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Composition of Trigonometric Functions and Inverse Trigonometric Functions

Let $ A - n =...

Question

Let $A_{n} = 2 n \sum r = 1 sin ({sin}^{- 1} x^{3 r - 2}), B_{n} = 2 n \sum r = 1 cos ({cos}^{- 1} x^{3 r - 1}), C_{n} = 2 n \sum r = 1 tan ({tan}^{- 1} x^{3 r}), n \in N, n \geq 3$ .

Which of the following option(s) is/are correct for $| x | \leq 1$ ?

A

A_{n} > C_{n} > B_{n}

for

x \in (- 1, 1)

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B

A_{n} = B_{n} = C_{n}

has two integral roots.

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C

A_{n} = B_{n} + C_{n}

has two irrational roots.

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D

B_{n} = C_{n} + A_{n}

has

6 n

complex roots.

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Solution

The correct options are
C $A_{n} = B_{n} + C_{n}$ has two irrational roots.
D $B_{n} = C_{n} + A_{n}$ has $6 n$ complex roots.
$A_{n} = 2 n \sum r = 1 sin ({sin}^{- 1} x^{3 r - 2}) \Rightarrow A_{n} = 2 n \sum r = 1 x^{3 r - 2} \Rightarrow A_{n} = x + x^{4} + x^{7} + \dots 2 n terms \Rightarrow A_{n} = \frac{x (1 - x^{6 n})}{1 - x^{3}}$

$B_{n} = 2 n \sum r = 1 cos ({cos}^{- 1} x^{3 r - 1}) \Rightarrow B_{n} = 2 n \sum r = 1 x^{3 r - 1} \Rightarrow B_{n} = x^{2} + x^{5} + x^{8} + \dots 2 n terms \Rightarrow B_{n} = \frac{x^{2} (1 - x^{6 n})}{1 - x^{3}}$

$C_{n} = 2 n \sum r = 1 tan ({tan}^{- 1} x^{3 r}) \Rightarrow C_{n} = 2 n \sum r = 1 x^{3 r} \Rightarrow C_{n} = x^{3} + x^{6} + x^{9} + \dots 2 n terms \Rightarrow C_{n} = \frac{x^{3} (1 - x^{6 n})}{1 - x^{3}}$

When $x = 0$
$A_{n} = B_{n} = C_{n} = 0$
When $x = - 1$
$A_{n} = B_{n} = C_{n} = 0$
When $x = 1$
$A_{n} = B_{n} = C_{n} = 2 n$
Therefore, $A_{n} = B_{n} = C_{n}$ has $3$ distinct integral roots.

For $0 < x < 1$ ,
$x^{3} < x^{2} < x$
Therefore, $A_{n} > B_{n} > C_{n}$

For $- 1 < x < 0$ ,
$x < x^{3} < x^{2}$
Therefore, $B_{n} > C_{n} > A_{n}$

$A_{n} = B_{n} + C_{n} \Rightarrow \frac{x (1 - x^{6 n})}{1 - x^{3}} = \frac{x^{2} (1 - x^{6 n})}{1 - x^{3}} + \frac{x^{3} (1 - x^{6 n})}{1 - x^{3}} \Rightarrow \frac{(1 - x^{6 n})}{1 - x^{3}} x [x^{2} + x - 1] = 0 \Rightarrow x = 0, - 1, \frac{- 1 \pm \sqrt{5}}{2}$
Therefore, two irrational roots.

$B_{n} = A_{n} + C_{n} \Rightarrow \frac{x^{2} (1 - x^{6 n})}{1 - x^{3}} = \frac{x (1 - x^{6 n})}{1 - x^{3}} + \frac{x^{3} (1 - x^{6 n})}{1 - x^{3}} \Rightarrow \frac{(1 - x^{6 n})}{1 - x^{3}} x [x^{2} - x + 1] = 0 \Rightarrow x^{6 n} = 1; x = 0; x = \frac{1 \pm i \sqrt{3}}{2}$

Therefore, the number of complex roots will be,
$(6 n - 2) + 2 = 6 n$

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