For non-negative integers n, let f(n) = (∑ₖ₌₀ⁿsin((k+1)/(n+2)π)sin((k+2)/(n+2)π))/(∑ₖ₌₀ⁿsin²((k+1)/(n+2)π)) Assuming cos⁻¹x takes values in [0,π], which of the following options is/are correct?
f ( n ) = ∑ k = 0 n sin ( k + 1 n + 2 π ) sin ( k + 2 n + 2 π ) ∑ k = 0 n sin 2 ( k + 1 n + 2 π ) f(n)=\dfrac{\sum_{k=0}^{n}\sin\left(\dfrac{k+1}{n+2}\pi\right)\sin\left(\dfrac{k+2}{n+2}\pi\right)}{\sum_{k=0}^{n}\sin^{2}\left(\dfrac{k+1}{n+2}\pi\right)} f ( n ) = ∑ k = 0 n sin 2 ( n + 2 k + 1 π ) ∑ k = 0 n sin ( n + 2 k + 1 π ) sin ( n + 2 k + 2 π )
To find: which of options ( 1 ) – ( 4 ) are correct \text{To find: which of options } (1)\text{--}(4) \text{ are correct} To find: which of options ( 1 ) – ( 4 ) are correct
m = n + 2 , θ = π m , j = k + 1 m=n+2,\quad \theta=\dfrac{\pi}{m},\quad j=k+1 m = n + 2 , θ = m π , j = k + 1
f ( n ) = ∑ j = 1 m − 1 sin ( j θ ) sin ( ( j + 1 ) θ ) ∑ j = 1 m − 1 sin 2 ( j θ ) f(n)=\dfrac{\sum_{j=1}^{m-1}\sin(j\theta)\sin((j+1)\theta)}{\sum_{j=1}^{m-1}\sin^{2}(j\theta)} f ( n ) = ∑ j = 1 m − 1 sin 2 ( j θ ) ∑ j = 1 m − 1 sin ( j θ ) sin (( j + 1 ) θ )
∑ j = 1 m − 1 sin 2 ( j θ ) = m − 1 2 − 1 2 ∑ j = 1 m − 1 cos ( 2 j θ ) \sum_{j=1}^{m-1}\sin^{2}(j\theta)=\dfrac{m-1}{2}-\dfrac{1}{2}\sum_{j=1}^{m-1}\cos(2j\theta) ∑ j = 1 m − 1 sin 2 ( j θ ) = 2 m − 1 − 2 1 ∑ j = 1 m − 1 cos ( 2 j θ )
∑ j = 1 m − 1 cos ( 2 j θ ) = − 1 ⇒ ∑ j = 1 m − 1 sin 2 ( j θ ) = m 2 \sum_{j=1}^{m-1}\cos(2j\theta)=-1 \;\Rightarrow\; \sum_{j=1}^{m-1}\sin^{2}(j\theta)=\dfrac{m}{2} ∑ j = 1 m − 1 cos ( 2 j θ ) = − 1 ⇒ ∑ j = 1 m − 1 sin 2 ( j θ ) = 2 m
sin ( j θ ) sin ( ( j + 1 ) θ ) = 1 2 [ cos θ − cos ( ( 2 j + 1 ) θ ) ] \sin(j\theta)\sin((j+1)\theta)=\dfrac{1}{2}\left[\cos\theta-\cos((2j+1)\theta)\right] sin ( j θ ) sin (( j + 1 ) θ ) = 2 1 [ cos θ − cos (( 2 j + 1 ) θ ) ]
∑ j = 1 m − 1 cos ( ( 2 j + 1 ) θ ) = − cos θ \sum_{j=1}^{m-1}\cos((2j+1)\theta)=-\cos\theta ∑ j = 1 m − 1 cos (( 2 j + 1 ) θ ) = − cos θ
∑ j = 1 m − 1 sin ( j θ ) sin ( ( j + 1 ) θ ) = 1 2 [ ( m − 1 ) cos θ + cos θ ] = m 2 cos θ \sum_{j=1}^{m-1}\sin(j\theta)\sin((j+1)\theta)=\dfrac{1}{2}\left[(m-1)\cos\theta+\cos\theta\right]=\dfrac{m}{2}\cos\theta ∑ j = 1 m − 1 sin ( j θ ) sin (( j + 1 ) θ ) = 2 1 [ ( m − 1 ) cos θ + cos θ ] = 2 m cos θ
f ( n ) = ( m / 2 ) cos θ m / 2 = cos θ = cos π n + 2 f(n)=\dfrac{(m/2)\cos\theta}{m/2}=\cos\theta=\cos\dfrac{\pi}{n+2} f ( n ) = m /2 ( m /2 ) cos θ = cos θ = cos n + 2 π
Plan: use f ( n ) = cos π n + 2 in options ( 1 ) – ( 4 ) \text{Plan: use } f(n)=\cos\dfrac{\pi}{n+2} \text{ in options }(1)\text{--}(4) Plan: use f ( n ) = cos n + 2 π in options ( 1 ) – ( 4 )
lim n → ∞ cos π n + 2 = cos 0 = 1 ≠ 1 2 ⇒ ( 1 ) FALSE \lim_{n\to\infty}\cos\dfrac{\pi}{n+2}=\cos 0=1\neq\dfrac{1}{2} \;\Rightarrow\; (1)\ \text{FALSE} lim n → ∞ cos n + 2 π = cos 0 = 1 = 2 1 ⇒ ( 1 ) FALSE
f ( 4 ) = cos π 6 = 3 2 ⇒ ( 2 ) TRUE f(4)=\cos\dfrac{\pi}{6}=\dfrac{\sqrt{3}}{2} \;\Rightarrow\; (2)\ \text{TRUE} f ( 4 ) = cos 6 π = 2 3 ⇒ ( 2 ) TRUE
f ( 6 ) = cos π 8 ⇒ α = tan ( cos − 1 ( cos π 8 ) ) = tan π 8 f(6)=\cos\dfrac{\pi}{8} \;\Rightarrow\; \alpha=\tan\left(\cos^{-1}\left(\cos\dfrac{\pi}{8}\right)\right)=\tan\dfrac{\pi}{8} f ( 6 ) = cos 8 π ⇒ α = tan ( cos − 1 ( cos 8 π ) ) = tan 8 π
tan π 8 = 2 − 1 \tan\dfrac{\pi}{8}=\sqrt{2}-1 tan 8 π = 2 − 1
α = 2 − 1 ⇒ α 2 + 2 α − 1 = ( 3 − 2 2 ) + ( 2 2 − 2 ) − 1 = 0 \alpha=\sqrt{2}-1 \;\Rightarrow\; \alpha^{2}+2\alpha-1=(3-2\sqrt{2})+(2\sqrt{2}-2)-1=0 α = 2 − 1 ⇒ α 2 + 2 α − 1 = ( 3 − 2 2 ) + ( 2 2 − 2 ) − 1 = 0
⇒ ( 3 ) TRUE \Rightarrow (3)\ \text{TRUE} ⇒ ( 3 ) TRUE
sin ( 7 cos − 1 f ( 5 ) ) = sin ( 7 ⋅ π 7 ) = sin π = 0 ⇒ ( 4 ) TRUE \sin\left(7\cos^{-1}f(5)\right)=\sin\left(7\cdot\dfrac{\pi}{7}\right)=\sin\pi=0 \;\Rightarrow\; (4)\ \text{TRUE} sin ( 7 cos − 1 f ( 5 ) ) = sin ( 7 ⋅ 7 π ) = sin π = 0 ⇒ ( 4 ) TRUE
Correct options: ( 2 ) , ( 3 ) , ( 4 ) \text{Correct options: } (2),\ (3),\ (4) Correct options: ( 2 ) , ( 3 ) , ( 4 )
