Question with Trigonometric functions

**holaboo** · May 2nd 2012, 05:45 AM

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?

**Prove It** · May 2nd 2012, 05:48 AM

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 ><

**holaboo** · May 2nd 2012, 05:50 AM

oh sorry, all the x are X, not times I'm not sure how to insert mathematic symbols...

**holaboo** · May 2nd 2012, 05:50 AM

Sin(2n+1)x . cosx - cos(2n+1)x . sinx = sin(2n+1)x - x = sin 2nx

**Prove It** · May 2nd 2012, 05:55 AM

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

**holaboo** · May 2nd 2012, 05:59 AM

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?

**Prove It** · May 2nd 2012, 06:14 AM

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.

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