Conceptual problems: least values and angle sums

Many exam­i­na­tion prob­lems in trigonom­e­try look long but rest on one short idea. This les­son col­lects sev­eral such ideas: least val­ues from the AM-GM inequal­ity, the care­ful han­dling of square roots, and sums that col­lapse because of com­ple­men­tary, sup­ple­men­tary or related angles. In each case, spot­ting the idea turns a page of cal­cu­la­tion into a few lines.

Con­cept 1: least val­ues from AM ≥ GM#

For pos­i­tive real num­bers, the arith­metic mean is at least the geo­met­ric mean (and the geo­met­ric mean is at least the har­monic mean): $\text{AM} \ge \text{GM} \ge \text{HM}$. For two pos­i­tive num­bers $a$ and $b$,

$$\displaystyle \frac{a + b}{2} \ge \sqrt{ab} \implies a + b \ge 2\sqrt{ab}$$

with equal­ity when $a = b$. Tak­ing $\displaystyle b = \frac{1}{a}$:

$$\displaystyle x + \frac{1}{x} \ge 2 \text{ for } x > 0, \qquad x + \frac{1}{x} \le -2 \text{ for } x < 0$$

Exam­ple 1#

Show that $\sin^2 x + \csc^2 x \ge 2$, and find the least value of $3\tan^2 x + 27\cot^2 x$.

$\sin^2 x$ is pos­i­tive (where $\csc x$ is defined) and $\displaystyle \csc^2 x = \frac{1}{\sin^2 x}$, so by the result above $\sin^2 x + \csc^2 x \ge 2$. In the same way $\cos^2 x + \sec^2 x \ge 2$.

For the sec­ond, both terms are pos­i­tive and their prod­uct is $3\tan^2 x \times 27\cot^2 x = 81$, so

$$3\tan^2 x + 27\cot^2 x \ge 2\sqrt{81} = 18$$

The least value $18$ occurs when the two terms are equal, $3\tan^2 x = 27\cot^2 x$, that is, $\tan^4 x = 9$.

Con­cept 2: the square root of a square#

For any real num­ber $x$,

$$\sqrt{x^2} = |x| = \begin{cases} x, & x \ge 0 \\ -x, & x < 0 \end{cases}$$

A square root is never neg­a­tive, so when we sim­plify $\sqrt{\cos^2\theta}$ we must check the sign of $\cos\theta$.

Exam­ple 2#

Find $\sqrt{1 - \sin^2 100^\circ} \cdot \sec 100^\circ$.

$\sqrt{1 - \sin^2 100^\circ} = \sqrt{\cos^2 100^\circ} = |\cos 100^\circ|$. Since $100^\circ$ is in the sec­ond quad­rant, $\cos 100^\circ < 0$, so $|\cos 100^\circ| = -\cos 100^\circ$.

$$\displaystyle -\cos 100^\circ \times \frac{1}{\cos 100^\circ} = -1$$

Exam­ple 3#

If $\pi < \alpha < 2\pi$, sim­plify $\displaystyle \frac{1}{\sin\alpha - \sqrt{\cot^2\alpha - \cos^2\alpha}}$.

$$\displaystyle \cot^2\alpha - \cos^2\alpha = \frac{\cos^2\alpha}{\sin^2\alpha} - \cos^2\alpha = \frac{\cos^2\alpha(1 - \sin^2\alpha)}{\sin^2\alpha} = \frac{\cos^4\alpha}{\sin^2\alpha}$$

So $\displaystyle \sqrt{\cot^2\alpha - \cos^2\alpha} = \frac{\cos^2\alpha}{|\sin\alpha|}$. For $\pi < \alpha < 2\pi$, $\sin\alpha < 0$, so $|\sin\alpha| = -\sin\alpha$, and the denom­i­na­tor becomes

$$\displaystyle \sin\alpha + \frac{\cos^2\alpha}{\sin\alpha} = \frac{\sin^2\alpha + \cos^2\alpha}{\sin\alpha} = \frac{1}{\sin\alpha}$$

The expres­sion equals $\sin\alpha$.

Exam­ple 4#

If $\displaystyle \sqrt{\frac{1 + \sin A}{1 - \sin A}} = \sec A + \tan A$, in which quad­rants can $A$ lie?

Mul­ti­ply top and bot­tom inside the root by $1 + \sin A$:

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

(The numer­a­tor $1 + \sin A$ is never neg­a­tive.) This equals $\displaystyle \frac{1 + \sin A}{\cos A} = \sec A + \tan A$ only when $|\cos A| = \cos A$, that is, $\cos A > 0$. So $A$ lies in the first or fourth quad­rant.

Con­cept 3: com­ple­men­tary angles#

If $A + B = 90^\circ$, then $\sin B = \cos A$, so $\sin^2 A + \sin^2 B = \sin^2 A + \cos^2 A = 1$. Like­wise $\tan A \tan B = \tan A \cot A = 1$. This makes long sums col­lapse in pairs.

Exam­ple 5#

Find $\sin^2 5^\circ + \sin^2 10^\circ + \sin^2 15^\circ + \dots + \sin^2 90^\circ$.

There are $18$ terms. Pair them: $(5^\circ, 85^\circ)$, $(10^\circ, 80^\circ)$, ..., $(40^\circ, 50^\circ)$ make $8$ pairs, each adding to $1$. Left over are $\displaystyle \sin^2 45^\circ = \frac{1}{2}$ and $\sin^2 90^\circ = 1$.

$$\displaystyle \text{Sum} = 8 + \frac{1}{2} + 1 = 9\frac{1}{2}$$

In the same way $\displaystyle \cos^2 5^\circ + \cos^2 10^\circ + \dots + \cos^2 90^\circ = 8 + \frac{1}{2} + 0 = 8\frac{1}{2}$.

Extend­ing to $180^\circ$, the angles from $95^\circ$ to $175^\circ$ repeat the val­ues from $85^\circ$ down to $5^\circ$, so $\displaystyle \sin^2 5^\circ + \dots + \sin^2 180^\circ = 8\frac{1}{2} + 1 + 8\frac{1}{2} + 0 = 18$, and sim­i­larly the cosine sum to $180^\circ$ is $18$. Up to $360^\circ$ each sum is $36$. (A quick check: the sine and cosine sums over the same angles must add to the num­ber of terms, since each pair $\sin^2\theta + \cos^2\theta$ is $1$.)

Con­cept 4: sup­ple­men­tary and related angles#

If $A + B = 180^\circ$, then $\sin A = \sin B$, $\cos A = -\cos B$ and $\tan A = -\tan B$. So $\cos A + \cos B = 0$ and $\tan A + \tan B = 0$.

In any quadri­lat­eral the angles add to $360^\circ$; in a cyclic quadri­lat­eral oppo­site angles are sup­ple­men­tary, $A + C = 180^\circ$ and $B + D = 180^\circ$. Hence $\cos A + \cos B + \cos C + \cos D = 0$ for a cyclic quadri­lat­eral.

If $A - B = 180^\circ$, then $A = 180^\circ + B$, so $\sin A + \sin B = 0$, $\cos A + \cos B = 0$ and $\tan A - \tan B = 0$.

If $A + B = 360^\circ$, then $\sin A = -\sin B$, $\cos A = \cos B$ and $\tan A = -\tan B$, so $\sin A + \sin B = 0$, $\cos A - \cos B = 0$ and $\tan A + \tan B = 0$.

Exam­ple 6#

Find $\displaystyle \cos\frac{\pi}{7} + \cos\frac{2\pi}{7} + \cos\frac{3\pi}{7} + \cos\frac{4\pi}{7} + \cos\frac{5\pi}{7} + \cos\frac{6\pi}{7}$.

The angles pair up to $\pi$: $\displaystyle \frac{\pi}{7} + \frac{6\pi}{7} = \pi$, $\displaystyle \frac{2\pi}{7} + \frac{5\pi}{7} = \pi$, $\displaystyle \frac{3\pi}{7} + \frac{4\pi}{7} = \pi$. Each pair of cosines adds to $0$, so the sum is $0$.

Exam­ple 7#

Find $\sin 24^\circ + \sin 32^\circ + \sin 204^\circ + \sin 212^\circ$.

$$\sin 204^\circ = \sin(180^\circ + 24^\circ) = -\sin 24^\circ, \qquad \sin 212^\circ = \sin(180^\circ + 32^\circ) = -\sin 32^\circ$$

So the sum is $\sin 24^\circ + \sin 32^\circ - \sin 24^\circ - \sin 32^\circ = 0$.

Prac­tice#

1. Find the least value of $4\tan^2 x + 9\cot^2 x$.
2. Find the least value of $\cos^2 x + \sec^2 x$.
3. Find $\sqrt{1 - \cos^2 200^\circ} \cdot \csc 200^\circ$.
4. If $\displaystyle \sqrt{\frac{1 - \sin A}{1 + \sin A}} = \sec A - \tan A$, in which quad­rants can $A$ lie?
5. Find $\sin^2 10^\circ + \sin^2 20^\circ + \dots + \sin^2 90^\circ$.
6. Find $\cos^2 1^\circ + \cos^2 2^\circ + \dots + \cos^2 89^\circ$.
7. Find $\tan 1^\circ \tan 2^\circ \tan 3^\circ \cdots \tan 89^\circ$.
8. Find $\cos 20^\circ + \cos 40^\circ + \cos 60^\circ + \dots + \cos 160^\circ$.
9. Find $\sin 10^\circ + \sin 20^\circ + \sin 190^\circ + \sin 200^\circ$.

Answers#