Question

Rewrite with only sin x and cos x.

sin 2x - cos x

Answer 1

Seems easy! What am I missing?

Answer 2

well sin(2x) = 2sin(x)cos(x)

substitute this and then perhaps factor the expression

Answer 3

if you need an expression in terms of an angle of a single x, then do what campbell says. use that rule.

If you have any "why" questions, or any other question, don't hesitate to ask.

Answer 4

I just don't understand how to substitute and factor when I'm only given sin2x and -cosx :(

Answer 5

lets do this again.
You have: \(\Large\color{black}{ \sin(2x)-\cos(x) }\)

and there is a rule,
\(\Large\color{blue}{ \sin(\color{red}{a}+\color{magenta}{b})=\sin(\color{red}{a})\cos(b)+\sin(\color{magenta}{b})\cos(\color{red}{a}) }\)

Answer 6

in your situation however,
\(\Large\color{black}{ \sin(2x)}\) is same as \(\Large\color{black}{ \sin(x+x)}\)
and that means that your a and b are the same, they are both x.

this is what you get,

\(\Large\color{blue}{ \color{black}{ \sin(2x)}= \sin(\color{red}{x}+\color{magenta}{x})=\sin(\color{red}{x})\cos(x)+\sin(\color{magenta}{x})\cos(\color{red}{x}) }\)

Answer 7

and this means that you have \(\Large\color{blue}{ \sin(\color{red}{x})\cos(x)+\sin(\color{magenta}{x})\cos(\color{red}{x}) }\)

which would be \(\Large\color{blue}{ 2\sin(\color{red}{x})\cos(x) }\)
just like
\(\Large\color{blue}{ C+C=2C }\)

Answer 8

So, so far, do you understand why
\(\Large\color{blue}{ \sin(2x)=2\sin(x)\cos(x) }\)

Answer 9

Yeah, I follow! Thank you for the explanation haha

Answer 10

So, when you have
\(\Large\color{blue}{ \sin(2x)-\cos(x) }\)
it becomes,
\(\Large\color{blue}{2 \sin(x) \cos(x)-\cos(x) }\).