Square roots, signs and componendo and dividendo

Expres­sions such as $\sqrt{1 + \sin 2A}$ sim­plify neatly, but only if you take care with the sign, because a square root is never neg­a­tive. This les­son shows how to remove such square roots, how to decide the sign of $\displaystyle \cos\frac{A}{2} \pm \sin\frac{A}{2}$ from the size of the angle, and how this pro­duces exact val­ues like $4\sin 27°$. It ends with the com­po­nendo and div­i­dendo rule applied to ratios of sines and tan­gents.

Square roots of $1 \pm \sin$#

Note.

$$\sqrt{1 + \sin 2A} = |\cos A + \sin A|, \qquad \sqrt{1 - \sin 2A} = |\cos A - \sin A|$$

$$\displaystyle \sqrt{1 + \sin A} = \left|\cos\frac{A}{2} + \sin\frac{A}{2}\right|, \qquad \sqrt{1 - \sin A} = \left|\cos\frac{A}{2} - \sin\frac{A}{2}\right|$$

Rea­son. Write $1 = \cos^2 A + \sin^2 A$ and $\sin 2A = 2\sin A\cos A$:

$$\sqrt{1 + \sin 2A} = \sqrt{\cos^2 A + \sin^2 A + 2\sin A\cos A} = \sqrt{(\cos A + \sin A)^2} = |\cos A + \sin A|$$

using $\sqrt{x^2} = |x|$. The oth­ers are the same with a minus sign, or with $\displaystyle \frac{A}{2}$ in place of $A$.

The sign of $C + S$ and $C - S$#

To remove the mod­u­lus you must know the sign of the expres­sion inside. Let $\displaystyle C = \cos\frac{A}{2}$ and $\displaystyle S = \sin\frac{A}{2}$. Divide the full turn by the lines at $\displaystyle \frac{\pi}{4}$, $\displaystyle \frac{3\pi}{4}$, $\displaystyle \frac{5\pi}{4}$ and $\displaystyle \frac{7\pi}{4}$ (that is, $45°$, $135°$, $225°$, $315°$). The signs depend on which part $\displaystyle \frac{A}{2}$ falls in:

The rea­son: by the com­pound angle for­mu­lae, $\cos\theta + \sin\theta = \sqrt{2}\sin(\theta + 45°)$ and $\cos\theta - \sin\theta = \sqrt{2}\cos(\theta + 45°)$. The first is pos­i­tive when $\theta + 45°$ is between $0°$ and $180°$, that is, for $\theta$ between $-45°$ and $135°$. The sec­ond is pos­i­tive when $\theta + 45°$ is between $-90°$ and $90°$, that is, for $\theta$ between $-135°$ and $45°$ (the same as $225°$ to $405°$). For exam­ple, $\cos 10° - \sin 10° > 0$ because $10°$ is in the first row, while $\cos 130° - \sin 130° < 0$ because $130°$ is in the sec­ond row.

Exam­ple 1#

Sim­plify $\sqrt{1 + \sin 200°}$ and $\sqrt{1 + \sin 400°}$.

For $\sqrt{1 + \sin 200°}$: here $A = 200°$, so $\displaystyle \frac{A}{2} = 100°$, which lies between $45°$ and $135°$. There $C + S > 0$, so

$$\sqrt{1 + \sin 200°} = \cos 100° + \sin 100°$$

For $\sqrt{1 + \sin 400°}$: $\displaystyle \frac{A}{2} = 200°$, between $135°$ and $225°$, where $C + S < 0$. So

$$\sqrt{1 + \sin 400°} = -(\cos 200° + \sin 200°)$$

Exam­ple 2#

Sim­plify $\sqrt{1 - \sin 20°}$.

$\displaystyle \frac{A}{2} = 10°$, where $C - S > 0$. So $\sqrt{1 - \sin 20°} = \cos 10° - \sin 10°$.

An exact value: $4\sin 27°$#

Exam­ple 3#

Find the value of $4\sin 27°$.

Let $A = 54°$, so $\displaystyle \frac{A}{2} = 27°$, which lies between $-45°$ and $45°$. There both $C + S$ and $C - S$ are pos­i­tive, so

$$\displaystyle \sqrt{1 + \sin A} = \cos\frac{A}{2} + \sin\frac{A}{2} \qquad (1)$$

$$\displaystyle \sqrt{1 - \sin A} = \cos\frac{A}{2} - \sin\frac{A}{2} \qquad (2)$$

Sub­tract (2) from (1):

$$\displaystyle \sqrt{1 + \sin A} - \sqrt{1 - \sin A} = 2\sin\frac{A}{2}$$

Put $A = 54°$ and use $\displaystyle \sin 54° = \cos 36° = \frac{\sqrt{5} + 1}{4}$:

$$\displaystyle \sqrt{1 + \frac{\sqrt{5} + 1}{4}} - \sqrt{1 - \frac{\sqrt{5} + 1}{4}} = 2\sin 27°$$

$$\displaystyle \frac{\sqrt{5 + \sqrt{5}}}{2} - \frac{\sqrt{3 - \sqrt{5}}}{2} = 2\sin 27°$$

$$4\sin 27° = \sqrt{5 + \sqrt{5}} - \sqrt{3 - \sqrt{5}}$$

Com­po­nendo and div­i­dendo#

Rule. If $\displaystyle \frac{a}{b} = \frac{c}{d}$, then

$$\displaystyle \frac{a + b}{a - b} = \frac{c + d}{c - d} \qquad \text{or equally} \qquad \frac{a - b}{a + b} = \frac{c - d}{c + d}$$

Applied to trigono­met­ric ratios, it gives two use­ful results.

Note 1. If $\displaystyle \frac{x}{y} = \frac{\sin A}{\sin B}$, then $\displaystyle \frac{x + y}{x - y} = \frac{\tan\frac{A + B}{2}}{\tan\frac{A - B}{2}}$.

Proof. By com­po­nendo and div­i­dendo, then the sum-to-prod­uct for­mu­lae:

$$\displaystyle \frac{x + y}{x - y} = \frac{\sin A + \sin B}{\sin A - \sin B} = \frac{2\sin\frac{A + B}{2}\cos\frac{A - B}{2}}{2\cos\frac{A + B}{2}\sin\frac{A - B}{2}} = \frac{\tan\frac{A + B}{2}}{\tan\frac{A - B}{2}}$$

Note 2. If $\displaystyle \frac{x}{y} = \frac{\tan A}{\tan B}$, then $\displaystyle \frac{x + y}{x - y} = \frac{\sin(A + B)}{\sin(A - B)}$.

Proof.

$$\displaystyle \frac{x + y}{x - y} = \frac{\tan A + \tan B}{\tan A - \tan B} = \frac{\frac{\sin A}{\cos A} + \frac{\sin B}{\cos B}}{\frac{\sin A}{\cos A} - \frac{\sin B}{\cos B}} = \frac{\sin A\cos B + \cos A\sin B}{\sin A\cos B - \cos A\sin B} = \frac{\sin(A + B)}{\sin(A - B)}$$

Exam­ple 4#

If $\tan A = 3\tan B$, show that $\sin(A + B) = 2\sin(A - B)$.

$\displaystyle \frac{\tan A}{\tan B} = \frac{3}{1}$, so by Note 2,

$$\displaystyle \frac{\sin(A + B)}{\sin(A - B)} = \frac{3 + 1}{3 - 1} = 2$$

That is, $\sin(A + B) = 2\sin(A - B)$.

Prac­tice#

1. Find $\sqrt{1 + \sin 60°}$ exactly.
2. Find $\sqrt{1 - \sin 120°}$ exactly.
3. Is $\cos 200° + \sin 200°$ pos­i­tive or neg­a­tive?
4. Is $\cos 160° - \sin 160°$ pos­i­tive or neg­a­tive?
5. Sim­plify $\sqrt{1 + \sin 50°}$.
6. Sim­plify $\sqrt{1 - \sin 300°}$, and find its exact value.
7. Prove that $4\sin 9° = \sqrt{3 + \sqrt{5}} - \sqrt{5 - \sqrt{5}}$.
8. If $\displaystyle \frac{\sin A}{\sin B} = \frac{7}{3}$, express $\displaystyle \tan\frac{A + B}{2}$ in terms of $\displaystyle \tan\frac{A - B}{2}$.
9. If $\tan A = 5\tan B$, show that $2\sin(A + B) = 3\sin(A - B)$.
10. If $\displaystyle \frac{a}{b} = \frac{2}{5}$, find $\displaystyle \frac{a + b}{a - b}$.

Answers#