Certainly! Let us solve each part of the given problem step-by-step.
### Part (a): Expand [tex]\(\sin(x + 2x)\)[/tex]
To expand [tex]\(\sin(x + 2x)\)[/tex], we use the angle addition formula for sine:
[tex]\[
\sin(a + b) = \sin(a)\cos(b) + \cos(a)\sin(b)
\][/tex]
Here, let [tex]\(a = x\)[/tex] and [tex]\(b = 2x\)[/tex]. Plugging these into the formula, we get:
[tex]\[
\sin(x + 2x) = \sin(x)\cos(2x) + \cos(x)\sin(2x)
\][/tex]
Next, we need to use the double angle formulas for [tex]\(\cos(2x)\)[/tex] and [tex]\(\sin(2x)\)[/tex]:
[tex]\[
\cos(2x) = 1 - 2\sin^2(x) \quad \text{and} \quad \sin(2x) = 2\sin(x)\cos(x)
\][/tex]
Substituting these back into our expression, we have:
[tex]\[
\sin(x + 2x) = \sin(x) [1 - 2\sin^2(x)] + \cos(x) [2\sin(x)\cos(x)]
\][/tex]
Simplifying each term, we get:
[tex]\[
\sin(x + 2x) = \sin(x) - 2\sin^3(x) + 2\sin(x)\cos^2(x)
\][/tex]
Notice that [tex]\(\cos^2(x) = 1 - \sin^2(x)\)[/tex]. Thus,
[tex]\[
2\sin(x)\cos^2(x) = 2\sin(x)(1 - \sin^2(x)) = 2\sin(x) - 2\sin^3(x)
\][/tex]
So, we can rewrite our expression as:
[tex]\[
\sin(x + 2x) = \sin(x) - 2\sin^3(x) + 2\sin(x) - 2\sin^3(x)
\][/tex]
Combine like terms:
[tex]\[
\sin(x + 2x) = 3\sin(x) - 4\sin^3(x)
\][/tex]
Therefore, we have:
[tex]\[
\sin(x + 2x) = 3\sin(x) - 4\sin^3(x)
\][/tex]
### Part (b): Express [tex]\(\sin(3x)\)[/tex] in terms of [tex]\(\sin x\)[/tex]
Since [tex]\(\sin(3x)\)[/tex] is simply [tex]\(\sin(x + 2x)\)[/tex], we can directly use our result from part (a):
[tex]\[
\sin(3x) = 3\sin(x) - 4\sin^3(x)
\][/tex]
Thus, the expression for [tex]\(\sin(3x)\)[/tex] in terms of [tex]\(\sin(x)\)[/tex] is:
[tex]\[
\boxed{\sin(3x) = 3\sin(x) - 4\sin^3(x)}
\][/tex]
And there it is, the step-by-step solution to the problem!
