Answer
a. The next two terms are \( 4 \sin x \cos^2 x \) and \( 8 \sin x \cos^3 x \).
b. The \( n \)th term is \( 2^{n-1} \sin x \cos^{n-1} x \).
c. The fifth term is \( 16 \sin x \cos^4 x \).
For the geometric sequence:
a. The value of \( x \) is \( 8 \) or \( -8 \).
Solution
Let's solve the problems step by step.
### Part 1: Sequence Analysis
#### a. Write down the next two terms.
The given sequence is:
1. \( 10 \sin x \)
2. \( \sin 2x \)
3. \( 2 \cos x \sin 2x \)
To find the next two terms, we need to identify a pattern.
1. The first term is \( 10 \sin x \).
2. The second term is \( \sin 2x \), which can be expressed using the double angle formula: \( \sin 2x = 2 \sin x \cos x \).
3. The third term is \( 2 \cos x \sin 2x = 2 \cos x (2 \sin x \cos x) = 4 \sin x \cos^2 x \).
Continuing this pattern, we can derive the next terms:
- The fourth term can be derived from the third term by applying the double angle formula again:
\[
2 \cos x \sin 2x = 4 \sin x \cos^2 x
\]
The fourth term would be \( 4 \sin x \cos^2 x \).
- The fifth term can be derived similarly:
\[
2 \cos x (4 \sin x \cos^2 x) = 8 \sin x \cos^3 x
\]
Thus, the next two terms are:
1. \( 4 \sin x \cos^2 x \)
2. \( 8 \sin x \cos^3 x \)
#### b. Find the \( n \)th term of the sequence.
From the pattern observed, we can express the \( n \)th term as:
\[
T_n = 2^{n-1} \sin x \cos^{n-1} x
\]
#### c. Determine the value of the fifth term in terms of \( \sin x \).
Using the formula for the \( n \)th term:
\[
T_5 = 2^{5-1} \sin x \cos^{5-1} x = 16 \sin x \cos^4 x
\]
### Part 2: Geometric Sequence Analysis
#### a. Determine the value(s) of \( x \) in the geometric sequence \( 2: x : 32 : \ldots \).
In a geometric sequence, the ratio between consecutive terms is constant. Let the common ratio be \( r \).
From the first two terms:
\[
\frac{x}{2} = r \implies x = 2r
\]
From the second and third terms:
\[
\frac{32}{x} = r \implies x = \frac{32}{r}
\]
Now we have two expressions for \( x \):
1. \( x = 2r \)
2. \( x = \frac{32}{r} \)
Setting these equal to each other:
\[
2r = \frac{32}{r}
\]
Multiplying both sides by \( r \) (assuming \( r \neq 0 \)):
\[
2r^2 = 32 \implies r^2 = 16 \implies r = 4 \text{ or } r = -4
\]
Now substituting back to find \( x \):
1. If \( r = 4 \):
\[
x = 2 \cdot 4 = 8
\]
2. If \( r = -4 \):
\[
x = 2 \cdot (-4) = -8
\]
Thus, the values of \( x \) are \( 8 \) and \( -8 \).
### Summary of Answers:
1. Next two terms: \( 4 \sin x \cos^2 x \) and \( 8 \sin x \cos^3 x \)
2. \( n \)th term: \( T_n = 2^{n-1} \sin x \cos^{n-1} x \)
3. Fifth term: \( 16 \sin x \cos^4 x \)
4. Values of \( x \): \( 8 \) and \( -8 \) in the geometric sequence.
Reviewed and approved by the UpStudy tutoring team
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