Trigonometric Formulae Confusion

Question

Confused among determining A & B in these trigonometric formulae..

$$\begin{align} 2 \sin A \cos B &= \sin (A+B)+ \sin (A-B)\\ 2 \cos A \sin B &= \sin (A+B)- \sin (A-B) \end{align}$$

I've got

$$ 2 \sin \left(\frac{x}{2}\right) \cos (nx) = \sin \left(\frac{1}{2}+n\right)x + \sin \left(\frac{1}{2}-n\right)x $$

Which is correct from the first formula..

But in my book it is given as,

$$ 2 \sin \left(\frac{x}{2}\right) \cos (nx)= \sin \left(n+\frac{1}{2}\right)x - \sin \left(n-\frac{1}{2}\right)x $$

Which is also correct...

then Why such a difference? Is there any method to determine A & B in these formulas

Answer 1

SO I guess the crucial point is to show that

$$\sin\left(\frac12-n\right)x~=~-\sin\left(n-\frac12\right)x$$

One could simply apply that the sine function is a odd function and therefore satisfies the relation

$$\sin(-x)=-\sin(x)$$

Wich becomes the above equation with $x=n-\frac12$. Another way is to use the addition theorem for the sine function, which is given by

$$\sin(A-B)~=~\sin A\cos B-\sin B\cos A$$

So we get

$$\begin{align} \sin\left(\frac12-n\right)x~&=~-\sin\left(n-\frac12\right)x\\ \sin\left(\frac{x}2\right)\cos\left(nx\right) -\sin\left(nx\right)\cos\left(\frac{x}2\right) ~&=~-\left(\sin\left(nx\right)\cos\left(\frac{x}2\right)- -\sin\left(\frac{x}2\right)\cos\left(nx\right)\right)\\ &=~\sin\left(\frac{x}2\right)\cos\left(nx\right)- \sin\left(nx\right)\cos\left(\frac{x}2\right) \end{align}$$

Answer 2

Note you've left off "$\sin$" on the second term in your original problem statement (typo).

The two are the same formula. It's just because $\sin \theta = -\sin(-\theta)$ since $\sin$ is an odd function. In particular, since $a-b=-(b-a)$, you can write $\sin(a-b)=-\sin(-(b-a))$.

Answer 3

Note via this graph that: $$\sin(x)\equiv-\sin(-x)$$

Hence: $$\sin[(\frac12 -n)x]\equiv-\sin[(n-\frac12)x]$$