Given sum
We need to solve, for 0 < θ < π 2 0<\theta<\dfrac{\pi}{2} 0 < θ < 2 π
∑ m = 1 6 csc ( θ + ( m − 1 ) π 4 ) csc ( θ + m π 4 ) = 4 2 . \sum_{m=1}^{6}\csc\left(\theta+\frac{(m-1)\pi}{4}\right)\csc\left(\theta+\frac{m\pi}{4}\right)=4\sqrt2. m = 1 ∑ 6 csc ( θ + 4 ( m − 1 ) π ) csc ( θ + 4 mπ ) = 4 2 . Let
A m = θ + ( m − 1 ) π 4 . A_m=\theta+\frac{(m-1)\pi}{4}. A m = θ + 4 ( m − 1 ) π . Then the sum becomes
∑ m = 1 6 csc A m csc ( A m + π 4 ) . \sum_{m=1}^6 \csc A_m\,\csc\left(A_m+\frac{\pi}{4}\right). m = 1 ∑ 6 csc A m csc ( A m + 4 π ) .
Use a standard identity
We use
cot x − cot y = sin ( y − x ) sin x sin y . \cot x-\cot y=\frac{\sin(y-x)}{\sin x\sin y}. cot x − cot y = sin x sin y sin ( y − x ) . Taking y = x + α y=x+\alpha y = x + α
cot x − cot ( x + α ) = sin α sin x sin ( x + α ) . \cot x-\cot(x+\alpha)=\frac{\sin\alpha}{\sin x\sin(x+\alpha)}. cot x − cot ( x + α ) = sin x sin ( x + α ) sin α . Hence,
csc x csc ( x + α ) = cot x − cot ( x + α ) sin α . \csc x\,\csc(x+\alpha)=\frac{\cot x-\cot(x+\alpha)}{\sin\alpha}. csc x csc ( x + α ) = sin α cot x − cot ( x + α ) . Here α = π 4 \alpha=\dfrac{\pi}{4} α = 4 π sin α = 1 2 \sin\alpha=\dfrac{1}{\sqrt2} sin α = 2 1
csc x csc ( x + π 4 ) = 2 [ cot x − cot ( x + π 4 ) ] . \csc x\,\csc\left(x+\frac{\pi}{4}\right)=\sqrt2\left[\cot x-\cot\left(x+\frac{\pi}{4}\right)\right]. csc x csc ( x + 4 π ) = 2 [ cot x − cot ( x + 4 π ) ] . So each term is telescoping.
Apply telescoping
Thus
∑ m = 1 6 csc ( θ + ( m − 1 ) π 4 ) csc ( θ + m π 4 ) = 2 ∑ m = 1 6 [ cot ( θ + ( m − 1 ) π 4 ) − cot ( θ + m π 4 ) ] . \sum_{m=1}^6 \csc\left(\theta+\frac{(m-1)\pi}{4}\right)\csc\left(\theta+\frac{m\pi}{4}\right)
=\sqrt2\sum_{m=1}^6\left[\cot\left(\theta+\frac{(m-1)\pi}{4}\right)-\cot\left(\theta+\frac{m\pi}{4}\right)\right]. m = 1 ∑ 6 csc ( θ + 4 ( m − 1 ) π ) csc ( θ + 4 mπ ) = 2 m = 1 ∑ 6 [ cot ( θ + 4 ( m − 1 ) π ) − cot ( θ + 4 mπ ) ] . This telescopes to
2 [ cot θ − cot ( θ + 6 π 4 ) ] = 2 [ cot θ − cot ( θ + 3 π 2 ) ] . \sqrt2\left[\cot\theta-\cot\left(\theta+\frac{6\pi}{4}\right)\right]
=\sqrt2\left[\cot\theta-\cot\left(\theta+\frac{3\pi}{2}\right)\right]. 2 [ cot θ − cot ( θ + 4 6 π ) ] = 2 [ cot θ − cot ( θ + 2 3 π ) ] . Now,
cot ( θ + 3 π 2 ) = cot ( θ + π 2 ) = − tan θ . \cot\left(\theta+\frac{3\pi}{2}\right)=\cot\left(\theta+\frac{\pi}{2}\right)=-\tan\theta. cot ( θ + 2 3 π ) = cot ( θ + 2 π ) = − tan θ . Therefore the sum is
2 ( cot θ + tan θ ) . \sqrt2\left(\cot\theta+\tan\theta\right). 2 ( cot θ + tan θ ) . Given this equals 4 2 4\sqrt2 4 2
cot θ + tan θ = 4. \cot\theta+\tan\theta=4. cot θ + tan θ = 4.
Solve the trigonometric equation
Recall
tan θ + cot θ n = sin θ cos θ + cos θ sin θ = sin 2 θ + cos 2 θ sin θ cos θ = 1 sin θ cos θ . \tan\theta+\cot\theta
n=\frac{\sin\theta}{\cos\theta}+\frac{\cos\theta}{\sin\theta}
=\frac{\sin^2\theta+\cos^2\theta}{\sin\theta\cos\theta}
=\frac{1}{\sin\theta\cos\theta}. tan θ + cot θ n = cos θ sin θ + sin θ cos θ = sin θ cos θ sin 2 θ + cos 2 θ = sin θ cos θ 1 . So
1 sin θ cos θ = 4 \frac{1}{\sin\theta\cos\theta}=4 sin θ cos θ 1 = 4 which gives
sin θ cos θ = 1 4 . \sin\theta\cos\theta=\frac14. sin θ cos θ = 4 1 . Hence,
sin 2 θ = 2 sin θ cos θ = 1 2 . \sin 2\theta=2\sin\theta\cos\theta=\frac12. sin 2 θ = 2 sin θ cos θ = 2 1 . Since 0 < θ < π 2 0<\theta<\dfrac{\pi}{2} 0 < θ < 2 π 0 < 2 θ < π 0<2\theta<\pi 0 < 2 θ < π
2 θ = π 6 or 2 θ = 5 π 6 . 2\theta=\frac{\pi}{6}\quad \text{or}\quad 2\theta=\frac{5\pi}{6}. 2 θ = 6 π or 2 θ = 6 5 π . Therefore,
θ = π 12 or θ = 5 π 12 . \theta=\frac{\pi}{12}\quad \text{or}\quad \theta=\frac{5\pi}{12}. θ = 12 π or θ = 12 5 π .
Check options
A: π 4 \dfrac{\pi}{4} 4 π sin 2 θ = sin π 2 = 1 ≠ 1 2 \sin 2\theta=\sin\dfrac{\pi}{2}=1\ne \dfrac12 sin 2 θ = sin 2 π = 1 = 2 1
B: π 6 \dfrac{\pi}{6} 6 π sin 2 θ = sin π 3 = 3 2 ≠ 1 2 \sin 2\theta=\sin\dfrac{\pi}{3}=\dfrac{\sqrt3}{2}\ne \dfrac12 sin 2 θ = sin 3 π = 2 3 = 2 1
C: π 12 \dfrac{\pi}{12} 12 π sin 2 θ = sin π 6 = 1 2 \sin 2\theta=\sin\dfrac{\pi}{6}=\dfrac12 sin 2 θ = sin 6 π = 2 1
D: 5 π 12 \dfrac{5\pi}{12} 12 5 π sin 2 θ = sin 5 π 6 = 1 2 \sin 2\theta=\sin\dfrac{5\pi}{6}=\dfrac12 sin 2 θ = sin 6 5 π = 2 1
So the correct options are C and D .
Comparison with stored answer
Stored correct answer: C, D
Our derived answer: C, D
They agree.