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Question: Let $S_n = \sum_{n=1}^{n} \sin^{-1}\left[\frac{(2n+1)}{n(n+1)(\sqrt{n(n+2)} + \sqrt{(n+1)(n-1)}}\rig...

Let Sn=∑n=1nsin⁡−1[(2n+1)n(n+1)(n(n+2)+(n+1)(n−1)]S_n = \sum_{n=1}^{n} \sin^{-1}\left[\frac{(2n+1)}{n(n+1)(\sqrt{n(n+2)} + \sqrt{(n+1)(n-1)}}\right]. Find the value of 100cos⁡(S99)100\cos(S_{99}).

Answer

1

Explanation

Solution

The problem asks us to find the value of 100cos⁡(S99)100\cos(S_{99}) where SnS_n is a given sum involving inverse sine functions.

The sum is given by: Sn=∑k=1nsin⁡−1[(2k+1)k(k+1)(k(k+2)+(k+1)(k−1)]S_n = \sum_{k=1}^{n} \sin^{-1}\left[\frac{(2k+1)}{k(k+1)(\sqrt{k(k+2)} + \sqrt{(k+1)(k-1)}}\right]

Let's analyze the general term inside the summation: Ak=(2k+1)k(k+1)(k(k+2)+(k+1)(k−1))A_k = \frac{(2k+1)}{k(k+1)(\sqrt{k(k+2)} + \sqrt{(k+1)(k-1)})}

We can rationalize the denominator involving the square roots: 1k(k+2)+(k+1)(k−1)=k(k+2)−(k+1)(k−1)k(k+2)−(k+1)(k−1)\frac{1}{\sqrt{k(k+2)} + \sqrt{(k+1)(k-1)}} = \frac{\sqrt{k(k+2)} - \sqrt{(k+1)(k-1)}}{k(k+2) - (k+1)(k-1)} =k2+2k−k2−1k2+2k−(k2−1)= \frac{\sqrt{k^2+2k} - \sqrt{k^2-1}}{k^2+2k - (k^2-1)} =k2+2k−k2−12k+1= \frac{\sqrt{k^2+2k} - \sqrt{k^2-1}}{2k+1}

Substitute this back into the expression for AkA_k: Ak=(2k+1)k(k+1)⋅k2+2k−k2−12k+1A_k = \frac{(2k+1)}{k(k+1)} \cdot \frac{\sqrt{k^2+2k} - \sqrt{k^2-1}}{2k+1} Ak=k2+2k−k2−1k(k+1)A_k = \frac{\sqrt{k^2+2k} - \sqrt{k^2-1}}{k(k+1)} Ak=k(k+2)k(k+1)−(k+1)(k−1)k(k+1)A_k = \frac{\sqrt{k(k+2)}}{k(k+1)} - \frac{\sqrt{(k+1)(k-1)}}{k(k+1)}

Now, let's try to express this in the form of sin⁡−1(x)−sin⁡−1(y)\sin^{-1}(x) - \sin^{-1}(y). The formula for this difference is sin⁡−1(x)−sin⁡−1(y)=sin⁡−1(x1−y2−y1−x2)\sin^{-1}(x) - \sin^{-1}(y) = \sin^{-1}(x\sqrt{1-y^2} - y\sqrt{1-x^2}), provided the arguments are within the valid range and the result is in [−π/2,π/2][-\pi/2, \pi/2].

Let x=1kx = \frac{1}{k} and y=1k+1y = \frac{1}{k+1}. Then 1−x2=1−1k2=k2−1k\sqrt{1-x^2} = \sqrt{1-\frac{1}{k^2}} = \frac{\sqrt{k^2-1}}{k}. And 1−y2=1−1(k+1)2=(k+1)2−1k+1=k2+2kk+1\sqrt{1-y^2} = \sqrt{1-\frac{1}{(k+1)^2}} = \frac{\sqrt{(k+1)^2-1}}{k+1} = \frac{\sqrt{k^2+2k}}{k+1}.

Now, calculate x1−y2−y1−x2x\sqrt{1-y^2} - y\sqrt{1-x^2}: x1−y2−y1−x2=1k⋅k2+2kk+1−1k+1⋅k2−1kx\sqrt{1-y^2} - y\sqrt{1-x^2} = \frac{1}{k} \cdot \frac{\sqrt{k^2+2k}}{k+1} - \frac{1}{k+1} \cdot \frac{\sqrt{k^2-1}}{k} =k2+2kk(k+1)−k2−1k(k+1)= \frac{\sqrt{k^2+2k}}{k(k+1)} - \frac{\sqrt{k^2-1}}{k(k+1)} =k2+2k−k2−1k(k+1)= \frac{\sqrt{k^2+2k} - \sqrt{k^2-1}}{k(k+1)}

This is exactly AkA_k. So, the general term of the sum is: sin⁡−1(Ak)=sin⁡−1(1k)−sin⁡−1(1k+1)\sin^{-1}(A_k) = \sin^{-1}\left(\frac{1}{k}\right) - \sin^{-1}\left(\frac{1}{k+1}\right).

Let's check the validity of the sin⁡−1\sin^{-1} difference formula. For k≥1k \ge 1, x=1/k∈(0,1]x=1/k \in (0, 1] and y=1/(k+1)∈(0,1/2]y=1/(k+1) \in (0, 1/2]. Both xx and yy are in [0,1][0,1]. The values sin⁡−1(1/k)\sin^{-1}(1/k) are in (0,π/2](0, \pi/2] and sin⁡−1(1/(k+1))\sin^{-1}(1/(k+1)) are in (0,π/2](0, \pi/2]. Since 1/k>1/(k+1)1/k > 1/(k+1), sin⁡−1(1/k)>sin⁡−1(1/(k+1))\sin^{-1}(1/k) > \sin^{-1}(1/(k+1)). The difference sin⁡−1(1/k)−sin⁡−1(1/(k+1))\sin^{-1}(1/k) - \sin^{-1}(1/(k+1)) will be in (0,π/2](0, \pi/2]. For example, for k=1k=1, sin⁡−1(1)−sin⁡−1(1/2)=π/2−π/6=π/3\sin^{-1}(1) - \sin^{-1}(1/2) = \pi/2 - \pi/6 = \pi/3. Since π/3∈[−π/2,π/2]\pi/3 \in [-\pi/2, \pi/2], the formula is valid for all terms in the sum.

Now, we can write the sum SnS_n as a telescoping series: Sn=∑k=1n[sin⁡−1(1k)−sin⁡−1(1k+1)]S_n = \sum_{k=1}^{n} \left[\sin^{-1}\left(\frac{1}{k}\right) - \sin^{-1}\left(\frac{1}{k+1}\right)\right] Sn=(sin⁡−1(1)−sin⁡−1(12))+(sin⁡−1(12)−sin⁡−1(13))+⋯+(sin⁡−1(1n)−sin⁡−1(1n+1))S_n = \left(\sin^{-1}(1) - \sin^{-1}\left(\frac{1}{2}\right)\right) + \left(\sin^{-1}\left(\frac{1}{2}\right) - \sin^{-1}\left(\frac{1}{3}\right)\right) + \dots + \left(\sin^{-1}\left(\frac{1}{n}\right) - \sin^{-1}\left(\frac{1}{n+1}\right)\right) All intermediate terms cancel out, leaving: Sn=sin⁡−1(1)−sin⁡−1(1n+1)S_n = \sin^{-1}(1) - \sin^{-1}\left(\frac{1}{n+1}\right) Since sin⁡−1(1)=π2\sin^{-1}(1) = \frac{\pi}{2}, we have: Sn=π2−sin⁡−1(1n+1)S_n = \frac{\pi}{2} - \sin^{-1}\left(\frac{1}{n+1}\right)

We need to find the value of 100cos⁡(S99)100\cos(S_{99}). First, calculate S99S_{99}: S99=π2−sin⁡−1(199+1)S_{99} = \frac{\pi}{2} - \sin^{-1}\left(\frac{1}{99+1}\right) S99=π2−sin⁡−1(1100)S_{99} = \frac{\pi}{2} - \sin^{-1}\left(\frac{1}{100}\right)

Now, calculate cos⁡(S99)\cos(S_{99}): cos⁡(S99)=cos⁡(π2−sin⁡−1(1100))\cos(S_{99}) = \cos\left(\frac{\pi}{2} - \sin^{-1}\left(\frac{1}{100}\right)\right) Using the trigonometric identity cos⁡(π2−θ)=sin⁡(θ)\cos\left(\frac{\pi}{2} - \theta\right) = \sin(\theta), with θ=sin⁡−1(1100)\theta = \sin^{-1}\left(\frac{1}{100}\right): cos⁡(S99)=sin⁡(sin⁡−1(1100))\cos(S_{99}) = \sin\left(\sin^{-1}\left(\frac{1}{100}\right)\right) cos⁡(S99)=1100\cos(S_{99}) = \frac{1}{100}

Finally, calculate 100cos⁡(S99)100\cos(S_{99}): 100cos⁡(S99)=100⋅1100=1100\cos(S_{99}) = 100 \cdot \frac{1}{100} = 1

The final answer is 1\boxed{1}.

Explanation of the solution:

  1. Simplify the general term: The term inside the sin⁡−1\sin^{-1} function is simplified by rationalizing the denominator involving square roots. (2k+1)k(k+1)(k(k+2)+(k+1)(k−1))=k2+2k−k2−1k(k+1)\frac{(2k+1)}{k(k+1)(\sqrt{k(k+2)} + \sqrt{(k+1)(k-1)})} = \frac{\sqrt{k^2+2k} - \sqrt{k^2-1}}{k(k+1)}.
  2. Recognize the inverse trigonometric identity: The simplified term is identified as x1−y2−y1−x2x\sqrt{1-y^2} - y\sqrt{1-x^2} where x=1kx = \frac{1}{k} and y=1k+1y = \frac{1}{k+1}. This corresponds to the expansion of sin⁡−1(x)−sin⁡−1(y)\sin^{-1}(x) - \sin^{-1}(y). So, sin⁡−1[(2k+1)k(k+1)(k(k+2)+(k+1)(k−1)]=sin⁡−1(1k)−sin⁡−1(1k+1)\sin^{-1}\left[\frac{(2k+1)}{k(k+1)(\sqrt{k(k+2)} + \sqrt{(k+1)(k-1)}}\right] = \sin^{-1}\left(\frac{1}{k}\right) - \sin^{-1}\left(\frac{1}{k+1}\right).
  3. Form a telescoping sum: The sum SnS_n becomes a telescoping series, where most terms cancel out. Sn=∑k=1n[sin⁡−1(1k)−sin⁡−1(1k+1)]=sin⁡−1(1)−sin⁡−1(1n+1)S_n = \sum_{k=1}^{n} \left[\sin^{-1}\left(\frac{1}{k}\right) - \sin^{-1}\left(\frac{1}{k+1}\right)\right] = \sin^{-1}(1) - \sin^{-1}\left(\frac{1}{n+1}\right).
  4. Evaluate S99S_{99}: Substitute n=99n=99 into the expression for SnS_n. S99=sin⁡−1(1)−sin⁡−1(1100)=π2−sin⁡−1(1100)S_{99} = \sin^{-1}(1) - \sin^{-1}\left(\frac{1}{100}\right) = \frac{\pi}{2} - \sin^{-1}\left(\frac{1}{100}\right).
  5. Calculate 100cos⁡(S99)100\cos(S_{99}): Use the identity cos⁡(π2−θ)=sin⁡(θ)\cos(\frac{\pi}{2} - \theta) = \sin(\theta). 100cos⁡(S99)=100cos⁡(π2−sin⁡−1(1100))=100sin⁡(sin⁡−1(1100))=100⋅1100=1100\cos(S_{99}) = 100\cos\left(\frac{\pi}{2} - \sin^{-1}\left(\frac{1}{100}\right)\right) = 100\sin\left(\sin^{-1}\left(\frac{1}{100}\right)\right) = 100 \cdot \frac{1}{100} = 1.