Hi guys, I need some help with a question from the IB HL maths papers.

Not sure how this works:

Sin(2n+1)xcosx - cos(2n+1)xsinx = sin(2n+1)x - x = sin 2nx

Could some1 show me more detailed steps or give me some hints on how equation holds?

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- May 2nd 2012, 05:45 AMholabooQuestion with Trigonometric functions
Hi guys, I need some help with a question from the IB HL maths papers.

Not sure how this works:

Sin(2n+1)xcosx - cos(2n+1)xsinx = sin(2n+1)x - x = sin 2nx

Could some1 show me more detailed steps or give me some hints on how equation holds? - May 2nd 2012, 05:48 AMProve ItRe: Question with Trigonometric functions
- May 2nd 2012, 05:50 AMholabooRe: Question with Trigonometric functions
oh sorry, all the x are X, not times I'm not sure how to insert mathematic symbols...

- May 2nd 2012, 05:50 AMholabooRe: Question with Trigonometric functions
Sin(2n+1)x . cosx - cos(2n+1)x . sinx = sin(2n+1)x - x = sin 2nx

- May 2nd 2012, 05:55 AMProve ItRe: Question with Trigonometric functions
Apply $\displaystyle \displaystyle \begin{align*} \sin{\left(a - b\right)} \equiv \sin{(a)}\cos{(b)} - \cos{(a)}\sin{(b)} \end{align*}$ with $\displaystyle \displaystyle \begin{align*} a = (2n + 1)x \end{align*}$ and $\displaystyle \displaystyle \begin{align*} b = x \end{align*}$.

- May 2nd 2012, 05:59 AMholabooRe: Question with Trigonometric functions
thanks I just worked it out myself too. Is the Sin A-B formula derived from the Sin A+B formula? Could you do the same for the Cos A+B formula?

- May 2nd 2012, 06:14 AMProve ItRe: Question with Trigonometric functions
Yes, the formula for $\displaystyle \displaystyle \begin{align*} \sin{(A - B)} \end{align*}$ is derived from the $\displaystyle \displaystyle \begin{align*} \sin{(A + B)} \end{align*}$ formula, just letting $\displaystyle \displaystyle \begin{align*} B = -B \end{align*}$ and remembering that $\displaystyle \displaystyle \begin{align*} \cos{(-B)} = \cos{(B)} \end{align*}$ and $\displaystyle \displaystyle \begin{align*} \sin{(-B)} = -\sin{(B)} \end{align*}$.

And yes, you can derive a formula for $\displaystyle \displaystyle \begin{align*} \cos{(A - B)} \end{align*}$ using the $\displaystyle \displaystyle \begin{align*} \cos{(A + B)} \end{align*}$ formula using the same reasoning.