# Question with Trigonometric functions

• May 2nd 2012, 06:45 AM
holaboo
Question 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, 06:48 AM
Prove It
Re: Question with Trigonometric functions
Quote:

Originally Posted by holaboo
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?

Hang on, is that x in the middle of each term a times or an \displaystyle \begin{align*} x \end{align*}? Some spacing and brackets would help ><
• May 2nd 2012, 06:50 AM
holaboo
Re: 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, 06:50 AM
holaboo
Re: Question with Trigonometric functions
Sin(2n+1)x . cosx - cos(2n+1)x . sinx = sin(2n+1)x - x = sin 2nx
• May 2nd 2012, 06:55 AM
Prove It
Re: Question with Trigonometric functions
Apply \displaystyle \begin{align*} \sin{\left(a - b\right)} \equiv \sin{(a)}\cos{(b)} - \cos{(a)}\sin{(b)} \end{align*} with \displaystyle \begin{align*} a = (2n + 1)x \end{align*} and \displaystyle \begin{align*} b = x \end{align*}.
• May 2nd 2012, 06:59 AM
holaboo
Re: 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, 07:14 AM
Prove It
Re: Question with Trigonometric functions
Quote:

Originally Posted by holaboo
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?

Yes, the formula for \displaystyle \begin{align*} \sin{(A - B)} \end{align*} is derived from the \displaystyle \begin{align*} \sin{(A + B)} \end{align*} formula, just letting \displaystyle \begin{align*} B = -B \end{align*} and remembering that \displaystyle \begin{align*} \cos{(-B)} = \cos{(B)} \end{align*} and \displaystyle \begin{align*} \sin{(-B)} = -\sin{(B)} \end{align*}.

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