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?
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?
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 $\displaystyle \displaystyle \begin{align*} b = x \end{align*}$.
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.