Trigonometry

1.Unit circle, radians and exact values

Every real angle $\theta$ determines a point $(\cos\theta,\sin\theta)$ on the unit circle $x^2+y^2=1$: start at $(1,0)$ and rotate counterclockwise by $\theta$. This definition is what makes trigonometry work for every real angle rather than only the acute angles of a triangle, and it is why the sign of each function is controlled by the quadrant.

Radians measure arc length on the unit circle. An arc of length $s$ subtends the angle $s$, so a full turn is $2\pi$ and $180^\circ=\pi$. The conversion $\theta_{\text{rad}}=\theta_{\text{deg}}\cdot\frac{\pi}{180}$ follows immediately. Contest answers are almost always written as multiples of $\pi$, so get fluent in converting both ways.

Exact values come from two special right triangles. The $45^\circ$-$45^\circ$-$90^\circ$ triangle with legs $1$ has hypotenuse $\sqrt2$, so $\sin\frac{\pi}{4}=\cos\frac{\pi}{4}=\frac{\sqrt2}{2}$. The $30^\circ$-$60^\circ$-$90^\circ$ triangle with hypotenuse $2$ has short leg $1$ and long leg $\sqrt3$, because $1^2+(\sqrt3)^2=2^2$. Reading it off gives $\sin\frac{\pi}{6}=\frac12$, $\cos\frac{\pi}{6}=\frac{\sqrt3}{2}$, and by complementarity $\sin\frac{\pi}{3}=\frac{\sqrt3}{2}$, $\cos\frac{\pi}{3}=\frac12$.

Reference angles and signs: the absolute values of $\sin\theta$ and $\cos\theta$ equal those at the reference angle, the acute angle between the terminal ray and the x-axis. The signs are the ASTC rule: all functions are positive in quadrant I, only sine in II, only tangent in III, and only cosine in IV.

Reflections give reduction formulas: $\sin(\pi-\theta)=\sin\theta$, $\cos(\pi-\theta)=-\cos\theta$, $\sin(\pi+\theta)=-\sin\theta$, and $\cos(\pi+\theta)=-\cos\theta$. Any angle can be brought into $[0,\frac{\pi}{2}]$ with a sign, which is the first step of nearly every computation in this topic.

• $(\cos\theta,\sin\theta)$ is the point at angle $\theta$ on $x^2+y^2=1$
• $1^\circ=\frac{\pi}{180}$ radians, so $150^\circ=\frac{5\pi}{6}$
• $\sin\frac{\pi}{6}=\frac12$, $\cos\frac{\pi}{6}=\frac{\sqrt3}{2}$, $\sin\frac{\pi}{4}=\frac{\sqrt2}{2}$
• $\cos(\pi+\theta)=-\cos\theta$ and $\sin(\pi-\theta)=\sin\theta$

• Mixing degrees and radians, for example using $\frac{\pi}{180}$ in the wrong direction.
• Reading the cosine sign from the reference angle and forgetting the quadrant, so $\cos\frac{7\pi}{6}$ is written as positive.

2.Pythagorean identities, signs and reduction

The identity $\sin^2\theta+\cos^2\theta=1$ is the equation $x^2+y^2=1$ written at the point $(\cos\theta,\sin\theta)$. Dividing by $\cos^2\theta$ when it is nonzero gives $\tan^2\theta+1=\sec^2\theta$, and dividing by $\sin^2\theta$ gives $1+\cot^2\theta=\csc^2\theta$. Memorize all three and practice rearranging them, because most problems give one of $\sin$, $\cos$ or $\tan$ and ask for another.

The Pythagorean identity fixes only the magnitude. If $\sin\theta=-\frac{5}{13}$ then $\cos^2\theta=\frac{144}{169}$, so $\cos\theta=\pm\frac{12}{13}$, and the quadrant decides the sign. The most common lost point in this topic is writing $\cos\theta=\sqrt{1-\sin^2\theta}$ with the positive root without checking the quadrant. Always write the sign of each function before combining them.

Squaring a sum is a powerful trick. Since $(\sin x+\cos x)^2=1+2\sin x\cos x$, knowing $\sin x+\cos x$ determines the product $\sin x\cos x$ without solving for $x$. The sign ambiguity disappears because the quantity is squared, and only the magnitude is needed. The derivation is one expansion followed by the Pythagorean identity to replace the $1$.

Parity gives the reduction rules: $\sin(-\theta)=-\sin\theta$ (odd), $\cos(-\theta)=\cos\theta$ (even), and $\tan$ is odd. Combined with the shifts by $\pi$ and $\frac{\pi}{2}$, every expression can be rewritten using a single acute angle, which keeps the sign bookkeeping honest. Carry out the reduction before you square or add anything.

• $\sin^2\theta+\cos^2\theta=1$, hence $\tan^2\theta+1=\sec^2\theta$ and $1+\cot^2\theta=\csc^2\theta$
• $\sin(-\theta)=-\sin\theta$ and $\cos(-\theta)=\cos\theta$
• $(\sin x+\cos x)^2=1+2\sin x\cos x$

• Taking the positive square root of $\cos^2\theta$ without using the quadrant.
• Applying $\sin^2 a+\cos^2 a=1$ to unequal angles, for example $\sin^2 10^\circ+\cos^2 20^\circ$.

3.Addition, double and half-angle formulas

The cleanest derivation of the difference formula compares two chords of the unit circle. Let $P=(\cos a,\sin a)$ and $Q=(\cos b,\sin b)$. They are separated by the central angle $a-b$, so by the law of cosines in the isosceles triangle with apex at the origin, $|PQ|^2=2-2\cos(a-b)$. Computing the same squared length in coordinates gives $(\cos a-\cos b)^2+(\sin a-\sin b)^2=2-2(\cos a\cos b+\sin a\sin b)$. Comparing the two yields $\cos(a-b)=\cos a\cos b+\sin a\sin b$.

Replacing $b$ by $-b$ gives $\cos(a+b)=\cos a\cos b-\sin a\sin b$. Using $\sin(a+b)=\cos\left(\frac{\pi}{2}-a-b\right)$ and then expanding gives $\sin(a+b)=\sin a\cos b+\cos a\sin b$, and $b\to -b$ gives the difference version. Dividing the sine formula by the cosine formula gives $\tan(a+b)=\frac{\tan a+\tan b}{1-\tan a\tan b}$, which is the workhorse for tangent problems.

Setting $a=b=x$ gives the double-angle formulas $\sin2x=2\sin x\cos x$ and $\cos2x=\cos^2x-\sin^2x=2\cos^2x-1=1-2\sin^2x$. Rearranging the last form gives the half-angle formulas $\sin^2\frac{x}{2}=\frac{1-\cos x}{2}$ and $\cos^2\frac{x}{2}=\frac{1+\cos x}{2}$. The sign of a half-angle root depends on the quadrant of $\frac{x}{2}$, so the squared versions are usually what you want.

Two simplifications recur in contests: $\frac{1-\cos2x}{\sin2x}=\tan x$ and $\frac{\sin2x}{1+\cos2x}=\tan x$, both immediate from the double-angle formulas. The triple-angle identities $\sin3x=3\sin x-4\sin^3x$ and $\cos3x=4\cos^3x-3\cos x$ follow from the sum formulas applied to $2x+x$.

• $\cos(a-b)=\cos a\cos b+\sin a\sin b$, $\sin(a+b)=\sin a\cos b+\cos a\sin b$
• $\tan(a+b)=\frac{\tan a+\tan b}{1-\tan a\tan b}$
• $\sin2x=2\sin x\cos x$, $\cos2x=2\cos^2x-1=1-2\sin^2x$
• $\sin^2\frac{x}{2}=\frac{1-\cos x}{2}$ and $\cos^2\frac{x}{2}=\frac{1+\cos x}{2}$
• $\sin3x=3\sin x-4\sin^3x$

• Writing $\sin(a+b)=\sin a+\sin b$, which fails for every pair except special cases.
• Dropping the constant term when converting $\cos 2x$ to $1-2\sin^2x$ or $2\cos^2x-1$.

4.Product-to-sum and sum-to-product

Product-to-sum identities come from adding and subtracting the sum formulas. Adding $\cos(a-b)$ and $\cos(a+b)$ gives $2\cos a\cos b$, so $\cos a\cos b=\frac12[\cos(a-b)+\cos(a+b)]$. Subtracting gives $\sin a\sin b=\frac12[\cos(a-b)-\cos(a+b)]$. Adding $\sin(a+b)$ and $\sin(a-b)$ gives $\sin a\cos b=\frac12[\sin(a+b)+\sin(a-b)]$. These three forms are all you need, and they are quick to rederive when you need them.

Sum-to-product identities come from the substitution $a=\frac{u+v}{2}$, $b=\frac{u-v}{2}$, which gives $a+b=u$ and $a-b=v$. For instance $\sin u+\sin v=2\sin\frac{u+v}{2}\cos\frac{u-v}{2}$, and similarly $\cos u+\cos v=2\cos\frac{u+v}{2}\cos\frac{u-v}{2}$ and $\cos u-\cos v=-2\sin\frac{u+v}{2}\sin\frac{u-v}{2}$. The minus sign in the last identity is the most frequent source of error on the exam.

The doubling trick is the classic payoff. Multiplying $\cos x\cos2x\cos4x\cdots\cos2^{n-1}x$ by $\sin x$ and repeatedly applying $2\sin y\cos y=\sin2y$ gives the product $\frac{\sin2^nx}{2^n\sin x}$. Problems that look like hard products of cosines in arithmetic or geometric progression often reduce to this shape after the substitution $x\to\frac{\pi}{n}$.

Telescoping sums are the second payoff. From the product-to-sum identity, $2\sin\frac{x}{2}\cos kx=\sin\left((k+\tfrac12)x\right)-\sin\left((k-\tfrac12)x\right)$. Multiplying a sum of cosines in arithmetic progression by $2\sin\frac{x}{2}$ therefore makes it collapse to two terms. This is the engine behind the sums in the next concept.

• $\cos a\cos b=\frac12[\cos(a-b)+\cos(a+b)]$
• $\sin a\cos b=\frac12[\sin(a+b)+\sin(a-b)]$
• $\cos u-\cos v=-2\sin\frac{u+v}{2}\sin\frac{u-v}{2}$
• $\cos x\cos2x\cdots\cos2^{n-1}x=\frac{\sin2^nx}{2^n\sin x}$

• Expecting $\cos a\cos b$ to simplify to a single function; it only becomes a sum of two cosines.
• Forgetting that $\cos u-\cos v$ carries a minus sign and a factor of $-2$.

5.Law of sines, law of cosines and area

Drop the altitude $h$ from $C$ to side $AB=c$ in triangle $ABC$. The altitude gives $h=b\sin A=a\sin B$, so $\frac{a}{\sin A}=\frac{b}{\sin B}$. Repeating with the altitude from $A$ yields the law of sines $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R$, where $R$ is the circumradius. The common ratio is the diameter of the circumcircle, a fact worth keeping in mind for problems about circles through the vertices.

The area is $\frac12ch$, and $h=a\sin B$, so $[ABC]=\frac12ca\sin B=\frac12ab\sin C=\frac12bc\sin A$. Heron's formula $\sqrt{s(s-a)(s-b)(s-c)}$ with $s=\frac{a+b+c}{2}$ is an equivalent route when all three sides are known.

For the law of cosines, place $C$ at the origin with $CA$ along the positive x-axis, so $A=(b,0)$ and $B=(a\cos C,a\sin C)$. Then $c^2=(a\cos C-b)^2+a^2\sin^2C=a^2+b^2-2ab\cos C$. This is the Pythagorean theorem with a correction term, and the correction vanishes exactly when $C=90^\circ$.

Choosing the right law is a matter of the given data. Use the law of sines when you know an angle together with its opposite side. Use the law of cosines for SAS and SSS. The SSA configuration is the classic trap: with sides $a,b$ and angle $A$, if $a<b\sin A$ there is no triangle, if $a=b\sin A$ there is one right triangle, if $b\sin A<a<b$ there are two triangles, and if $a\ge b$ there is exactly one.

• $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R$
• $c^2=a^2+b^2-2ab\cos C$
• $[ABC]=\frac12ab\sin C$
• $\cos C=\frac{a^2+b^2-c^2}{2ab}$

• Using the law of sines in SSA without checking for the ambiguous case.
• Forgetting the factor $\frac12$ in the area formula, or using the sine of the wrong angle.

6.Solving trig equations and counting solutions

Trigonometric equations have infinitely many solutions on the real line, so contest questions ask for solutions in an interval, the number of solutions, or a sum of solutions. The method is always the same: reduce to a single function of a single angle, solve the basic equation using the unit circle, and then add the period.

For $\sin x=c$ with $|c|<1$, the solutions in $[0,2\pi)$ are $\alpha$ and $\pi-\alpha$, where $\alpha=\arcsin c$. For $\cos x=c$ they are $\alpha$ and $2\pi-\alpha$ with $\alpha=\arccos c$. Each pair is symmetric about an axis of the circle, and that symmetry is the fastest way to list them all.

When an equation is $\sin(kx)=c$ on $[0,L)$, count the solutions in one period of $\sin(kx)$, which is $\frac{2\pi}{k}$ and contains two solutions when $|c|<1$, then multiply by the number of periods in the interval. Be careful at the endpoints: a half-open interval includes its left endpoint and excludes its right.

Factoring is the usual next step. The equation $\sin2x=\cos x$ becomes $\cos x(2\sin x-1)=0$, a product of two simpler equations. Solutions from different factors can coincide, so take the union of the solution sets rather than adding the counts. A double root such as $(\sin x-1)^2=0$ contributes one solution.

Tangency needs care. The equation $2\sin^2x-3\sin x+1=0$ factors as $(2\sin x-1)(\sin x-1)=0$, and the root $\sin x=1$ touches the graph without crossing it. Sign-change methods miss such roots, so count geometric solutions directly from the factored form.

• $\sin x=c$ has solutions $\arcsin c$ and $\pi-\arcsin c$ per period
• $\sin A-\sin B=2\cos\frac{A+B}{2}\sin\frac{A-B}{2}$ turns difference equations into products
• $\sin(kx)=c$ has $2$ solutions per period $\frac{2\pi}{k}$ when $|c|<1$

• Adding counts from two factors when their solution sets overlap.
• Forgetting that a half-open interval $[0,2\pi)$ excludes $2\pi$.

7.Inverse trigonometric functions

The inverse functions need restricted domains to be well defined: $\arcsin:[-1,1]\to[-\frac{\pi}{2},\frac{\pi}{2}]$, $\arccos:[-1,1]\to[0,\pi]$, and $\arctan:\mathbb R\to(-\frac{\pi}{2},\frac{\pi}{2})$. The value $\arcsin(\sin\theta)$ is not always $\theta$. It is the unique angle in $[-\frac{\pi}{2},\frac{\pi}{2}]$ with the same sine, so $\arcsin\left(\sin\frac{5\pi}{4}\right)=-\frac{\pi}{4}$, not $\frac{5\pi}{4}$.

The complementary identity $\arcsin x+\arccos x=\frac{\pi}{2}$ holds on $[-1,1]$. To see it, put $\alpha=\arcsin x\in[-\frac{\pi}{2},\frac{\pi}{2}]$. Then $\cos(\frac{\pi}{2}-\alpha)=\sin\alpha=x$, and $\frac{\pi}{2}-\alpha$ lies in $[0,\pi]$, so it equals $\arccos x$.

Compositions are the workhorse. Because $\arccos u\in[0,\pi]$, its sine is nonnegative, and $\sin(\arccos u)=\sqrt{1-u^2}$. A right triangle with legs $u$ and $\sqrt{1-u^2}$ gives the same answer picturally, and similarly $\tan(\arcsin u)=\frac{u}{\sqrt{1-u^2}}$. The sign comes from the principal range of the inner function, not from the original angle.

Arctangent sums follow from the tangent addition formula. For $a,b>0$ with $ab<1$, the sum $\arctan a+\arctan b$ lies in $(0,\frac{\pi}{2})$, so it equals $\arctan\frac{a+b}{1-ab}$. If $ab>1$, the sum lies in $(\frac{\pi}{2},\pi)$ and one must add $\pi$ to the arctangent of the fraction. For example $\arctan2+\arctan3=\pi+\arctan\frac{5}{-5}=\frac{3\pi}{4}$, a value that a naive formula gets wrong by $\pi$.

• $\arcsin x+\arccos x=\frac{\pi}{2}$ for $x\in[-1,1]$
• $\sin(\arccos u)=\sqrt{1-u^2}$ and $\tan(\arcsin u)=\frac{u}{\sqrt{1-u^2}}$
• $\arctan a+\arctan b=\arctan\frac{a+b}{1-ab}$ when $ab<1$, and $+\pi$ when $ab>1$ with $a,b>0$

• Concluding $\arcsin(\sin\theta)=\theta$ without checking the principal range.
• Applying the arctangent sum formula when $ab>1$ and forgetting the $\pi$ correction.

8.Linear combinations and sums of sines and cosines

Any expression $a\sin x+b\cos x$ is a single sinusoid. Let $R=\sqrt{a^2+b^2}$ and choose $\varphi$ with $\cos\varphi=\frac aR$ and $\sin\varphi=\frac bR$, which is possible because $\left(\frac aR\right)^2+\left(\frac bR\right)^2=1$. Then $a\sin x+b\cos x=R(\cos\varphi\sin x+\sin\varphi\cos x)=R\sin(x+\varphi)$. The maximum is $R$ and the minimum is $-R$, which answers "find the maximum" questions at once.

Expressions quadratic in $\sin$ and $\cos$ reduce to this form. For example $\sin^2x+\sqrt3\sin x\cos x=\frac{1-\cos2x}{2}+\frac{\sqrt3}{2}\sin2x=\frac12+\left(\frac{\sqrt3}{2}\sin2x-\frac12\cos2x\right)$. The bracket has amplitude $\sqrt{\frac34+\frac14}=1$, so the maximum is $\frac12+1=\frac32$. Forgetting the constant $\frac12$ is the characteristic error.

For $\theta\notin2\pi\mathbb Z$, the sum of cosines in arithmetic progression is $\sum_{k=1}^{n}\cos k\theta=\frac{\sin\left((n+\frac12)\theta\right)-\sin\frac{\theta}{2}}{2\sin\frac{\theta}{2}}$. Derivation: multiply by $2\sin\frac\theta2$ and telescope with the product-to-sum identity from the previous concept. The sine analogue is proved the same way.

Roots of unity give a clean special case. The real parts of the $N$th roots of unity sum to zero. Taking $\theta=\frac{2\pi}{N}$ in the cosine sum, the terms $k$ and $N-k$ are equal, so half of the total is determined. For instance $\sum_{k=1}^{20}\cos\frac{2k\pi}{41}=-\frac12$, because the full sum over $k=0,\ldots,40$ is zero and contains the term $1$ at $k=0$.

• $a\sin x+b\cos x=\sqrt{a^2+b^2}\,\sin(x+\varphi)$ with $\cos\varphi=\frac{a}{R}$, $\sin\varphi=\frac{b}{R}$
• $\sum_{k=1}^{n}\cos k\theta=\frac{\sin((n+\frac12)\theta)-\sin\frac\theta2}{2\sin\frac\theta2}$
• $\sum_{k=0}^{N-1}\cos\frac{2\pi k}{N}=0$ for $N\ge2$

• Taking the amplitude as $a+b$ instead of $\sqrt{a^2+b^2}$.
• Dropping the constant term when a squared trig expression is rewritten.

9.Periodicity and graphs

A function $f$ has period $T$ when $f(x+T)=f(x)$ for all $x$. The functions $\sin(ax+b)$ and $\cos(ax+b)$ have period $\frac{2\pi}{|a|}$, and $\tan(ax+b)$ has period $\frac{\pi}{|a|}$. These follow from substituting $u=ax+b$: the base function completes a full cycle exactly when $u$ increases by its own period, which means $x$ increases by that period divided by $|a|$.

For a sum $f+g$ where $f$ has period $p$ and $g$ has period $q$ with $p/q$ rational, the period of the sum is the least common multiple of $p$ and $q$: the least positive $T$ that is an integer multiple of both. Concretely, for $p=\pi$ and $q=\frac{2\pi}{3}$ we need $m\pi=n\cdot\frac{2\pi}{3}$, so $3m=2n$, giving $m=2$, $n=3$ and $T=2\pi$. The least common multiple can be larger than either period, so always solve the integer equation rather than guessing.

Absolute values and squares change the period. The functions $|\sin x|$ and $\sin^2x$ have period $\pi$, not $2\pi$, because the negative lobes of $\sin$ are reflected to positive ones. This is the most common error when combining periods: $\sin^2x+\cos3x$ has period $2\pi$, not $\pi$ or $\frac{2\pi}{3}$.

Graphs are used to count intersections. To count the intersections of $y=\sin x$ with a line such as $y=\frac{x}{10}$, examine one lobe at a time. On a lobe where $\sin$ is positive, the second derivative $-\sin x$ is negative, so the graph is concave and meets a line at most twice. Endpoint sign checks establish existence. Combining the lobe counts with the bound $|\sin x|\le1$ finishes the count.

• $\sin(ax+b)$ and $\cos(ax+b)$ have period $\frac{2\pi}{|a|}$; $\tan(ax+b)$ has period $\frac{\pi}{|a|}$
• $\sin^2x$ and $|\sin x|$ have period $\pi$
• The period of $f+g$ is the least common multiple of the periods of $f$ and $g$

• Reporting the period of $\sin^2x$ as $2\pi$ rather than $\pi$.
• Taking the least common period as the product of the two periods.

10.Tangent identities in triangles

If $A+B+C=\pi$ then $\tan C=-\tan(A+B)=-\frac{\tan A+\tan B}{1-\tan A\tan B}$. Clearing the denominator gives $\tan C\,(1-\tan A\tan B)=-(\tan A+\tan B)$, which rearranges to the key identity $\tan A+\tan B+\tan C=\tan A\tan B\tan C$. It holds in every triangle with no right angle, because then all three tangents are defined. The identity turns a sum of tangents into a product and back, and it is often the fastest route through a problem about a triangle's angles.

The cotangent analogue is $\cot A\cot B+\cot B\cot C+\cot C\cot A=1$. It follows from the tangent identity by replacing each tangent with the reciprocal of its cotangent and multiplying through by the product of the cotangents. Use whichever form matches the expression in the problem.

For half-angles, the three angles $\frac A2,\frac B2,\frac C2$ sum to $\frac\pi2$. Therefore $\tan\frac A2\tan\frac B2+\tan\frac B2\tan\frac C2+\tan\frac C2\tan\frac A2=1$, which follows from the tangent-of-sum identity applied to $\frac A2+\frac B2=\frac\pi2-\frac C2$.

The half-angle substitution $t=\tan\frac x2$ rationalizes the trigonometric functions: $\sin x=\frac{2t}{1+t^2}$ and $\cos x=\frac{1-t^2}{1+t^2}$. These come from the double-angle formulas divided by $\cos^2\frac x2$, using $1+\tan^2=\sec^2$. Once a messy expression has been rewritten in $t$, it is an ordinary rational function, which is why this substitution appears in hard problems about the extremes of triangle expressions.

• If $A+B+C=\pi$ then $\tan A+\tan B+\tan C=\tan A\tan B\tan C$
• $\cot A\cot B+\cot B\cot C+\cot C\cot A=1$ in any triangle
• $\sin x=\frac{2t}{1+t^2}$ and $\cos x=\frac{1-t^2}{1+t^2}$ where $t=\tan\frac{x}{2}$

• Applying $\tan A+\tan B+\tan C=\tan A\tan B\tan C$ to a right triangle, where one tangent is undefined.
• Forgetting the sign: an obtuse angle has a negative tangent.

Example 1: Complementary values on the circle

Evaluate $\sin\frac{11\pi}{6}+\cos\frac{5\pi}{3}$.

Example 2: A half-angle ratio

If $\tan x=2$, find $\dfrac{\sin 2x}{1+\cos 2x}$.

Example 3: Square the sum

If $\sin x+\cos x=\frac12$, find $\sin x\cos x$.

Example 4: Angles and sides in an obtuse triangle

In triangle $ABC$, $\cos A=\frac35$ and $\sin B=\frac{5}{13}$, where $\angle B$ is acute. Find $\cos C$.

Example 5: A telescoping cosine sum

Find $\cos\frac{\pi}{7}+\cos\frac{3\pi}{7}+\cos\frac{5\pi}{7}$.

Example 6: Minimizing a sum of fourth powers

Find the minimum value of $\sec^4x+\csc^4x$ for $0<x<\frac{\pi}{2}$.