find an expression for sin (5 theta) as a fifth degree polynomial in the variable sin(theta) ?

One way -- do the algebra involved in the following:

sin 5x = sin 3x cos 2x + sin 2x cos 3x
where
sin 3x = sin 2x cos x + cos 2x sin x
cos 3x = cos 2x cos x - sin 2x sin x
cos 2x = 1 - 2 sin^2 x
sin 2x = 2 sin x cos x
and cos^2 x = 1 - sin^2 x

substitute these up the line, simplify and get:

sin 5 x =16 sin^5 x - 20sin^3 x +5sin x

Another way -- combine Demoivre and the binomial theorem :

sin^5 theta is the imaginary part of ( cos theta + i sin theta)^5

(a+bi)^5 = a^5 +5a^4bi -10a^3b^2 -10a^2b^3i +5ab^4 +b^5i etc

Well, aren't you a math aficionado asking such a question! Okay, let's dive right into it. To find an expression for sin(5θ) as a fifth degree polynomial in the variable sin(θ), let's use the good old power-reducing identity for sin(5θ):

sin(5θ) = 5sin(θ) - 20sin³(θ) + 16sin⁵(θ)

Ta-da! There you have it. sin(5θ) as a fifth degree polynomial in the variable sin(θ). Now go out there and rock those trigonometric functions like a mathematician with style!

To express sin(5 theta) as a fifth-degree polynomial in sin(theta), we can use the identity for the sine of multiple angles, which is:

sin(n theta) = (sin(theta))^n * (sum from k = 0 to (n-1) of) [cos((n-2k) theta) * (sin(theta))^2k],

where ^ denotes exponentiation.

Using this identity, we can express sin(5 theta) as:

sin(5 theta) = (sin(theta))^5 * (cos(3 theta) * (sin(theta))^0 + cos(theta) * (sin(theta))^2 + cos(-theta) * (sin(theta))^4).

Simplifying the coefficients of this expression, we have:

sin(5 theta) = sin^5(theta) * cos(3 theta) + sin^3(theta) * cos(theta) + sin(theta) * cos(-theta).

To find an expression for sin(5θ) as a fifth-degree polynomial in sin(θ), we can use the multiple angle formula for sine:

sin(nθ) = (sin(θ))^n * C(n,0) * cos^(n-0)(θ) * (-1)^0 + (sin(θ))^(n-2) * C(n,2) * cos^(n-2)(θ) * (-1)^1 + (sin(θ))^(n-4) * C(n,4) * cos^(n-4)(θ) * (-1)^2 + ... + (sin(θ))^0 * C(n,n) * cos^(n-n)(θ) * (-1)^(n/2)

In this case, n = 5, so we can plug it into the formula and simplify:

sin(5θ) = (sin(θ))^5 * 1 * cos^0(θ) * (-1)^0 + (sin(θ))^3 * 5 * cos^2(θ) * (-1)^1 + (sin(θ))^1 * 10 * cos^4(θ) * (-1)^2

= (sin(θ))^5 + 5*(sin(θ))^3 * cos^2(θ) - 10*(sin(θ))*^1 * cos^4(θ)

Thus, sin(5θ) as a fifth-degree polynomial in sin(θ) is:

sin(5θ) = (sin(θ))^5 + 5*(sin(θ))^3 * (1 - (sin(θ))^2) - 10*(sin(θ))^1 * (1 - (sin(θ))^2)^2

This expression is a fifth-degree polynomial in sin(θ). Note that you can further expand and simplify this expression if desired.