Trigonometric equations and inequalities

Proof

1. 1
   $$\dato{a}{-\frac{\pi}{2} \leq x \leq \frac{\pi}{2} \rightarrow \left( \sin x = c \leftrightarrow x = \alpha \right)}$$

   Let $\alpha = \arcsin c$, which lies in $\icc{-\frac{\pi}{2}}{\frac{\pi}{2}}$ and satisfies $\sin \alpha = c$. Since the sine is strictly increasing on that interval, it does not take the same value twice there: $\alpha$ is the only solution in the first half of the turn. On the unit circle, $P(\alpha)$ is the point of the right half at height $c$.

2. 2
   $$\sin x = \resaltar{\sin(\pi - x)} = c \leftrightarrow \resaltar{\pi - x = \alpha} \leftrightarrow \dato{b}{x = \pi - \alpha}$$

   If $x$ lies in $\ioo{\frac{\pi}{2}}{\frac{3\pi}{2}}$, the angle $\pi - x$ lies in $\ioo{-\frac{\pi}{2}}{\frac{\pi}{2}}$ and has the same sine as $x$. By the preceding step, $x$ is a solution exactly when $\pi - x = \alpha$: the horizontal line at height $c$ cuts the left half at the symmetric point $P(\pi - \alpha)$.

3. 3
   $$\sin x = \resaltar{\sin x_0} = c \leftrightarrow \dato{g}{\resaltar{x = \alpha + 2k\pi \vee x = \pi - \alpha + 2k\pi}}$$

   Now let $x$ be any real number. By the lemma, with $p = 2\pi$ and $a = -\frac{\pi}{2}$, we write $x = x_0 + 2k\pi$ with $x_0$ in the turn studied, and by periodicity $\sin x = \sin x_0$. Therefore, $x$ is a solution if and only if $x_0$ is $\alpha$ or $\pi - \alpha$.

4. 4
   $$\alpha = \pm \frac{\pi}{2} \rightarrow \resaltar{\pi - \alpha = \alpha} \vee \resaltar{\pi - \alpha = \alpha + 2\pi}$$

   Conversely, every number of those two families is a solution, again by periodicity. On the graph of the sine, the line $y = c$ cuts the curve at two points in each period. If $c = 1$ or $c = -1$, the two families coincide, because $\pi - \alpha$ differs from $\alpha$ by a multiple of $2\pi$, and there is a single intersection point per period.

5. 5
   $$\frac{\pi}{2} - x = \resaltar{\frac{\pi}{2} - \arccos c} + 2k\pi \vee \frac{\pi}{2} - x = \resaltar{\frac{\pi}{2} + \arccos c} + 2k\pi$$

   For the cosine we use the symmetry of complementary angles, $\cos x = \sin\left( \frac{\pi}{2} - x \right)$, and the theorem on complementary angles from the lesson on the inverse functions, $\arcsin c = \frac{\pi}{2} - \arccos c$. Applying the preceding step to the angle $\frac{\pi}{2} - x$ yields two families.

6. 6
   $$\cos x = c \leftrightarrow \resaltar{x = \pm \arccos c + 2k\pi}$$

   Solving for $x$, the families are $\arccos c - 2k\pi$ and $-\arccos c - 2k\pi$; since $-k$ ranges over the integers when $k$ does, they are those of the statement. On the circle, the points $P(\beta)$ and $P(-\beta)$, with $\beta = \arccos c$, are the two intersections of the vertical line with abscissa $c$.

7. 7
   $$\tan x = \resaltar{\tan x_0} = c \leftrightarrow \resaltar{x = \arctan c + k\pi}$$

   For the tangent, with period $\pi$, the lemma is applied with $p = \pi$ and $a = -\frac{\pi}{2}$: if $x = x_0 + k\pi$ with $x_0$ in $\ioo{-\frac{\pi}{2}}{\frac{\pi}{2}}$, where the tangent is a bijection onto $\mathbb{R}$, the only solution is $x_0 = \arctan c$; the point $-\frac{\pi}{2}$ does not count, because the tangent is not defined there. If $\abs{c} > 1$, neither the sine nor the cosine reaches the value $c$, because their values lie in $\icc{-1}{1}$.

Proof

1. 1
   $$\dato{u}{x_1 + 2j\pi \in \ico{a}{a + 2\pi} \leftrightarrow j = -k}$$

   Let $x_1$ be a real number. By the lemma, with $p = 2\pi$, there is a unique integer $k$ such that $x_1 - 2k\pi$ lies in $\ico{a}{a + 2\pi}$: the family of the numbers $x_1 + 2j\pi$, with $j$ an integer, has exactly one member in the turn.

2. 2
   $$\alpha + 2k\pi = \pi - \alpha + 2j\pi \leftrightarrow \resaltar{\alpha = \frac{\pi}{2} + (j - k)\pi} \leftrightarrow \alpha = \pm \frac{\pi}{2} \leftrightarrow \resaltar{c = \pm 1}$$

   Let $\alpha = \arcsin c$, in $\icc{-\frac{\pi}{2}}{\frac{\pi}{2}}$. If the two sine families have a common member, they are the same family, and this happens exactly when $\alpha$ is $\frac{\pi}{2}$ or $-\frac{\pi}{2}$, that is, when $c = \pm 1$. Therefore, if $\abs{c} < 1$ the two families contribute two distinct solutions to the turn, and if $\abs{c} = 1$, only one.

3. 3
   $$\beta + 2k\pi = -\beta + 2j\pi \leftrightarrow \resaltar{\beta = (j - k)\pi} \leftrightarrow \beta \in \left\{ 0, \pi \right\} \leftrightarrow \resaltar{c = \pm 1}$$

   Let $\beta = \arccos c$, in $\icc{0}{\pi}$. The two cosine families have a common member exactly when $\beta$ is an integer multiple of $\pi$, that is, when $\beta$ is $0$ or $\pi$, which is equivalent to $c = \pm 1$; the count is the same as for the sine.

4. 4
   $$-\frac{\pi}{2} < \arctan c < \frac{\pi}{2} \rightarrow \resaltar{\arctan c + k\pi \neq \frac{\pi}{2} + j\pi}$$

   For the tangent there is a single family, with spacing $\pi$, and the lemma, with $p = \pi$, gives it exactly one member in $\ico{a}{a + \pi}$. That member is a solution, because $\arctan c + k\pi$ is never of the form $\frac{\pi}{2} + j\pi$, where the tangent is not defined: $\arctan c$ lies in $\ioo{-\frac{\pi}{2}}{\frac{\pi}{2}}$.