Question

If \(\mathrm{A}=\sin ^{20} \theta+\cos ^{48} \theta\) then (a) \(\quad A \geq 1\) (b) \(0<\mathrm{A} \leq 1\) (c) \(1<\mathrm{A}<3\) (d) \(A \geq 3\)

Short answer

Answer: (c) \(1 < A < 3\)

Step-by-step solution

Examine the individual terms

As noted before, both \(\sin^{20}\theta\) and \(\cos^{48}\theta\) must lie between 0 and 1. The minimum value of both terms will naturally occur when either \(\sin\theta\) or \(\cos\theta\) is 0.

Find the minimum and maximum values

Since both \(\sin^{20}\theta\) and \(\cos^{48}\theta\) are even powers, their minimum value occurs when either \(\sin\theta = 0\) or \(\cos\theta = 0\). Therefore, the minimum value for each individual term is 0. The maximum value of either term will occur when the other term is 0. So, when \(\sin\theta = 1\) or \(\cos\theta = 1\), then either \(\sin^{20}\theta = 1\) or \(\cos^{48}\theta = 1\). Thus, the individual terms can have minimum and maximum values of 0 and 1, respectively.

Determine the range of A

Since both individual terms can lie between 0 and 1, the possible range for A will be the sum of the minimum and maximum values for each term: Minimum of A: \(0 + 0 = 0\) Maximum of A: \(1 + 1 = 2\)

Match the range of A to the given intervals

Now we can observe which of the given intervals matches the range we found in Step 3: (a) \(A \geq 1\) -- This interval doesn't include 0, our minimum value. (b) \(0 < A \leq 1\) -- This interval includes our maximum value but not our minimum value. (c) \(1 < A < 3\) -- This interval starts from 1 and goes up to 3 (excluding), which includes our entire range of A. (d) \(A \geq 3\) -- This interval doesn't include any part of our range. So, the answer is (c) \(1< A < 3\).

Key concepts

Understanding Inequalities

Inequalities allow us to compare the sizes or values of expressions. In this exercise, we use inequalities to determine the possible range of the expression \( A = \sin^{20}\theta + \cos^{48}\theta \). We want to know where \( A \) falls within given intervals.

Let's break down common symbols in inequalities:

• \( \leq \) means "less than or equal to."
• \( \geq \) means "greater than or equal to."
• \( < \) means "less than."
• \( > \) means "greater than."

In our case, we discover that \( A \) must be greater than 1 but less than 3. This understanding helps us conclude that choice (c) is correct. It's because \( A \) can take any value between 1 and 2 non-inclusive, which matches the condition \(1 < A < 3\).

Inequalities are crucial in determining boundaries in mathematics, especially when working with functions that vary over different values.

Power and Properties of Even Powers

When dealing with trigonometric functions raised to an even power, there are specific patterns that help simplify the problem. Both \( \sin^{20}\theta \) and \( \cos^{48}\theta \) are examples of even powers.

A key property of any term raised to an even power is that the result is always non-negative. This is because squaring a number, whether it's positive or negative, always yields a positive outcome. For example, \( (-1)^2 = 1 \) like \( 1^2 = 1 \).

Knowing this, both \( \sin^{20}\theta \) and \( \cos^{48}\theta \) will always yield results between 0 and 1. This range helps us in determining the possible outcomes for the value of \( A \).

• The minimum value will occur when one of the individual terms is 0.
• The maximum value is 1 when the corresponding base trigonometric functions equal 1.

These properties help us understand why \( A \) falls within the range provided in choice (c).

Exploring the Range of Functions

The range of a function represents all possible output values. Here, we are looking at the range of \( A = \sin^{20}\theta + \cos^{48}\theta \).

Since each trigonometric function when raised to an even power varies between 0 and 1, the summation creates a new function \( A \) with its own range. Here’s how it works:

• The smallest possible value of \( A \) is 0, which occurs when both terms are zero.
• The largest value is 2, happening when both terms equal 1.

Thus, the entire range \( [0, 2] \) is possible for \( A \). However, solving the problem requires finding the interval that includes these possibilities exclusively between 1 and 2.

This includes choice (c) \( 1 < A < 3 \), capturing the appropriate segment of the range for \( A \). Understanding ranges helps in mapping outputs in calculus and other various fields of math, making it a vital concept for grasping the behavior of complex expressions.