Write down the expansion of (cosx + isinx)^3. Hence, by using De Moivre's theorem, find cos3x in terms of powers of cosx.

First, we expand by using binomial expansion (it is always helpful to draw Pascal's triangle to find each term's coefficients): 1 1
1 2 1
1 3 3 1
Since the function is raised to the power of 3, we use the underlined coefficients. Therefore, the expansion will be:
(cosx + isinx)³ = 1cos³x + 3cos²x*(isinx)+ 3cosx(isinx)² + 1*(isinx)³
Since i² = -1, we can substitute the i² terms in the equation, giving:
(cosx + isinx)³ = cos³x + 3cos²x(isinx) - 3cosxsin²x - isin³x
Now that we have our expansion, we can use De Moivre's theorem to obtain the cos3x term in the question (we simply convert the power into a multiple of the angle:
(cosx + isinx)³ = cos3x + isin3x
Equating the expansion to the above equation,
cos3x + isin3x = cos³x + 3cos²x(isinx) - 3cosxsin²x - isin³x
The question only asks for the real terms, so we can simple ignore the imaginary ones, since these will be the isin3x terms:
cos3x = cos³x - 3cosxsin²x
Although the above equation seems complete, the question asked for power of cosx. Therefore, we have to find a way to replace the sin²x term with cosx powers. Remembering the identity:
sin²x + cos²x = 1
We can replace the sin term:cos3x = cos³x - 3cosx(1 - cos²x)
Expanding, we get the answer:
cos3x = 4cos³x - 3cosx