4. Trigonometry and inverse branches

Learning goals and geometric definitions

Measure angles in radians, derive addition formulas from rotations, solve periodic equations, and distinguish reciprocal from inverse trigonometric functions. A radian angle is arc length divided by radius. On the unit circle, the point reached by oriented angle $\theta$ has coordinates $(\cos\theta,\sin\theta)$, giving $\sin^2\theta+\cos^2\theta=1$.

Sine and cosine have period $2\pi$. Define $\tan\theta=\sin\theta/\cos\theta$ where $\cos\theta\ne0$; tangent has period $\pi$. Negative angles give $\cos(-\theta)=\cos\theta$ and $\sin(-\theta)=-\sin\theta$ by reflection. The notation $\sin^{-1}x$ is often used for arcsine, not $1/\sin x$; use $\arcsin$ and $\csc$ to avoid that ambiguity.

Theorem: addition formulas

For all real $a,b$,

$$\cos(a+b)=\cos a\cos b-\sin a\sin b,\qquad \sin(a+b)=\sin a\cos b+\cos a\sin b.$$

Proof

Take as a geometric fact that a rotation about the origin preserves distances and orientation and is linear in coordinate components. Let $R_\alpha$ rotate counterclockwise through $\alpha$. On the unit circle, it sends the first and second coordinate basis vectors to

$$R_\alpha e_1=(\cos\alpha,\sin\alpha)^T,\qquad R_\alpha e_2=(-\sin\alpha,\cos\alpha)^T.$$

By linearity, its action on the point at angle $\beta$ is

$$R_\alpha(\cos\beta,\sin\beta)^T =(\cos\alpha\cos\beta-\sin\alpha\sin\beta,\ \sin\alpha\cos\beta+\cos\alpha\sin\beta)^T.$$

Geometrically, successive rotations add their angles, so this same point is $(\cos(\alpha+\beta),\sin(\alpha+\beta))^T$. Comparing coordinates proves both identities without assuming them. Setting $\beta=\alpha=\theta$ gives $\sin2\theta=2\sin\theta\cos\theta$ and $\cos2\theta=\cos^2\theta-\sin^2\theta$. An identity holds for every input in its domain; an equation asks which inputs make a particular equality true.

Inverse functions and equation solving

Restrict sine to $[-\pi/2,\pi/2]$, where it is increasing onto $[-1,1]$; its inverse is $\arcsin$. Restrict cosine to $[0,\pi]$ and tangent to $(-\pi/2,\pi/2)$ to define $\arccos$ and $\arctan$. Thus $\arcsin(\sin\theta)$ need not equal $\theta$ outside the chosen branch. By contrast $\sin(\arcsin x)=x$ for every $x\in[-1,1]$.

If $\sin\theta=c$ with $|c|\le1$ and $\alpha=\arcsin c$, all solutions are $\theta=\alpha+2k\pi$ or $\theta=\pi-\alpha+2k\pi$, $k\in\mathbb Z$. An identity holds wherever both expressions are defined; an equation holds only at selected values. Squaring or multiplying by a possibly zero trigonometric factor can change the solution set.

Proof: inverse branches and complete periodic solutions

We use the unit-circle geometric definition of sine and cosine and the already proved addition formulas. The basic geometric convention is that one counterclockwise traversal of the unit circle has angle $2\pi$; each point has a unique angle modulo $2\pi$, and each point of a specified semicircle has the corresponding angle in its stated half-turn interval. The signs of coordinates are their ordinary quadrant signs. These are geometric prerequisites, not derivative formulas.

Strict monotonicity. Subtracting the addition formulas gives

$$\sin y-\sin x=2\cos\frac{x+y}{2}\sin\frac{y-x}{2},\qquad \cos x-\cos y=2\sin\frac{x+y}{2}\sin\frac{y-x}{2}.$$

For $-\pi/2\le x<y\le\pi/2$, the cosine factor in the first identity is positive, as is its sine factor, so sine is strictly increasing on that interval. For $0\le x<y\le\pi$, both sine factors in the second identity are positive, so cosine is strictly decreasing. If $-\pi/2<x<y<\pi/2$, then

$$\tan y-\tan x=\frac{\sin(y-x)}{\cos x\cos y}>0,$$

because $0<y-x<\pi$ and both cosines are positive. Thus tangent is strictly increasing on this open interval. These arguments include the finite endpoints for sine and cosine without dividing by a vanishing factor.

Surjectivity and exact ranges. For any $c\in[-1,1]$, the unit-circle point $(\sqrt{1-c^2},c)$ lies on the right semicircle and hence has an angle in $[-\pi/2,\pi/2]$ with sine $c$. The point $(c,\sqrt{1-c^2})$ lies on the upper semicircle and has an angle in $[0,\pi]$ with cosine $c$. No sine or cosine value can lie outside $[-1,1]$ because the point lies on the unit circle. For any real $t$, the point $(1,t)/\sqrt{1+t^2}$ lies on the open right semicircle, so its angle $\theta\in(-\pi/2,\pi/2)$ satisfies $\tan\theta=t$. Consequently the restricted functions are bijections onto $[-1,1]$, $[-1,1]$ and $\mathbb R$, respectively. Their inverse functions are precisely $\arcsin$, $\arccos$ and $\arctan$ with these domains and branch ranges.

All solutions, not just principal values. If $|c|\le1$ and $\alpha=\arcsin c$, a circle point with ordinate $c$ has abscissa either $\sqrt{1-c^2}$ or $-\sqrt{1-c^2}$. These are the points at angles $\alpha$ and $\pi-\alpha$. Uniqueness of angle modulo one full turn gives exactly $\theta=\alpha+2k\pi$ or $\theta=\pi-\alpha+2k\pi$, $k\in\mathbb Z$. At $c=\pm1$ the two families coincide, which creates no extra solutions. If $|c|>1$, there is no solution. Similarly, for $\beta=\arccos c$ all solutions of $\cos\theta=c$ are $\theta=\pm\beta+2k\pi$. For real $t$ and $\gamma=\arctan t$, a unit-circle point of slope $t$ is one of the antipodal pair $\pm(1,t)/\sqrt{1+t^2}$; hence all solutions of $\tan\theta=t$ are $\theta=\gamma+k\pi$. The tangent poles are never admitted. This proves the branch restrictions and completeness of each periodic solution family without calculus.

Worked examples

An equation with a potentially lost branch. Solve $\sin(2x)=\sin x$ on $[0,2\pi)$. Factor to obtain $\sin x(2\cos x-1)=0$. The first factor gives $x=0,\pi$; the second gives $x=\pi/3,5\pi/3$. Dividing by $\sin x$ would have lost the first pair.

Amplitude and phase. Rewrite $3\sin x+4\cos x$ as $5\sin(x+\phi)$, where $\cos\phi=3/5$, $\sin\phi=4/5$ and $0<\phi<\pi/2$. The addition formula verifies the expression. Its range is $[-5,5]$; this is an exact bound, not $[-7,7]$ from bounding the two summands separately.

Graded practice

1. Foundation. Convert $150^\circ$ to radians and find its sine and cosine.
2. Core. Evaluate $\arcsin(\sin(5\pi/6))$ and $\arccos(\cos(5\pi/3))$.
3. Core. Solve $2\cos^2 x-3\cos x+1=0$ on $[0,2\pi)$.
4. Proof; optional extension requiring integration. Derive $\sin^2x=(1-\cos2x)/2$, and use it to show that the average of $\sin^2x$ over one period is $1/2$. The identity needs only this chapter; the average-value calculation uses elementary definite integration and is not a prerequisite for the core precalculus material.

Full solutions

1. Multiply degrees by $\pi/180$: the angle is $5\pi/6$. Its sine is $1/2$ and cosine is $-\sqrt3/2$ because it lies in quadrant II.
2. The first is $\pi/6$, the unique angle in the arcsine branch with sine $1/2$. The second is $\pi/3$, the unique angle in $[0,\pi]$ with cosine $1/2$.
3. Factoring gives $(2\cos x-1)(\cos x-1)=0$. Thus $\cos x=1/2$ or $1$, giving $x=\pi/3,5\pi/3,0$. The endpoint $2\pi$ is excluded.
4. The cosine double-angle formula is $\cos2x=\cos^2x-\sin^2x=1-2\sin^2x$. Rearranging gives the identity. Then $(2\pi)^{-1}\int_0^{2\pi}\sin^2x\,dx=(2\pi)^{-1}[x/2-\sin(2x)/4]_0^{2\pi}=1/2$.

Summary and term definitions

Periodicity repeats values under a fixed input shift. A principal branch is a restricted interval making a trigonometric function invertible. Amplitude is the nonnegative size of sinusoidal oscillation, while phase shifts the angle. Radians make later derivative formulas natural. Check domain restrictions before manipulating tangent or reciprocals, and report every solution in the stated interval.