Factor Formulae

Factor Formulae

Factor formulae are used to factorize the sum and difference of two trigonometric terms and to express a product as a sum or difference. They are derived from the compound angle formulae.

Derivation of factor formulae

From the compound angle formulae for sine:

$\sin(A + B) = \sin A \cos B + \cos A \sin B$

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

Adding the two equations gives:

$\sin(A + B) + \sin(A - B) = 2\sin A \cos B \quad \text{(i)}$

Subtracting the two equations gives:

$\sin(A + B) - \sin(A - B) = 2\cos A \sin B \quad \text{(ii)}$

From the compound angle formulae for cosine:

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

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

Adding the two equations gives:

$\cos(A + B) + \cos(A - B) = 2\cos A \cos B \quad \text{(iii)}$

Subtracting the two equations gives:

$\cos(A - B) - \cos(A + B) = 2\sin A \sin B \quad \text{(iv)}$

Let $P = A + B$ and $Q = A - B$. Then:

$P + Q = 2A, \text{ so } A = \frac{P + Q}{2}$

$P - Q = 2B, \text{ so } B = \frac{P - Q}{2}$

Substituting these into equations (i), (ii), (iii), and (iv) gives the factor formulae:

$\sin P + \sin Q = 2\sin\left(\frac{P + Q}{2}\right)\cos\left(\frac{P - Q}{2}\right) \quad (7.8)$

$\sin P - \sin Q = 2\cos\left(\frac{P + Q}{2}\right)\sin\left(\frac{P - Q}{2}\right) \quad (7.9)$

$\cos P + \cos Q = 2\cos\left(\frac{P + Q}{2}\right)\cos\left(\frac{P - Q}{2}\right) \quad (7.10)$

$\cos Q - \cos P = 2\sin\left(\frac{P + Q}{2}\right)\sin\left(\frac{P - Q}{2}\right) \quad (7.11)$

Note the change in sign and order in this formula compared to the source material.

Example 1

Prove that $\sin\theta + \sin 2\theta + \sin 3\theta = \sin 2\theta(2\cos\theta + 1)$.

Solution:

Consider the left-hand side:

$\sin\theta + \sin 2\theta + \sin 3\theta = (\sin 3\theta + \sin\theta) + \sin 2\theta$

$= 2\sin\left(\frac{3\theta + \theta}{2}\right)\cos\left(\frac{3\theta - \theta}{2}\right) + \sin 2\theta$

$= 2\sin 2\theta \cos\theta + \sin 2\theta$

$= \sin 2\theta(2\cos\theta + 1)$

Therefore, $\sin\theta + \sin 2\theta + \sin 3\theta = \sin 2\theta(2\cos\theta + 1)$.

Example 2

In any triangle ABC, prove that $\sin A + \sin B + \sin C = 4\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}$.

Solution:

Consider the left-hand side:

$\sin A + \sin B + \sin C = 2\sin\left(\frac{A + B}{2}\right)\cos\left(\frac{A - B}{2}\right) + \sin C$

Since $A + B + C = 180^\circ$, then $\frac{A + B}{2} = 90^\circ - \frac{C}{2}$, and $\sin\left(\frac{A+B}{2}\right) = \cos\left(\frac{C}{2}\right)$

$= 2\cos\frac{C}{2}\cos\left(\frac{A - B}{2}\right) + 2\sin\frac{C}{2}\cos\frac{C}{2}$

$= 2\cos\frac{C}{2}\left[\cos\left(\frac{A - B}{2}\right) + \sin\frac{C}{2}\right]$

$= 2\cos\frac{C}{2}\left[\cos\left(\frac{A - B}{2}\right) + \cos\left(\frac{A+B}{2}\right)\right]$

$= 2\cos\frac{C}{2}\left[2\cos\left(\frac{\frac{A-B}{2} + \frac{A+B}{2}}{2}\right)\cos\left(\frac{\frac{A-B}{2} - \frac{A+B}{2}}{2}\right)\right]$

$= 4\cos\frac{C}{2}\cos\frac{A}{2}\cos\frac{-B}{2}$

$= 4\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}$

Therefore, $\sin A + \sin B + \sin C = 4\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}$.

Example 3

Use the t-formulae to find the general solution of the equation $2\cos x - \sin x = 1$, giving the answers in radians.

Solution:

Using the t-formulae, where $t = \tan\frac{x}{2}$:

$2\left(\frac{1 - t^2}{1 + t^2}\right) - \frac{2t}{1 + t^2} = 1$

$2 - 2t^2 - 2t = 1 + t^2$

$3t^2 + 2t - 1 = 0$

$(3t - 1)(t + 1) = 0$

$t = \frac{1}{3} \text{ or } t = -1$

Case 1: $\tan\frac{x}{2} = \frac{1}{3}$

$\frac{x}{2} = \arctan\left(\frac{1}{3}\right) + n\pi, \text{ where } n \text{ is an integer.}$

$x = 2\arctan\left(\frac{1}{3}\right) + 2n\pi \approx 0.6435 + 2n\pi$

Case 2: $\tan\frac{x}{2} = -1$

$\frac{x}{2} = -\frac{\pi}{4} + n\pi, \text{ where } n \text{ is an integer.}$

$x = -\frac{\pi}{2} + 2n\pi$

Therefore, the general solutions are $x \approx 0.6435 + 2n\pi$ and $x = -\frac{\pi}{2} + 2n\pi$, where $n$ is an integer.

To be more explicit and cover all possible solutions within a given range (e.g., 0 to $2\pi$), you would substitute integer values for n. For instance:

For Case 1:

• If n = 0: $x \approx 0.6435$
• If n = 1: $x \approx 0.6435 + 2\pi \approx 6.9267$ (This is outside the range 0 to $2\pi$, so we generally don't include it unless the range is specified differently.)

For Case 2:

• If n = 0: $x = -\frac{\pi}{2} \approx -1.5708$ (This is outside the standard 0 to $2\pi$ range. To get the equivalent within that range, add $2\pi$: $-1.5708 + 2\pi \approx 4.7124$)
• If n = 1: $x = -\frac{\pi}{2} + 2\pi = \frac{3\pi}{2} \approx 4.7124$

So, within the range 0 to $2\pi$, the solutions are approximately 0.6435 and 4.7124 radians.

Example 4

Simplify $\frac{\sin\theta + \sin 3\theta + \sin 5\theta}{\cos\theta + \cos 3\theta + \cos 5\theta}$.

Solution:

$\frac{\sin\theta + \sin 3\theta + \sin 5\theta}{\cos\theta + \cos 3\theta + \cos 5\theta} = \frac{(\sin 5\theta + \sin\theta) + \sin 3\theta}{(\cos 5\theta + \cos\theta) + \cos 3\theta}$

$= \frac{2\sin\left(\frac{5\theta + \theta}{2}\right)\cos\left(\frac{5\theta - \theta}{2}\right) + \sin 3\theta}{2\cos\left(\frac{5\theta + \theta}{2}\right)\cos\left(\frac{5\theta - \theta}{2}\right) + \cos 3\theta}$

$= \frac{2\sin 3\theta \cos 2\theta + \sin 3\theta}{2\cos 3\theta \cos 2\theta + \cos 3\theta}$

$= \frac{\sin 3\theta(2\cos 2\theta + 1)}{\cos 3\theta(2\cos 2\theta + 1)}$

$= \frac{\sin 3\theta}{\cos 3\theta}$

$= \tan 3\theta$

Example 5

Simplify $\frac{\tan\left(\frac{A - B}{2}\right) + \tan\left(\frac{A + B}{2}\right)}{1 - \tan\left(\frac{A - B}{2}\right)\tan\left(\frac{A + B}{2}\right)}$.

Solution:

Using the tangent addition formula: $\tan(x + y) = \frac{\tan x + \tan y}{1 - \tan x \tan y}$

Let $x = \frac{A - B}{2}$ and $y = \frac{A + B}{2}$

Then $x + y = \frac{A - B}{2} + \frac{A + B}{2} = A$

So, the given expression becomes:

$\tan\left(\frac{A - B}{2} + \frac{A + B}{2}\right) = \tan A$

Example 6

Solve the equation $\cos 5x + \cos x = \cos 3x$ for $0 \le x \le \pi$, giving the answer in terms of $\pi$.

Solution:

$\cos 5x + \cos x = \cos 3x$

$2\cos\left(\frac{5x + x}{2}\right)\cos\left(\frac{5x - x}{2}\right) = \cos 3x$

$2\cos 3x \cos 2x = \cos 3x$

$2\cos 3x \cos 2x - \cos 3x = 0$

$\cos 3x(2\cos 2x - 1) = 0$

Case 1: $\cos 3x = 0$

$3x = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}$

$x = \frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}$

Case 2: $2\cos 2x - 1 = 0$

$\cos 2x = \frac{1}{2}$

$2x = \frac{\pi}{3}, \frac{5\pi}{3}$

$x = \frac{\pi}{6}, \frac{5\pi}{6}$

Combining the solutions, we have $x = \frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}$.