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Trigonometric Functions and Equations question

2019 · Shift 2 · Q19
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Trigonometric Functions and Equations question

2019 · Shift 2 · Q19

JEE AdvancedMathematicsTrigonometric Functions and EquationsMultiple correct+4 / −1
For non-negative integers n, let f(n)=∑k=0nsin⁡(k+1n+2π)sin⁡(k+2n+2π)∑k=0nsin⁡2(k+1n+2π)f(n) = {{\sum\limits_{k = 0}^n {\sin \left( {{{k + 1} \over {n + 2}}\pi } \right)} \sin \left( {{{k + 2} \over {n + 2}}\pi } \right)} \over {\sum\limits_{k = 0}^n {{{\sin }^2}\left( {{{k + 1} \over {n + 2}}\pi } \right)} }}f(n)=k=0∑n​sin2(n+2k+1​π)k=0∑n​sin(n+2k+1​π)sin(n+2k+2​π)​ Assuming cos −1-1−1 x takes values in [0, π\piπ], which of the following options is/are correct?
  1. A
    If α\alphaα = tan(cos −-− 1 f(6)), then α\alphaα 2 + 2 α−\alpha -α− 1 = 0
  2. B
    f(4)=32f(4) = {{\sqrt 3 } \over 2}f(4)=23​​
  3. C
    sin(7 cos −-− 1 f(5)) = 0
  4. D
    lim⁡n→∞ f(n)=12\mathop {\lim }\limits_{n \to \infty } \,f(n) = {1 \over 2}n→∞lim​f(n)=21​
View written solutionFree

Correct answer: A, B, C

  1. Rewrite the sum in a cleaner form

Let

θ=πn+2,xk=(k+1)θ(k=0,1,…,n).\theta = \frac{\pi}{n+2}, \qquad x_k=(k+1)\theta \quad (k=0,1,\dots,n).θ=n+2π​,xk​=(k+1)θ(k=0,1,…,n).

Then

f(n)=∑k=0nsin⁡xk sin⁡(xk+θ)∑k=0nsin⁡2xk.f(n)=\frac{\sum_{k=0}^n \sin x_k\,\sin(x_k+\theta)}{\sum_{k=0}^n \sin^2 x_k}.f(n)=∑k=0n​sin2xk​∑k=0n​sinxk​sin(xk​+θ)​.

Since xk=(k+1)θx_k=(k+1)\thetaxk​=(k+1)θ, this becomes

f(n)=∑j=1n+1sin⁡(jθ)sin⁡((j+1)θ)∑j=1n+1sin⁡2(jθ).f(n)=\frac{\sum_{j=1}^{n+1} \sin(j\theta)\sin((j+1)\theta)}{\sum_{j=1}^{n+1}\sin^2(j\theta)}.f(n)=∑j=1n+1​sin2(jθ)∑j=1n+1​sin(jθ)sin((j+1)θ)​.

Also, because (n+2)θ=π(n+2)\theta=\pi(n+2)θ=π, we have the standard identities

∑j=1n+1sin⁡2(jθ)=n+22.\sum_{j=1}^{n+1}\sin^2(j\theta)=\frac{n+2}{2}.j=1∑n+1​sin2(jθ)=2n+2​.

We now simplify the numerator.

  1. Simplify the numerator

Use

sin⁡Asin⁡B=12[cos⁡(A−B)−cos⁡(A+B)].\sin A\sin B=\frac{1}{2}[\cos(A-B)-\cos(A+B)].sinAsinB=21​[cos(A−B)−cos(A+B)].

So,

sin⁡(jθ)sin⁡((j+1)θ)=12(cos⁡θ−cos⁡((2j+1)θ)).\sin(j\theta)\sin((j+1)\theta)=\frac12\big(\cos\theta-\cos((2j+1)\theta)\big).sin(jθ)sin((j+1)θ)=21​(cosθ−cos((2j+1)θ)).

Hence

∑j=1n+1sin⁡(jθ)sin⁡((j+1)θ)=12((n+1)cos⁡θ−∑j=1n+1cos⁡((2j+1)θ)).\sum_{j=1}^{n+1}\sin(j\theta)\sin((j+1)\theta) =\frac12\left((n+1)\cos\theta-\sum_{j=1}^{n+1}\cos((2j+1)\theta)\right).j=1∑n+1​sin(jθ)sin((j+1)θ)=21​((n+1)cosθ−j=1∑n+1​cos((2j+1)θ)).

Now,

∑j=1n+1cos⁡((2j+1)θ)=cos⁡3θ+cos⁡5θ+⋯+cos⁡((2n+3)θ).\sum_{j=1}^{n+1}\cos((2j+1)\theta)=\cos3\theta+\cos5\theta+\cdots+\cos((2n+3)\theta).j=1∑n+1​cos((2j+1)θ)=cos3θ+cos5θ+⋯+cos((2n+3)θ).

This is an AP in angle with first term 3θ3\theta3θ, common difference 2θ2\theta2θ, and n+1n+1n+1 terms. Its sum is

sin⁡((n+1)θ)sin⁡θcos⁡((n+3)θ).\frac{\sin((n+1)\theta)}{\sin\theta}\cos((n+3)\theta).sinθsin((n+1)θ)​cos((n+3)θ).

Since

(n+2)θ=π  ⟹  (n+1)θ=π−θ,(n+2)\theta=\pi \implies (n+1)\theta=\pi-\theta,(n+2)θ=π⟹(n+1)θ=π−θ,

we get

sin⁡((n+1)θ)=sin⁡θ,\sin((n+1)\theta)=\sin\theta,sin((n+1)θ)=sinθ,

so the sum becomes

cos⁡((n+3)θ)=cos⁡(π+θ)=−cos⁡θ.\cos((n+3)\theta)=\cos(\pi+\theta)=-\cos\theta.cos((n+3)θ)=cos(π+θ)=−cosθ.

Therefore numerator

=12((n+1)cos⁡θ−(−cos⁡θ))=12(n+2)cos⁡θ.=\frac12\big((n+1)\cos\theta-(-\cos\theta)\big) =\frac12(n+2)\cos\theta.=21​((n+1)cosθ−(−cosθ))=21​(n+2)cosθ.

Thus

f(n)=12(n+2)cos⁡θ12(n+2)=cos⁡θ=cos⁡(πn+2).f(n)=\frac{\frac12(n+2)\cos\theta}{\frac12(n+2)}=\cos\theta=\cos\left(\frac{\pi}{n+2}\right).f(n)=21​(n+2)21​(n+2)cosθ​=cosθ=cos(n+2π​).

So the key result is

f(n)=cos⁡(πn+2).\boxed{f(n)=\cos\left(\frac{\pi}{n+2}\right)}.f(n)=cos(n+2π​)​.
  1. Check option A

For n=6n=6n=6,

f(6)=cos⁡(π8).f(6)=\cos\left(\frac{\pi}{8}\right).f(6)=cos(8π​).

Since cos⁡−1\cos^{-1}cos−1 takes values in [0,π][0,\pi][0,π],

cos⁡−1(f(6))=cos⁡−1(cos⁡π8)=π8.\cos^{-1}(f(6))=\cos^{-1}\left(\cos\frac\pi8\right)=\frac\pi8.cos−1(f(6))=cos−1(cos8π​)=8π​.

Therefore

α=tan⁡(cos⁡−1f(6))=tan⁡π8=2−1.\alpha=\tan\left(\cos^{-1}f(6)\right)=\tan\frac\pi8=\sqrt2-1.α=tan(cos−1f(6))=tan8π​=2​−1.

Now check:

α2+2α−1=(2−1)2+2(2−1)−1.\alpha^2+2\alpha-1=(\sqrt2-1)^2+2(\sqrt2-1)-1.α2+2α−1=(2​−1)2+2(2​−1)−1.

Compute:

(2−1)2=2+1−22=3−22.(\sqrt2-1)^2=2+1-2\sqrt2=3-2\sqrt2.(2​−1)2=2+1−22​=3−22​.

So

α2+2α−1=(3−22)+(22−2)−1=0.\alpha^2+2\alpha-1=(3-2\sqrt2)+(2\sqrt2-2)-1=0.α2+2α−1=(3−22​)+(22​−2)−1=0.

Hence A is correct.

  1. Check option B

For n=4n=4n=4,

f(4)=cos⁡(π6)=32.f(4)=\cos\left(\frac{\pi}{6}\right)=\frac{\sqrt3}{2}.f(4)=cos(6π​)=23​​.

Hence B is correct.

  1. Check option C

For n=5n=5n=5,

f(5)=cos⁡(π7).f(5)=\cos\left(\frac{\pi}{7}\right).f(5)=cos(7π​).

Thus

cos⁡−1(f(5))=π7.\cos^{-1}(f(5))=\frac\pi7.cos−1(f(5))=7π​.

So

sin⁡(7cos⁡−1f(5))=sin⁡(7⋅π7)=sin⁡π=0.\sin\big(7\cos^{-1}f(5)\big)=\sin\left(7\cdot\frac\pi7\right)=\sin\pi=0.sin(7cos−1f(5))=sin(7⋅7π​)=sinπ=0.

Hence C is correct.

  1. Check option D

We have

f(n)=cos⁡(πn+2).f(n)=\cos\left(\frac{\pi}{n+2}\right).f(n)=cos(n+2π​).

As n→∞n\to\inftyn→∞,

πn+2→0,\frac{\pi}{n+2}\to 0,n+2π​→0,

therefore

lim⁡n→∞f(n)=cos⁡0=1.\lim_{n\to\infty} f(n)=\cos 0=1.n→∞lim​f(n)=cos0=1.

So the claim that the limit is 12\frac1221​ is false. Hence D is incorrect.

  1. Final conclusion

The correct options are

A, B, C.\boxed{A,\ B,\ C}.A, B, C​.

This matches the stored correct answer.

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