Elaborate Solution for Term Independent of x in Binomial Expansion

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1. Key Concept: The Binomial Theorem and General Term

The Binomial Theorem states that for any positive integer $n$, the expansion of $(a+b)^n$ is given by: $(a+b)^n = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^r$ The general term, or the $(r+1)^{th}$ term, in the expansion of $(a+b)^n$ is given by: $T_{r+1} = \binom{n}{r} a^{n-r} b^r$ A term is said to be "independent of $x$" if the power of $x$ in that term is zero. This means we need to find the value of $r$ for which the $x$ terms cancel out.

2. Step-by-Step Working

2.1. Finding the General Term ($T_{r+1}$)

Given the expression ${\left( {{x \over {\cos \theta }} + {1 \over {x\sin \theta }}} \right)^{16}}$, we can identify $a = \frac{x}{\cos \theta}$, $b = \frac{1}{x\sin \theta}$, and $n=16$.

Using the general term formula $T_{r+1} = \binom{n}{r} a^{n-r} b^r$: $T_{r+1} = \binom{16}{r} \left( \frac{x}{\cos \theta} \right)^{16-r} \left( \frac{1}{x\sin \theta} \right)^r$

Now, we separate the terms involving $x$ from the terms involving $\theta$: $T_{r+1} = \binom{16}{r} \frac{x^{16-r}}{(\cos \theta)^{16-r}} \frac{1}{x^r (\sin \theta)^r}$ $T_{r+1} = \binom{16}{r} x^{16-r-r} \frac{1}{(\cos \theta)^{16-r} (\sin \theta)^r}$ $T_{r+1} = \binom{16}{r} x^{16-2r} \frac{1}{(\cos \theta)^{16-r} (\sin \theta)^r}$ Explanation: We applied the binomial theorem to find the general term. The powers of $x$ are combined to simplify the expression, and the trigonometric terms are grouped together.

2.2. Finding the Term Independent of $x$

For the term to be independent of $x$, the exponent of $x$ must be zero. So, we set $16-2r = 0$. $16 - 2r = 0 \implies 2r = 16 \implies r = 8$

Since $r=8$ is a valid integer between $0$ and $16$, there is a term independent of $x$. This term is the $T_{8+1} = T_9$ term.

Substitute $r=8$ back into the expression for $T_{r+1}$: $T_9 = \binom{16}{8} x^{16-2(8)} \frac{1}{(\cos \theta)^{16-8} (\sin \theta)^8}$ $T_9 = \binom{16}{8} x^0 \frac{1}{(\cos \theta)^8 (\sin \theta)^8}$ $T_9 = \binom{16}{8} \frac{1}{(\cos \theta \sin \theta)^8}$

We know the identity $\sin 2\theta = 2 \sin \theta \cos \theta$, which means $\sin \theta \cos \theta = \frac{\sin 2\theta}{2}$. Substituting this into the expression for $T_9$: $T_9 = \binom{16}{8} \frac{1}{\left( \frac{\sin 2\theta}{2} \right)^8}$ $T_9 = \binom{16}{8} \frac{1}{\frac{(\sin 2\theta)^8}{2^8}}$ $T_9 = \binom{16}{8} \frac{2^8}{(\sin 2\theta)^8}$ Explanation: By setting the power of $x$ to zero, we found the specific value of $r$ that yields the term independent of $x$. Then, we simplified the expression using the double-angle identity for sine to make it easier to analyze its least value.

2.3. Analyzing for Least Value

The term independent of $x$ is $T_9 = \binom{16}{8} \frac{2^8}{(\sin 2\theta)^8}$. Since $\binom{16}{8}$ and $2^8$ are positive constants, to find the least value of $T_9$, we need to maximize the denominator $(\sin 2\theta)^8$. This is because for a fixed positive numerator, a fraction is minimized when its denominator is maximized. Since the power is an even number (8), $(\sin 2\theta)^8$ will always be non-negative. Maximizing $(\sin 2\theta)^8$ is equivalent to maximizing $|\sin 2\theta|$. As $\sin 2\theta$ can be positive, we aim to maximize $\sin 2\theta$ itself. The maximum value of $\sin y$ is $1$.

2.4. Calculating $\ell_1$

We are given the range for $\theta$: $\frac{\pi}{8} \le \theta \le \frac{\pi}{4}$. To find the range for $2\theta$, we multiply the inequality by 2: $2 \times \frac{\pi}{8} \le 2\theta \le 2 \times \frac{\pi}{4}$ $\frac{\pi}{4} \le 2\theta \le \frac{\pi}{2}$ In the interval $\left[ \frac{\pi}{4}, \frac{\pi}{2} \right]$, the sine function is increasing. Therefore, $\sin 2\theta$ is maximized when $2\theta$ is at its maximum value in this interval, which is $\frac{\pi}{2}$. The maximum value of $\sin 2\theta$ in this range is $\sin\left(\frac{\pi}{2}\right) = 1$.

Substitute this maximum value into the expression for $T_9$ to find $\ell_1$: $\ell_1 = \binom{16}{8} \frac{2^8}{(1)^8}$ $\ell_1 = \binom{16}{8} 2^8$ Explanation: We determined the range of $2\theta$ based on the given range of $\theta$. By analyzing the behavior of the sine function within this specific range, we found its maximum value, which in turn helps us find the minimum value of $T_9$.

2.5. Calculating $\ell_2$

We are given the range for $\theta$: $\frac{\pi}{16} \le \theta \le \frac{\pi}{8}$. To find the range for $2\theta$, we multiply the inequality by 2: $2 \times \frac{\pi}{16} \le 2\theta \le 2 \times \frac{\pi}{8}$ $\frac{\pi}{8} \le 2\theta \le \frac{\pi}{4}$ In the interval $\left[ \frac{\pi}{8}, \frac{\pi}{4} \right]$, the sine function is also increasing. Therefore, $\sin 2\theta$ is maximized when $2\theta$ is at its maximum value in this interval, which is $\frac{\pi}{4}$. The maximum value of $\sin 2\theta$ in this range is $\sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$.

Substitute this maximum value into the expression for $T_9$ to find $\ell_2$: $\ell_2 = \binom{16}{8} \frac{2^8}{\left( \frac{1}{\sqrt{2}} \right)^8}$ $\ell_2 = \binom{16}{8} \frac{2^8}{\frac{1}{2^{8/2}}}$ $\ell_2 = \binom{16}{8} \frac{2^8}{\frac{1}{2^4}}$ $\ell_2 = \binom{16}{8} 2^8 \times 2^4$ $\ell_2 = \binom{16}{8} 2^{12}$ Explanation: Similar to calculating $\ell_1$, we established the range for $2\theta$ and identified the maximum value of $\sin 2\theta$ within this new range to calculate $\ell_2$.

2.6. Finding the Ratio $\ell_2 : \ell_1$

Now we find the ratio $\frac{\ell_2}{\ell_1}$: $\frac{\ell_2}{\ell_1} = \frac{\binom{16}{8} 2^{12}}{\binom{16}{8} 2^8}$ $\frac{\ell_2}{\ell_1} = \frac{2^{12}}{2^8}$ $\frac{\ell_2}{\ell_1} = 2^{12-8}$ $\frac{\ell_2}{\ell_1} = 2^4$ $\frac{\ell_2}{\ell_1} = 16$ So, the ratio $\ell_2 : \ell_1$ is $16:1$.

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3. Tips and Common Mistakes to Avoid

• Trigonometric Ranges: Always carefully determine the range of the argument for the trigonometric function ($2\theta$ in this case) based on the given range of $\theta$. A common mistake is to consider only the range of $\theta$ itself.
• Maximizing vs. Minimizing: When a term is in the denominator of a fraction, minimizing the fraction requires maximizing the denominator (assuming positive values). Pay close attention to powers; even powers make the base's sign irrelevant, focusing on magnitude.
• Simplification: Use trigonometric identities like $\sin 2\theta = 2\sin\theta\cos\theta$ to simplify expressions and make them easier to analyze.
• Binomial Coefficient: Remember that $\binom{n}{r}$ is a constant for a given $n$ and $r$, and does not affect the optimization process unless $r$ itself depends on the variable being optimized.

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4. Summary and Key Takeaway

This problem effectively combines concepts from the Binomial Theorem and Trigonometry. The key steps involved:

1. Deriving the general term of the binomial expansion.
2. Identifying the term independent of $x$ by setting the power of $x$ to zero.
3. Simplifying the resulting trigonometric expression.
4. Analyzing the range of the angle and the behavior of the sine function to find its maximum value within the specified intervals.
5. Using these maximum values to calculate the least values of the term independent of $x$ ($\ell_1$ and $\ell_2$).
6. Finally, calculating the required ratio.

The core takeaway is that finding the least value of an expression often involves maximizing its denominator, especially when the expression is a fraction with a constant positive numerator. Careful domain analysis for trigonometric functions is crucial.