Complementary angles and power identities

This les­son col­lects a set of results that turn up again and again in prob­lems. Com­ple­men­tary angles (adding to $90°$) and sup­ple­men­tary angles (adding to $180°$) give quick sim­pli­fi­ca­tions and neat sums. Iden­ti­ties for $\sin^4\theta + \cos^4\theta$ and $\sin^6\theta + \cos^6\theta$ reduce higher pow­ers. The inequal­ity between the arith­metic and geo­met­ric means gives the least val­ues of expres­sions such as $\tan^2 x + \cot^2 x$. Two small alge­bra tools, the square root of a square and the rules of pro­por­tion, com­plete the set.

Com­ple­men­tary angles#

Result. If $A$ and $B$ are com­ple­men­tary, that is, $A + B = 90°$, then

$$\sin A = \cos B, \quad \cos A = \sin B, \quad \tan A = \cot B, \quad \cot A = \tan B, \quad \sec A = \csc B, \quad \csc A = \sec B$$

and also

$$\sin^2 A + \sin^2 B = 1, \quad \cos^2 A + \cos^2 B = 1, \quad \tan A \tan B = 1, \quad \cot A \cot B = 1$$

Proof. $B = 90° - A$, so $\sin A = \sin(90° - B) = \cos B$, and the oth­ers are sim­i­lar. Then

$$\sin^2 A + \sin^2 B = \cos^2 B + \sin^2 B = 1$$

and $\tan A \tan B = \cot B \tan B = 1$.

For exam­ple: $\sin 81° = \cos 9°$, $\cos 70° = \sin 20°$, $\sin 85° = \cos 5°$, $\cos 89° = \sin 1°$, $\tan 80° = \cot 10°$, $\cot 80° = \tan 10°$.

Exam­ple 1#

Find $\sin^2 5° + \sin^2 10° + \sin^2 15° + \cdots + \sin^2 85°$.

The angles go up in steps of $5°$ from $5°$ to $85°$, which is $17$ terms. Pair them from the two ends, each pair adding to $90°$:

$$(5°, 85°), \ (10°, 80°), \ (15°, 75°), \ (20°, 70°), \ (25°, 65°), \ (30°, 60°), \ (35°, 55°), \ (40°, 50°)$$

That is $8$ pairs, with $45°$ left in the mid­dle. Each pair gives $\sin^2 A + \sin^2 B = 1$. So the sum is

$$\displaystyle 8(1) + \sin^2 45° = 8 + \left(\frac{1}{\sqrt{2}}\right)^2 = 8 + \frac{1}{2} = \frac{17}{2}$$

Sup­ple­men­tary angles#

Result. If $A$ and $B$ are sup­ple­men­tary, that is, $A + B = 180°$, then

$$\cos A + \cos B = 0 \qquad \text{and} \qquad \tan A + \tan B = 0$$

Proof. $B = 180° - A$, so $\cos A + \cos B = \cos A + \cos(180° - A) = \cos A - \cos A = 0$. In the same way $\tan A + \tan(180° - A) = \tan A - \tan A = 0$.

(For sines the oppo­site hap­pens: $\sin A = \sin B$.)

Note. In a cyclic quadri­lat­eral $ABCD$ (one whose four ver­tices lie on a cir­cle), oppo­site angles are sup­ple­men­tary: $A + C = 180°$ and $B + D = 180°$.

Exam­ple 2#

If $ABCD$ is a cyclic quadri­lat­eral, find $\cos A + \cos B + \cos C + \cos D$.

$$\cos A + \cos B + \cos(180° - A) + \cos(180° - B) = \cos A + \cos B - \cos A - \cos B = 0$$

Iden­ti­ties for higher pow­ers#

Results.

1. $\sin^2\theta + \cos^4\theta = \cos^2\theta + \sin^4\theta = 1 - \sin^2\theta\cos^2\theta$
2. $\sin^4\theta + \cos^4\theta = 1 - 2\sin^2\theta\cos^2\theta$
3. $\sin^6\theta + \cos^6\theta = 1 - 3\sin^2\theta\cos^2\theta$

Proof of 1.

$$\begin{aligned} \sin^2\theta + \cos^4\theta &= \sin^2\theta + \cos^2\theta \cdot \cos^2\theta \\ &= \sin^2\theta + \cos^2\theta(1 - \sin^2\theta) \\ &= \sin^2\theta + \cos^2\theta - \sin^2\theta\cos^2\theta \\ &= 1 - \sin^2\theta\cos^2\theta \end{aligned}$$

Proof of 2. Use $a^2 + b^2 = (a + b)^2 - 2ab$ with $a = \sin^2\theta$, $b = \cos^2\theta$:

$$\sin^4\theta + \cos^4\theta = (\sin^2\theta + \cos^2\theta)^2 - 2\sin^2\theta\cos^2\theta = 1 - 2\sin^2\theta\cos^2\theta$$

Proof of 3. Use $a^3 + b^3 = (a + b)^3 - 3ab(a + b)$ with the same $a$ and $b$. Since $a + b = 1$, this gives $1 - 3\sin^2\theta\cos^2\theta$.

Exam­ple 3#

Check result 2 at $\theta = 30°$.

Directly: $\displaystyle \sin^4 30° + \cos^4 30° = \frac{1}{16} + \frac{9}{16} = \frac{10}{16} = \frac{5}{8}$.

By the for­mula: $\displaystyle 1 - 2 \cdot \frac{1}{4} \cdot \frac{3}{4} = 1 - \frac{3}{8} = \frac{5}{8}$.

Least val­ues from A.M. and G.M.#

Note. For pos­i­tive num­bers, the arith­metic mean is at least the geo­met­ric mean, which is at least the har­monic mean: $\text{A.M.} \geq \text{G.M.} \geq \text{H.M.}$ For two pos­i­tive num­bers $a$ and $b$, $\displaystyle \frac{a + b}{2} \geq \sqrt{ab}$, with equal­ity only when $a = b$.

Result. If $x > 0$ then $\displaystyle x + \frac{1}{x} \geq 2$. If $x < 0$ then $\displaystyle x + \frac{1}{x} \leq -2$.

Proof. For $x > 0$, take the pos­i­tive num­bers $x$ and $\displaystyle \frac{1}{x}$. A.M. $\geq$ G.M. gives

$$\displaystyle \frac{x + \frac{1}{x}}{2} \geq \sqrt{x \cdot \frac{1}{x}} = 1 \implies x + \frac{1}{x} \geq 2$$

For $x < 0$, the num­bers $-x$ and $\displaystyle -\frac{1}{x}$ are pos­i­tive, so

$$\displaystyle \frac{-x - \frac{1}{x}}{2} \geq \sqrt{(-x)\left(-\frac{1}{x}\right)} = 1 \implies -x - \frac{1}{x} \geq 2 \implies x + \frac{1}{x} \leq -2$$

Result. For every $x$ where both terms are defined,

$$\sin^2 x + \csc^2 x \geq 2, \qquad \cos^2 x + \sec^2 x \geq 2, \qquad \tan^2 x + \cot^2 x \geq 2$$

Proof. $\sin^2 x$ and $\csc^2 x$ are pos­i­tive, and $\sin^2 x \cdot \csc^2 x = 1$. So

$$\displaystyle \frac{\sin^2 x + \csc^2 x}{2} \geq \sqrt{\sin^2 x \cdot \csc^2 x} = 1 \implies \sin^2 x + \csc^2 x \geq 2$$

The other two are proved the same way. So the least value of $\tan^2 x + \cot^2 x$ is $2$ (reached when $\tan^2 x = 1$, for exam­ple at $x = 45°$).

The square root of a square#

Note. For any real $x$, $\sqrt{x^2} = |x|$: it equals $x$ if $x \geq 0$ and $-x$ if $x < 0$.

Exam­ple 4#

Sim­plify $\sqrt{1 - \sin^2 100°}$.

$$\sqrt{1 - \sin^2 100°} = \sqrt{\cos^2 100°} = |\cos 100°| = -\cos 100°$$

because $100°$ is in the sec­ond quad­rant, so $\cos 100° < 0$. Writ­ing $\cos 100°$ as the answer would give a neg­a­tive num­ber for a square root, which is wrong.

Ratio and pro­por­tion#

Two rules about equal ratios will be used later.

1. If $\displaystyle \frac{a}{b} = \frac{c}{d}$, then $\displaystyle \frac{a}{b} = \frac{c}{d} = \frac{al + cm}{bl + dm}$ for any $l$, $m$ (with $bl + dm \neq 0$).
2. Com­po­nendo and div­i­dendo. If $\displaystyle \frac{a}{b} = \frac{c}{d}$, then $\displaystyle \frac{a + b}{a - b} = \frac{c + d}{c - d}$, or equally $\displaystyle \frac{a - b}{a + b} = \frac{c - d}{c + d}$.

For exam­ple, if $\displaystyle \frac{x}{y} = \frac{5}{3}$ then $\displaystyle \frac{x + y}{x - y} = \frac{5 + 3}{5 - 3} = 4$.

Prac­tice#

1. Write $\sin 72°$ as a cosine.
2. Find $\tan 65° \tan 25°$.
3. Find $\cos^2 20° + \cos^2 70°$.
4. Find $\sin^2 10° + \sin^2 20° + \cdots + \sin^2 80°$.
5. Find $\cos^2 1° + \cos^2 2° + \cdots + \cos^2 89°$.
6. Show that $\cos 40° + \cos 140° = 0$.
7. $ABCD$ is a cyclic quadri­lat­eral. Find $\tan A + \tan C$.
8. Find $\sin^4\theta + \cos^4\theta$ at $\theta = 30°$ using the for­mula.
9. Find $\sin^6 45° + \cos^6 45°$ in two ways.
10. Find the least value of $\sec^2 x + \cos^2 x$.
11. Sim­plify $\sqrt{1 - \cos^2 200°}$.
12. If $\displaystyle \frac{a}{b} = \frac{7}{2}$, find $\displaystyle \frac{a + b}{a - b}$.

Answers#