1.1 Equation structure and trigonometric identities

Motivation

Trigonometric calculation is reliable only when the logical status of every line is clear. An identity asserts that two expressions have the same value for every input in a stated common domain. An equation asks for the inputs in a stated domain at which two expressions happen to agree. Confusing these roles is dangerous: an identity may be used to rewrite an equation, but solving one equation does not prove an identity.

Domain information is part of the mathematics. For instance, the statement $1+\tan^2\theta=\sec^2\theta$ is an identity on the angles for which $\cos\theta\ne0$; neither side is defined outside that common domain. Likewise, clearing a denominator in an equation is reversible only after its zeros have been excluded. A sound solution therefore begins by declaring the domain, records every exclusion, produces complete periodic families, and finally checks candidates in the original equation.

The geometric viewpoint explains why the algebra works. Radians measure a signed rotation by comparing arc and radius. The terminal point on the unit circle supplies the signed coordinates $\cos\theta$ and $\sin\theta$. Reflections give symmetry, repeated rotations give periodicity, and the circle equation gives the Pythagorean identity. Compound-angle formulas then organize more elaborate rewrites, while inverse trigonometric functions select one principal representative from an entire periodic family.

This chapter develops those ideas as one connected method. The aim is not to memorize a formula sheet. It is to know where each formula comes from, where it is defined, what it preserves, and how it helps solve an equation without losing or inventing roots.

Definitions

Definition

Equation, identity, and declared domain

Let $D$ be a set of admissible real inputs. An equation $F(x)=G(x)$ asks for the solution set

$$S=\{x\in D:F(x)=G(x)\}.$$

An identity on $D$ states that $F(x)=G(x)$ for every $x\in D$ for which both expressions are defined. The phrase “on $D$” cannot be omitted when a formula contains a quotient, tangent, cotangent, secant, or cosecant.

Two equations are equivalent on $D$ when they have exactly the same solution set there. Adding the same defined expression to both sides, applying an identity on its common domain, or multiplying by a quantity known to be nonzero preserves equivalence. Squaring, or multiplying by an expression that may vanish, usually gives only a necessary forward implication and can create extra candidates. Dividing by an expression that may vanish can discard a whole case. Such a division requires a case split.

A useful written device is a domain ledger. First list the inputs for which the original statement exists. Beside each transformation, record any new condition used to make the step reversible. At the end, compare the candidate families with the original ledger rather than with a later simplified line. This separates three questions that are often blurred together: whether an expression is defined, whether a transformation preserves both directions of implication, and whether a candidate actually satisfies the starting equation. The ledger is especially important for periodic answers, because a single forbidden tangent or denominator value represents an infinite family of excluded angles rather than one isolated number.

Definition

Radian measure and oriented geometry

For a central angle with radius $r\gt0$ and directed arc displacement $s$, its signed radian measure is

$$\theta=\frac{s}{r}.$$

Counterclockwise rotation is positive and clockwise rotation is negative. One full counterclockwise turn is $2\pi$ radians, so

$$\pi\text{ radians}=180^\circ, \qquad 1^\circ=\frac{\pi}{180}\text{ radians}.$$

With this oriented convention, $s=r\theta$ is a directed arc displacement and $A_{\text{signed}}=\tfrac12r^2\theta$ is signed swept area. The corresponding unsigned traversal length and swept area are $r|\theta|$ and $\tfrac12r^2|\theta|$. When $|\theta|\gt2\pi$, these quantities count repeated passes with multiplicity rather than the boundary length or area of one simple sector region.

An angle is in standard position when its vertex is the origin and its initial ray is the positive $x$-axis. Unless stated otherwise, every angle here is measured in radians and may represent any real signed rotation, not only an angle between zero and one revolution.

Definition

Trigonometric functions from coordinates

If the terminal ray of $\theta$ meets a circle of radius $r$ at $P(x,y)$, where $r=\sqrt{x^2+y^2}$, then

$$\sin\theta=\frac{y}{r},\qquad \cos\theta=\frac{x}{r},\qquad \tan\theta=\frac{y}{x},$$

and

$$\csc\theta=\frac{r}{y},\qquad \sec\theta=\frac{r}{x},\qquad \cot\theta=\frac{x}{y}.$$

Each quotient is defined only when its denominator is nonzero. On the unit circle, the terminal point is exactly $(\cos\theta,\sin\theta)$.

The selected standard-angle values are

Definition

Principal inverse trigonometric functions

The inverse sine is the inverse of sine restricted to $[-\pi/2,\pi/2]$; hence

$$\sin^{-1}:[-1,1]\longrightarrow[-\pi/2,\pi/2].$$

The inverse cosine uses the restriction $[0,\pi]$, and inverse tangent uses $(-\pi/2,\pi/2)$:

$$\cos^{-1}:[-1,1]\longrightarrow[0,\pi], \qquad \tan^{-1}:\mathbb R\longrightarrow(-\pi/2,\pi/2).$$

Here the exponent $-1$ means inverse function, not reciprocal; for example, $\sin^{-1}k$ is not $\csc k$.

Theorem / Proposition

Theorem

Equivalence-preserving equation work

Fix a declared domain $D$. Replacing either side by an identity valid on $D$, adding or subtracting the same defined quantity, and multiplying or dividing by a quantity proved nonzero on $D$ preserve the solution set. If a step is only one-way, its output must be labelled as a candidate set and tested in the original equation.

Theorem

Radian geometry

For radius $r\gt0$ and signed angle $\theta$,

$$s=r\theta, \qquad A_{\text{signed}}=\frac12r^2\theta.$$

The unit-circle terminal point is $(\cos\theta,\sin\theta)$. Thus signs of the trigonometric functions come from the quadrant, while the reference angle determines their magnitudes.

Theorem

Symmetry and periodicity

For every real $\theta$,

$$\sin(-\theta)=-\sin\theta, \qquad \cos(-\theta)=\cos\theta,$$$$\sin(\pi-\theta)=\sin\theta, \qquad \cos(\pi-\theta)=-\cos\theta,$$$$\sin(\pi+\theta)=-\sin\theta, \qquad \cos(\pi+\theta)=-\cos\theta,$$

and the cofunction relations are

$$\sin\left(\frac\pi2-\theta\right)=\cos\theta, \qquad \cos\left(\frac\pi2-\theta\right)=\sin\theta,$$$$\sin\left(\frac\pi2+\theta\right)=\cos\theta, \qquad \cos\left(\frac\pi2+\theta\right)=-\sin\theta.$$$$\sin(\theta+2\pi)=\sin\theta, \qquad \cos(\theta+2\pi)=\cos\theta.$$

Also $\tan(-\theta)=-\tan\theta$, $\tan(\pi-\theta)=-\tan\theta$, and $\tan(\theta+\pi)=\tan\theta$ wherever both sides are defined. The cofunction tangent values are $\cot\theta$ and $-\cot\theta$, again only on their common domains. These relations reduce an arbitrary signed rotation to a reference angle while preserving its quadrant sign. Secant and cosecant have period $2\pi$, while cotangent has period $\pi$, on their respective domains.

Theorem

Pythagorean identities on their common domains

For every real $\theta$,

$$\sin^2\theta+\cos^2\theta=1.$$

Where $\cos\theta\ne0$,

$$1+\tan^2\theta=\sec^2\theta.$$

Where $\sin\theta\ne0$,

$$1+\cot^2\theta=\csc^2\theta.$$

Theorem

Compound-angle formulas

For all real $\alpha,\beta$,

$$\sin(\alpha\mathbin{\pm}\beta) =\sin\alpha\cos\beta\mathbin{\pm}\cos\alpha\sin\beta,$$$$\cos(\alpha\mathbin{\pm}\beta) =\cos\alpha\cos\beta\mathbin{\mp}\sin\alpha\sin\beta.$$

The tangent versions are

$$\tan(\alpha\mathbin{\pm}\beta) =\frac{\tan\alpha\mathbin{\pm}\tan\beta} {1\mathbin{\mp}\tan\alpha\tan\beta},$$

but only on the common domain where $\tan\alpha$, $\tan\beta$, and $\tan(\alpha\mathbin{\pm}\beta)$ are all defined. On that domain the displayed denominator is necessarily nonzero.

Theorem

Double-angle formulas

For all real $\alpha$,

$$\sin 2\alpha=2\sin\alpha\cos\alpha,$$$$\cos 2\alpha=\cos^2\alpha-\sin^2\alpha =2\cos^2\alpha-1=1-2\sin^2\alpha.$$

Furthermore,

$$\tan 2\alpha=\frac{2\tan\alpha}{1-\tan^2\alpha}$$

only where $\tan\alpha$ and $\tan 2\alpha$ are both defined; equivalently the quotient on the right must also have a nonzero denominator.

Theorem

Product-to-sum formulas

For all real $\alpha,\beta$,

$$\sin\alpha\cos\beta =\frac12\{\sin(\alpha+\beta)+\sin(\alpha-\beta)\},$$$$\cos\alpha\sin\beta =\frac12\{\sin(\alpha+\beta)-\sin(\alpha-\beta)\},$$$$\sin\alpha\sin\beta =-\frac12\{\cos(\alpha+\beta)-\cos(\alpha-\beta)\},$$$$\cos\alpha\cos\beta =\frac12\{\cos(\alpha+\beta)+\cos(\alpha-\beta)\}.$$

Theorem

Sum-to-product formulas

For all real $A,B$,

$$\sin A+\sin B=2\sin\frac{A+B}{2}\cos\frac{A-B}{2},$$$$\sin A-\sin B=2\cos\frac{A+B}{2}\sin\frac{A-B}{2},$$$$\cos A+\cos B=2\cos\frac{A+B}{2}\cos\frac{A-B}{2},$$$$\cos A-\cos B=-2\sin\frac{A+B}{2}\sin\frac{A-B}{2}.$$

Theorem

Complete general solution families

Let $n\in\mathbb Z$. If $k\in[-1,1]$ and $\alpha=\sin^{-1}k$, then

$$\sin\theta=k \quad\Longleftrightarrow\quad \theta=\alpha+2n\pi \ \text{or}\ \theta=\pi-\alpha+2n\pi.$$

If $k\in[-1,1]$ and $\alpha=\cos^{-1}k$, then

$$\cos\theta=k \quad\Longleftrightarrow\quad \theta=\mathbin{\pm}\alpha+2n\pi.$$

If $k\in\mathbb R$ and $\alpha=\tan^{-1}k$, then

$$\tan\theta=k \quad\Longleftrightarrow\quad \theta=\alpha+n\pi.$$

At endpoint values the two sine or cosine descriptions may coincide; this is duplication, not an extra family.

Theorem

Subsidiary-angle form

For real $a,b$, not both zero, set $R=\sqrt{a^2+b^2}$. There is a unique $\alpha\in[0,2\pi)$ satisfying both

$$R\cos\alpha=a, \qquad R\sin\alpha=b.$$

Then

$$a\sin\theta+b\cos\theta=R\sin(\theta+\alpha).$$

The quotient $\tan\alpha=b/a$, when available, determines only a line of possible directions; the two signed coefficient equations determine the correct quadrant. Similarly, $a\sin\theta+b\cos\theta=R\cos(\theta-\beta)$ when $R\sin\beta=a$ and $R\cos\beta=b$.

Proof Sketch or Proof Idea

Proof

Why radians control arc and signed sector area

The fraction of a full turn represented by a signed angle $\theta$ is $\theta/(2\pi)$. Multiplying this fraction by the circumference $2\pi r$ gives $s=r\theta$. Multiplying it by the disk area $\pi r^2$ gives $A_{\text{signed}}=\tfrac12r^2\theta$. A clockwise sweep has negative orientation, so these directed quantities are negative. Taking absolute values gives total traversal length and unsigned swept area; repeated turns are counted with multiplicity.

Proof

Why symmetry, periodicity, and Pythagoras follow from the circle

The point at angle $\theta$ is $(\cos\theta,\sin\theta)$. Reflection in the $x$-axis sends it to $(\cos\theta,-\sin\theta)$, which is the point for $-\theta$. A full turn returns to the same point, and a half-turn changes both coordinate signs; the ratio $y/x$ is therefore unchanged after $\pi$ whenever it is defined. Finally, every unit-circle point satisfies $x^2+y^2=1$, giving $\sin^2\theta+\cos^2\theta=1$. Division by $\cos^2\theta$ or $\sin^2\theta$ gives the other two identities only on the corresponding nonzero-denominator domains.

Proof

Deriving compound-angle identities from cosine difference

Take $A=(\cos\alpha,\sin\alpha)$ and $B=(\cos\beta,\sin\beta)$ on the unit circle. Let $\phi\in[0,\pi]$ be the smaller angle between $OA$ and $OB$. The directed difference $\alpha-\beta$ is congruent modulo $2\pi$ to either $\phi$ or $-\phi$; evenness and periodicity of cosine therefore give $\cos\phi=\cos(\alpha-\beta)$. The cosine rule applied to triangle $OAB$ now gives

$$AB^2=2-2\cos(\alpha-\beta).$$

The coordinate distance formula gives the same squared distance as

$$AB^2=(\cos\alpha-\cos\beta)^2 +(\sin\alpha-\sin\beta)^2 =2-2\cos\alpha\cos\beta-2\sin\alpha\sin\beta.$$

Equating them proves the cosine-difference formula. Replacing $\beta$ by $-\beta$, and using the unit-circle symmetry rules, gives cosine addition. Cofunction substitutions give the sine formulas. Dividing a sine formula by the matching cosine formula yields tangent only after every required cosine has been declared nonzero.

Proof

How double-angle and product-sum identities are generated

Set $\beta=\alpha$ in the compound-angle formulas to obtain the double-angle formulas. Next add the sine addition and sine subtraction formulas to isolate $2\sin\alpha\cos\beta$; subtract them to isolate $2\cos\alpha\sin\beta$. Adding and subtracting the two cosine formulas gives the remaining products. Conversely, in the product formulas set $\alpha=(A+B)/2$ and $\beta=(A-B)/2$; then $\alpha+\beta=A$ and $\alpha-\beta=B$, which yields the four sum-to-product formulas.

Proof

Why the inverse formulas give every solution

The principal inverse value identifies one unit-circle point. A horizontal line meets the circle at two points with angles $\alpha$ and $\pi-\alpha$, so sine uses those two representatives and then adds full turns. A vertical line meets at angles $\alpha$ and $-\alpha$, so cosine uses those representatives. Tangent is a slope: the same slope returns after a half-turn, giving the single family $\alpha+n\pi$. These geometric descriptions also explain why the principal ranges must be fixed before inverse notation is meaningful.

Worked Examples

Worked example

1. Signed radians, coordinates, arc, and area

Convert $-45^\circ$ to radians and consider a circle of radius $3$. Since

$$-45^\circ=-45\frac{\pi}{180}=-\frac{\pi}{4},$$

the unit-circle terminal point is $(\sqrt2/2,-\sqrt2/2)$. On the radius-three circle, the directed arc displacement and signed swept area are

$$s=3\left(-\frac\pi4\right)=-\frac{3\pi}{4}, \qquad A_{\text{signed}}=\frac12(3)^2\left(-\frac\pi4\right) =-\frac{9\pi}{8}.$$

The corresponding ordinary arc length and geometric area are $3\pi/4$ and $9\pi/8$. The negative signs record clockwise orientation, not negative physical size.

Worked example

2. An exact compound-angle value

Express $\sin15^\circ$ in surd form. This example uses degree measure. Use $15^\circ=45^\circ-30^\circ$:

$$\begin{aligned} \sin15^\circ &=\sin45^\circ\cos30^\circ-\cos45^\circ\sin30^\circ\\ &=\frac{\sqrt2}{2}\frac{\sqrt3}{2} -\frac{\sqrt2}{2}\frac12 =\frac{\sqrt6-\sqrt2}{4}. \end{aligned}$$

The sign is consistent with a small first-quadrant angle, so the positive answer is also a useful reasonableness check.

Worked example

3. A triangle tangent identity

Let $A,B,C$ be the angles of a triangle and assume all three tangents are defined. Since $A+B+C=\pi$,

$$\tan C=\tan(\pi-A-B)=-\tan(A+B) =-\frac{\tan A+\tan B}{1-\tan A\tan B}.$$

The common-domain assumption rules out a right angle and therefore also makes the displayed compound-angle denominator nonzero. Rearranging gives

$$\tan A+\tan B+\tan C =\tan A\tan B\tan C.$$

Without the domain clause, the statement would incorrectly appear to include right triangles, where one of its terms is undefined.

Worked example

4. The half-angle substitution

Let $t=\tan(\theta/2)$. For finite $t$, meaning $\theta\not\equiv\pi\pmod{2\pi}$, double-angle and Pythagorean formulas give

$$\sin\theta =\frac{2\tan(\theta/2)}{1+\tan^2(\theta/2)} =\frac{2t}{1+t^2},$$$$\cos\theta =\frac{1-\tan^2(\theta/2)}{1+\tan^2(\theta/2)} =\frac{1-t^2}{1+t^2}.$$

Consequently,

$$\tan\theta=\frac{2t}{1-t^2}$$

only when $t\ne\pm1$, exactly the cases in which $\cos\theta\ne0$. The substitution misses $\theta\equiv\pi\pmod{2\pi}$ because $\tan(\theta/2)$ is then undefined. Any equation solved by this substitution must test that missing family separately and retain every original denominator restriction.

Worked example

5. A rational tangent equation

Solve

$$\frac{1+\tan x}{1-\tan x}=1+\sin2x.$$

The original domain requires $\cos x\ne0$ and $\tan x\ne1$. Put $t=\tan x$, so $t\ne1$, and use $\sin2x=2t/(1+t^2)$. Since both clearing factors are nonzero on this domain,

$$\frac{1+t}{1-t}=1+\frac{2t}{1+t^2} \quad\Longleftrightarrow\quad (1+t)(1+t^2)=(1-t)(1+t^2)+2t(1-t).$$

Thus $2t^2(1+t)=0$, so $t=0$ or $t=-1$. Both satisfy the original equation. The complete answer is

$$x=n\pi \quad\text{or}\quad x=-\frac\pi4+n\pi, \qquad n\in\mathbb Z.$$

Worked example

6. Factor first, then restrict the interval

First show that

$$\cos2x+2\cos3x+\cos4x=4\cos^2\frac{x}{2}\cos3x.$$

Indeed, sum-to-product and the half-angle form of $1+\cos x$ give

$$\begin{aligned} \cos2x+\cos4x+2\cos3x &=2\cos3x\cos(-x)+2\cos3x\\ &=2\cos3x(\cos x+1) =4\cos3x\cos^2\frac{x}{2}. \end{aligned}$$

On $[0,2\pi]$, a zero therefore satisfies $\cos3x=0$ or $\cos(x/2)=0$. The first condition gives $x=\pi/6+n\pi/3$; the second adds $x=\pi$. After enforcing the interval, the solutions are

$$\left\{\frac\pi6,\frac\pi2,\frac{5\pi}6,\pi, \frac{7\pi}6,\frac{3\pi}2,\frac{11\pi}6\right\}.$$

Worked example

7. Choose the subsidiary angle by signs

Express $\sqrt3\sin\theta-\cos\theta$ in both standard subsidiary forms. For $R\sin(\theta+\alpha)$, coefficient comparison requires

$$R\cos\alpha=\sqrt3, \qquad R\sin\alpha=-1.$$

Hence $R=2$. The cosine condition is positive and the sine condition is negative, so $\alpha$ lies in quadrant IV; both equations select $\alpha=11\pi/6$. Therefore

$$\sqrt3\sin\theta-\cos\theta =2\sin\left(\theta+\frac{11\pi}{6}\right).$$

For $R\cos(\theta-\beta)$, comparison instead gives $R\sin\beta=\sqrt3$ and $R\cos\beta=-1$. Thus $\beta$ is in quadrant II, so $\beta=2\pi/3$ and

$$\sqrt3\sin\theta-\cos\theta =2\cos\left(\theta-\frac{2\pi}{3}\right).$$

A tangent value alone would not distinguish either accepted angle from the angle opposite it.

Worked example

8. Correcting the branch in a single-sine equation

Write

$$\sqrt3\cos x-\sin x =2\sin\left(x+\frac{2\pi}{3}\right).$$

Then $\sqrt3\cos x-\sin x=1$ becomes $\sin(x+2\pi/3)=1/2$. The two sine families give

$$x+\frac{2\pi}{3}=\frac\pi6+2n\pi \quad\text{or}\quad x+\frac{2\pi}{3}=\frac{5\pi}{6}+2n\pi.$$

Therefore

$$x=-\frac\pi2+2n\pi \quad\text{or}\quad x=\frac\pi6+2n\pi.$$

The second branch is $x=\pi/6+2n\pi$, not $-\pi/6+2n\pi$; direct substitution at $x=\pi/6$ gives $3/2-1/2=1$.

Worked example

9. Denominators and candidate verification

Solve

$$8\cos x=\frac1{\cos x}-\frac{\sqrt3}{\sin x}.$$

The declared domain is $\sin x\ne0$ and $\cos x\ne0$. Multiplication by $\sin x\cos x$ is equivalent on that domain. After dividing the resulting equality by two and using product-to-sum,

$$2\sin2x\cos x=\frac12\sin x-\frac{\sqrt3}{2}\cos x,$$$$\sin3x+\sin x=\frac12\sin x-\frac{\sqrt3}{2}\cos x,$$

so

$$\sin3x=\sin\left(x+\frac{4\pi}{3}\right).$$

The two sine branches yield

$$x=\frac{2\pi}{3}+n\pi \quad\text{or}\quad x=-\frac\pi{12}+\frac{n\pi}{2}, \qquad n\in\mathbb Z.$$

Neither family ever makes sine or cosine zero, so every candidate respects the original exclusions. Substitution through the equivalent cleared equation, together with the nonzero denominators, confirms both complete families.

Common Mistakes

Common mistake

Treating an equation as an identity

An equation such as $\sin x=1/2$ is true only at its solutions. An identity such as $\sin^2x+\cos^2x=1$ is true throughout its declared domain. State which claim you are making before manipulating it.

Common mistake

Writing only a principal inverse value

$\sin^{-1}k$, $\cos^{-1}k$, and $\tan^{-1}k$ provide principal representatives, not all solutions. Use symmetry and periodicity to write the complete families, then impose any requested interval.

Common mistake

Using tangent identities outside their common domain

The sine and cosine compound formulas hold for all real angles. Their tangent quotients do not. Check the cosines of the input angles and of the sum or difference before claiming equality.

Common mistake

Dividing away a possible solution

From $F(x)G(x)=0$, division by $F(x)$ deletes the entire case $F(x)=0$. Use the zero-product rule or split into cases unless non-vanishing has already been proved.

Common mistake

Forgetting exclusions after clearing denominators

Write denominator exclusions before multiplying. Clearing produces an equivalent equation only on that restricted domain; the final answer must still be filtered and checked against the original rational equation.

Common mistake

Assuming the half-angle substitution covers the circle

Finite $t=\tan(\theta/2)$ omits $\theta\equiv\pi\pmod{2\pi}$. Test that family separately, and remember that the rational formula for tangent also requires $1-t^2\ne0$.

Common mistake

Choosing a subsidiary angle from tangent alone

The ratio fixes an angle only modulo $\pi$. Use both signed coefficient conditions, such as $R\cos\alpha=a$ and $R\sin\alpha=b$, to select the quadrant.

Common mistake

Reading signed area as physical area

$\tfrac12r^2\theta$ is signed when $\theta$ is signed. Unsigned swept area is nonnegative and uses $|\theta|$; for a multi-turn angle it includes every repeated sweep rather than describing one simple sector region.

Summary

Verification is not a ceremonial final line: it confirms that the domain, every branch, and every equivalence condition survived the transformation chain.

• Begin an equation by declaring its real domain and every denominator exclusion. Mark one-way steps, retain candidates, and verify them in the original equation.
• Radians connect signed rotation to directed arc displacement and signed swept area. Unit-circle coordinates control signs, symmetries, and periods.
• Pythagorean, compound-angle, double-angle, product-to-sum, and sum-to-product formulas form a derivable system rather than an unrelated list. Quotient identities always carry common-domain restrictions.
• Principal inverse trigonometric values choose representatives. Symmetry and periodicity expand them into complete sine, cosine, and tangent families.
• A subsidiary angle compresses $a\sin\theta+b\cos\theta$ into one sinusoid; its radius comes from $\sqrt{a^2+b^2}$ and its quadrant comes from both signed coefficient equations.
• Rational trigonometric equations demand the same discipline as rational algebra: exclude denominator zeros first, transform only on the allowed domain, solve completely, and back-substitute.

Exercises

1. On the common domain of the two sides, prove $$\tan(A+B)+\tan(A-B) =\frac{2\sin2A}{\cos2A+\cos2B}.$$
2. Derive both triple-angle identities $$\sin3\theta=3\sin\theta-4\sin^3\theta, \qquad \cos3\theta=4\cos^3\theta-3\cos\theta.$$
3. Prove $$4\cos\theta\cos\left(\frac{2\pi}{3}+\theta\right) \cos\left(\frac{2\pi}{3}-\theta\right)=\cos3\theta.$$
4. Prove $$\sin^2\theta+\sin^2\phi-\sin^2(\theta-\phi) =2\sin\theta\sin\phi\cos(\theta-\phi).$$
5. If $A,B,C$ are the angles of a triangle, prove $$\tan\frac A2\tan\frac B2 +\tan\frac B2\tan\frac C2 +\tan\frac C2\tan\frac A2=1.$$
6. If $A,B,C$ are the angles of a triangle, prove $$\cos^2A+\cos^2B+\cos^2C =1-2\cos A\cos B\cos C.$$
7. For real $a,b$ with $b\ne0$, derive $a\sin\theta+b\cos\theta=R\cos(\theta-\alpha)$ with $R\gt0$, and state conditions that determine $\alpha\in[0,2\pi)$ without quadrant ambiguity.
8. Give complete real solution families for $\sin\theta=-1/2$, $\cos\theta=\sqrt2/2$, and $\tan\theta=-\sqrt3$.

Solutions

Solution · Solution 1

Where all displayed tangents and quotients are defined,

$$\tan(A+B)+\tan(A-B) =\frac{\sin(A+B)}{\cos(A+B)} +\frac{\sin(A-B)}{\cos(A-B)}.$$

Combining the fractions, the numerator is $\sin(A+B)\cos(A-B)+\sin(A-B)\cos(A+B)=\sin2A$. Also,

$$2\cos(A+B)\cos(A-B)=\cos2A+\cos2B.$$

Substitution gives the required right-hand side. The common-domain clause is essential because both denominators must be nonzero.

Solution · Solution 2

Apply the compound formulas to $2\theta+\theta$:

$$\begin{aligned} \sin3\theta &=2\sin\theta\cos^2\theta +(\cos^2\theta-\sin^2\theta)\sin\theta\\ &=3\sin\theta-4\sin^3\theta, \end{aligned}$$

where $\cos^2\theta=1-\sin^2\theta$. Similarly,

$$\begin{aligned} \cos3\theta &=(\cos^2\theta-\sin^2\theta)\cos\theta -2\sin^2\theta\cos\theta\\ &=4\cos^3\theta-3\cos\theta. \end{aligned}$$

Solution · Solution 3

Product-to-sum gives

$$2\cos\left(\frac{2\pi}{3}+\theta\right) \cos\left(\frac{2\pi}{3}-\theta\right) =\cos\frac{4\pi}{3}+\cos2\theta =-\frac12+\cos2\theta.$$

Multiplying by $2\cos\theta$ yields

$$2\cos\theta\left(2\cos^2\theta-\frac32\right) =4\cos^3\theta-3\cos\theta=\cos3\theta.$$

Solution · Solution 4

Expand $\sin(\theta-\phi)$ and simplify the left-hand side:

$$\begin{aligned} &\sin^2\theta+\sin^2\phi -(\sin\theta\cos\phi-\cos\theta\sin\phi)^2\\ &=2\sin^2\theta\sin^2\phi +2\sin\theta\cos\theta\sin\phi\cos\phi\\ &=2\sin\theta\sin\phi (\sin\theta\sin\phi+\cos\theta\cos\phi)\\ &=2\sin\theta\sin\phi\cos(\theta-\phi). \end{aligned}$$

Solution · Solution 5

Put $u=A/2$, $v=B/2$, and $w=C/2$. Then $u+v+w=\pi/2$, while each half-angle lies in $(0,\pi/2)$. Expanding the cosine of the sum gives

$$0=\cos(u+v+w) =\cos u\cos v\cos w \left(1-\tan u\tan v-\tan v\tan w-\tan w\tan u\right).$$

The three cosine factors are positive, so the parenthesis is zero. Rearranging proves the stated identity.

Solution · Solution 6

Since $C=\pi-(A+B)$,

$$\cos C=\sin A\sin B-\cos A\cos B,$$

so $\cos C+\cos A\cos B=\sin A\sin B$. Now

$$\begin{aligned} &\cos^2A+\cos^2B+\cos^2C +2\cos A\cos B\cos C\\ &=(\cos C+\cos A\cos B)^2 +\cos^2A+\cos^2B-\cos^2A\cos^2B\\ &=\sin^2A\sin^2B+\cos^2A+\cos^2B-\cos^2A\cos^2B=1. \end{aligned}$$

Move the product term to the other side.

Solution · Solution 7

Expand

$$R\cos(\theta-\alpha) =R\cos\theta\cos\alpha+R\sin\theta\sin\alpha.$$

Coefficient comparison requires

$$R\sin\alpha=a, \qquad R\cos\alpha=b.$$

Squaring and adding gives $R=\sqrt{a^2+b^2}$. The ratio gives $\tan\alpha=a/b$, but the signs of both coefficient equations choose the unique $\alpha\in[0,2\pi)$. This proves the form without a quadrant guess.

Solution · Solution 8

For sine, $\alpha=\sin^{-1}(-1/2)=-\pi/6$, hence

$$\theta=-\frac\pi6+2n\pi \quad\text{or}\quad \theta=\frac{7\pi}{6}+2n\pi.$$

For cosine, $\alpha=\cos^{-1}(\sqrt2/2)=\pi/4$, hence

$$\theta=\mathbin{\pm}\frac\pi4+2n\pi.$$

For tangent, $\alpha=\tan^{-1}(-\sqrt3)=-\pi/3$, hence

$$\theta=-\frac\pi3+n\pi, \qquad n\in\mathbb Z.$$