Question

Rewrite with only sin x and cos x.

cos 3x

Answer 1

\(\bf cos(3x)\implies cos(2x+x)\implies cos(2x)cos(x)-sin(2x)sin(x)\)

expand the 2x angles, check your trig identites for the "double-angle identities"

Answer 2

thanks for actually answering but its the whole double angle thing that I don't understand. like I can expand it and all but from there im lost

Answer 3

ok, so \(\bf cos(3x)\implies cos(2x+x)\implies cos(2x)cos(x)-sin(2x)sin(x)\\ \quad \\
\textit{let us use the cos(2x) of }\quad 1-2sin^2(x)\qquad thus\\ \quad \\\
[\quad 1-2sin^2(x)\quad ]cos(x)-[\quad 2sin(x)cos(x)\quad ]sin(x)\\ \quad \\\
[cos(x)-2sin^2(x)cos(x)]-[2sin^2(x)cos(x)]\implies cos(x)-4sin^2(x)cos(x)\\ \quad \\
cos(x)[1-4sin^2(x)]\implies cos(x)[1^2-4sin^2(x)]\\ \quad \\
\textit{recall that}\qquad a^2-b^2 = (a-b)(a+b)\\ \quad \\
cos(x)[1^2-2^2sin^2(x)]\implies cos(x)[1^2-(2sin(x))^2]\\ \quad \\
cos(x)[1^2-(2sin(x))^2]\implies cos(x)[1-2sin(x)][1+2sin(x)]\)

Answer 4

hmm got a bit truncated

\(\bf cos(3x)\implies cos(2x+x)\implies cos(2x)cos(x)-sin(2x)sin(x)\\ \quad \\
\textit{let us use the cos(2x) of }\quad 1-2sin^2(x)\qquad thus\\ \quad \\\
[\quad 1-2sin^2(x)\quad ]cos(x)-[\quad 2sin(x)cos(x)\quad ]sin(x)\\ \quad \\\
[cos(x)-2sin^2(x)cos(x)]-[2sin^2(x)cos(x)]\\ \quad \\\implies cos(x)-4sin^2(x)cos(x)\\ \quad \\
cos(x)[1-4sin^2(x)]\implies cos(x)[1^2-4sin^2(x)]\\ \quad \\
\textit{recall that}\qquad a^2-b^2 = (a-b)(a+b)\\ \quad \\
cos(x)[1^2-2^2sin^2(x)]\implies cos(x)[1^2-(2sin(x))^2]\\ \quad \\
cos(x)[1^2-(2sin(x))^2]\implies cos(x)[1-2sin(x)][1+2sin(x)]\)

Answer 5

is there any way you can break that down into English? O.o

Answer 6

hehe

Answer 7

\(\bf cos(3x)\implies cos(2x+x)\implies cos(2x)cos(x)-sin(2x)sin(x)\\ \quad \\
\color{red}{\textit{let us use the cos(2x) of }\quad 1-2sin^2(x)\qquad thus}\\ \quad \\\
[\quad 1-2sin^2(x)\quad ]cos(x)-[\quad 2sin(x)cos(x)\quad ]sin(x)\\ \quad \\\
[cos(x)-2sin^2(x)cos(x)]-[2sin^2(x)cos(x)]\\ \quad \\\implies cos(x)-4sin^2(x)cos(x)\\ \quad \\
cos(x)[1-4sin^2(x)]\implies cos(x)[1^2-4sin^2(x)]\\ \quad \\
\color{red}{\textit{recall that}\qquad a^2-b^2 = (a-b)(a+b)}\\ \quad \\
cos(x)[1^2-2^2sin^2(x)]\implies cos(x)[1^2-(2sin(x))^2]\\ \quad \\
cos(x)[1^2-(2sin(x))^2]\implies cos(x)[1-2sin(x)][1+2sin(x)]\)

Answer 8

is pretty straightforward, we're just using the \(1-2sin^2(x)\) identity for the cos(2x)

then we get 2 values added that give us a difference of squares, which we then expand into 2 binomials

Answer 9

ohhhh.....I think I get it a little better now. thanks!

Answer 10

yw