Trigonometric Sum-to-Product and Product-to-Sum Identities

Lesson · Intermediate

Trigonometry

\sin\alpha+\sin\beta = 2\sin\left(\frac{\alpha+\beta}{2}\right) \cos\left(\frac{\alpha-\beta}{2}\right)

\sin\alpha-\sin\beta = 2\cos\left(\frac{\alpha+\beta}{2}\right) \sin\left(\frac{\alpha-\beta}{2}\right)

\cos\alpha+\cos\beta = 2\cos\left(\frac{\alpha+\beta}{2}\right) \cos\left(\frac{\alpha-\beta}{2}\right)

\cos\alpha-\cos\beta = -2\sin\left(\frac{\alpha+\beta}{2}\right) \sin\left(\frac{\alpha-\beta}{2}\right)

\sin p\cos q = \frac12 \left( \sin(p+q)+\sin(p-q) \right)

\cos p\cos q = \frac12 \left( \cos(p+q)+\cos(p-q) \right)

\sin p\sin q = \frac12 \left( \cos(p-q)-\cos(p+q) \right)

\sin\alpha+\sin\beta = 2\sin\left(\frac{\alpha+\beta}{2}\right) \cos\left(\frac{\alpha-\beta}{2}\right)

\sin\alpha-\sin\beta = 2\cos\left(\frac{\alpha+\beta}{2}\right) \sin\left(\frac{\alpha-\beta}{2}\right)

\cos\alpha+\cos\beta = 2\cos\left(\frac{\alpha+\beta}{2}\right) \cos\left(\frac{\alpha-\beta}{2}\right)

\cos\alpha-\cos\beta = -2\sin\left(\frac{\alpha+\beta}{2}\right) \sin\left(\frac{\alpha-\beta}{2}\right)

\sin(x+y) = \sin x\cos y+\cos x\sin y

and

\sin(x-y) = \sin x\cos y-\cos x\sin y.

Adding these equations gives

\sin(x+y)+\sin(x-y) = 2\sin x\cos y.

Let

x=\frac{\alpha+\beta}{2}, \qquad y=\frac{\alpha-\beta}{2}.

Then $x+y=\alpha$ and $x-y=\beta$ . Therefore,

\sin\alpha+\sin\beta = 2\sin\left(\frac{\alpha+\beta}{2}\right) \cos\left(\frac{\alpha-\beta}{2}\right).

Replacing $\beta$ by $-\beta$ gives

\sin\alpha-\sin\beta = 2\cos\left(\frac{\alpha+\beta}{2}\right) \sin\left(\frac{\alpha-\beta}{2}\right).

Also,

\cos(x+y) = \cos x\cos y-\sin x\sin y

and

\cos(x-y) = \cos x\cos y+\sin x\sin y.

Adding gives

\cos(x-y)+\cos(x+y) = 2\cos x\cos y.

Using the same substitutions,

\cos\alpha+\cos\beta = 2\cos\left(\frac{\alpha+\beta}{2}\right) \cos\left(\frac{\alpha-\beta}{2}\right).

Subtracting the two cosine equations gives

\cos(x+y)-\cos(x-y) = -2\sin x\sin y.

Therefore,

\cos\alpha-\cos\beta = -2\sin\left(\frac{\alpha+\beta}{2}\right) \sin\left(\frac{\alpha-\beta}{2}\right).

\sin p\cos q = \frac12 \left( \sin(p+q)+\sin(p-q) \right)

\cos p\cos q = \frac12 \left( \cos(p+q)+\cos(p-q) \right)

\sin p\sin q = \frac12 \left( \cos(p-q)-\cos(p+q) \right)

\sin x+\sin2x+\sin3x+\cdots+\sin nx = \frac{ \sin\left(\frac{nx}{2}\right) \sin\left(\frac{(n+1)x}{2}\right) }{ \sin\left(\frac{x}{2}\right) }, \qquad \sin\left(\frac{x}{2}\right)\ne0.

2\sin\frac{x}{2}\sin kx = \cos\left(kx-\frac{x}{2}\right) - \cos\left(kx+\frac{x}{2}\right)

= \cos\left(\frac{(2k-1)x}{2}\right) - \cos\left(\frac{(2k+1)x}{2}\right).

\begin{aligned} 2\sin\frac{x}{2} \sum_{k=1}^{n}\sin kx ={}& \cos\left(\frac{(2n-1)x}{2}\right) - \cos\left(\frac{(2n+1)x}{2}\right) \\[8pt] &+ \cos\left(\frac{(2n-3)x}{2}\right) - \cos\left(\frac{(2n-1)x}{2}\right) \\[4pt] &\qquad\vdots \\[4pt] &+ \cos\left(\frac{3x}{2}\right) - \cos\left(\frac{5x}{2}\right) \\[8pt] &+ \cos\left(\frac{x}{2}\right) - \cos\left(\frac{3x}{2}\right) \\[8pt] &+ \cos\left(-\frac{x}{2}\right) - \cos\left(\frac{x}{2}\right). \end{aligned}

2\sin\frac{x}{2} \sum_{k=1}^{n}\sin kx = \cos\left(-\frac{x}{2}\right) - \cos\left(\frac{(2n+1)x}{2}\right)

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

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

2\sin\frac{x}{2} \sum_{k=1}^{n}\sin kx = 2 \sin\left(\frac{nx}{2}\right) \sin\left(\frac{(n+1)x}{2}\right).

\boxed{ \sum_{k=1}^{n}\sin(kx) = \frac{ \sin\left(\frac{nx}{2}\right) \sin\left(\frac{(n+1)x}{2}\right) }{ \sin\left(\frac{x}{2}\right) } }