \cdots

= 1 + 1 2 e 2 i θ + 1 ⋅ 3 2 ⋅ 4 e 4 i θ + 1 ⋅ 3 ⋅ 5 2 ⋅ 4 ⋅ 6 e 6 i θ + ⋯ =1+\frac 12 e^{2i \theta} +\frac {1\cdot3}{2\cdot4} e^{4i\theta}+ \frac {1\cdot3\cdot5}{2\cdot4\cdot6} e^{6i\theta}+ \cdots

= 1 + ( 1 2 ) 1 ! e 2 i θ + ( 1 2 ) ( 1 + 1 2 ) 2 ! ( e 2 i θ ) 2 + ( 1 2 ) ( 1 + 1 2 ) ( 2 + 1 2 ) 3 ! ( e 2 i θ ) 3 + ⋯ =1+\frac{\left(\frac 12\right)}{1!} e^{2i \theta} +\frac {\left(\frac 12\right)\left(1+\frac 12\right)}{2!} \left(e^{2i\theta}\right)^{2}+ \frac {\left(\frac 12\right)\left(1+\frac 12\right)\left(2+\frac 12\right)}{3!}\left(e^{2i\theta}\right)^{3}+ \cdots

= ( 1 − e 2 i θ ) − 1 / 2 =\left(1-e^{2i\theta}\right)^{-1/2 }

= ( 1 − cos ⁡ 2 θ − i sin ⁡ 2 θ ) − 1 / 2 =\left(1-\cos 2\theta-i\sin 2\theta\right)^{-1/2 }

= ( 2 sin ⁡ 2 θ − i ⋅ 2 sin ⁡ θ cos ⁡ θ ) − 1 / 2 =\left(2\sin ^{2} \theta-i\cdot2\sin \theta\cos\theta\right)^{-1/2 }

= ( 2 sin ⁡ θ ) − 1 / 2 ( sin ⁡ θ − i cos ⁡ θ ) − 1 / 2 =\left(2\sin\theta\right)^{-1/2}\left(\sin\theta-i\cos\theta\right)^{-1/2 }

= ( 2 sin ⁡ θ ) − 1 / 2 [ cos ⁡ ( π 2 − θ ) − i sin ⁡ ( π 2 − θ ) ] − 1 / 2 =\left(2\sin\theta\right)^{-1/2}\left[\cos\left(\frac{\pi}{2}-\theta\right)-i\sin\left(\frac{\pi}{2}-\theta\right)\right]^{-1/2 }

= ( 2 sin ⁡ θ ) − 1 / 2 [ cos ⁡ ( π 4 − θ 2 ) − i sin ⁡ ( π 4 − θ 2 ) ] =\left(2\sin\theta\right)^{-1/2}\left[\cos\left(\frac{\pi}{4}-\frac{\theta}{2}\right)-i\sin\left(\frac{\pi}{4}-\frac{\theta}{2}\right)\right]

Equating real parts on both sides, We get C = ( 2 sin ⁡ θ ) − 1 / 2 cos ⁡ ( π 4 − θ 2 ) C=\left(2\sin\theta\right)^{-1/2}\cos\left(\frac{\pi}{4}-\frac{\theta}{2}\right)

So, a = 2 , b = − 1 2 , c = 4 , d = 2 a=2, b=-\frac 12, c=4, d=2

And a + b + c + d = 7.5 a+b+c+d=7.5

2 Helpful 0 Interesting 2 Brilliant 0 Confused

I Did using Catalan Generating function.

Nikola Alfredi - 9 months, 2 weeks ago

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