Linear algebra leans on trigonometry for exactly one job, over and over: turning two directions into a single number that says how aligned they are. That number is always a cosine. Chapter 1 wraps a whole formula around it — \cos\theta = \dfrac{\mathbf u\cdot\mathbf v}{|\mathbf u||\mathbf v|} — but every piece of the trig that formula needs is already sitting in your toolbox from an earlier course. This section collects exactly that piece: what an angle is measured in, what cosine means as a coordinate, and the Law of Cosines, the same computation that formula is secretly running under a change of notation.
Measuring an angle: degrees and radians
Degrees chop a circle into 360 pieces for no mathematical reason. That number comes from ancient Babylonian base-60 counting, not from anything about circles themselves. Radians ask the circle to measure itself instead. Walk exactly one radius's worth of distance along the rim, and the angle you just swept out is defined to be 1 radian. Walk all the way around and you cover a distance of 2\pi radii, since that's the circle's circumference. So you've swept 2\pi radians, a full turn. It's a ruler built out of the circle's own radius, not an outside convention.
A radian measures an angle by the arc length it cuts on a circle of radius 1, rather than by dividing a circle into 360 arbitrary pieces. One full trip around the unit circle has circumference 2\pi, so a full turn is 2\pi radians — exactly 360°. That single equivalence gives the conversion in both directions:
180° = \pi \text{ radians} \quad\Longrightarrow\quad \text{degrees} \to \text{radians}: \ \times\frac{\pi}{180}, \qquad \text{radians} \to \text{degrees}: \ \times\frac{180}{\pi}.Degrees are the everyday unit; radians are the one every formula in this course (and every formula involving calculus, later) is written in, because a radian angle and the arc length it produces are numerically the same, with no leftover conversion factor cluttering the algebra. Either unit measures the same thing the rest of this section is building toward: cosine is the horizontal coordinate the swept angle lands on, once that angle is drawn starting from a circle of radius 1.
Cosine as a coordinate
Theorem
The Unit Circle Definition of Sine and Cosine
Recalled from College Algebra · The unit circle definitions — first taught there.
Let \theta be an angle in standard position, and let (x, y) be the point where its terminal side intersects the unit circle. Then \cos\theta = x \qquad\text{and}\qquad \sin\theta = y.
Take an acute angle \theta, so its terminal side crosses the unit circle at a point (x,y) in the first quadrant. Drop a perpendicular segment straight down from (x,y) to the x-axis. That segment, the x-axis, and the terminal side form a right triangle with:
- hypotenuse = 1 (the radius of the unit circle),
- the side adjacent to \theta running along the x-axis, with length x,
- the side opposite \theta standing vertically, with length y.
By the right-triangle definition already in hand, \cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{x}{1} = x, \qquad \sin\theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{y}{1} = y. So for every acute angle, the coordinate definition and the right-triangle definition hand back the exact same numbers — the unit circle isn't replacing the old definition, it's the same idea with the hypotenuse fixed at 1 so "adjacent" and "opposite" collapse into plain x and y.
For \theta in any other quadrant, drop the same perpendicular from (x,y) to the x-axis and let \theta' be the resulting reference angle — the acute angle between the terminal side and the x-axis. That perpendicular still forms a right triangle with hypotenuse 1, legs |x| and |y|, and angle \theta', so the acute case already proven gives \cos\theta' = |x| and \sin\theta' = |y|. The sign is supplied automatically, not separately argued: x and y are the point's actual (signed) coordinates, negative exactly in the quadrants where the terminal side puts them, so \cos\theta = x and \sin\theta = y hold with their correct sign built in — nothing about the coordinate definition ever stops applying, only the right-triangle language ("opposite," "adjacent") stops making sense once \theta passes 90°.
Picture sunlight shining straight down onto the circle, casting the shadow of the moving point onto the ground below. That shadow's position along the ground is \cos\theta, the cosine of whatever angle the point has swept out so far. However far the point travels around the rim, its shadow can never fall past the edges of the circle's own diameter. There's simply no point on the circle whose shadow lands outside it. That physical ceiling on the shadow is the same ceiling that keeps every cosine value trapped between -1 and 1, no exceptions.
Because (\cos\theta, \sin\theta) is a point on a circle of radius 1, \cos\theta can never fall outside [-1, 1] — no angle, no matter how it's chosen, produces a coordinate past the rim of that circle. Two consequences of the definition are worth carrying forward: the sign of \cos\theta flips with the quadrant the terminal side lands in (positive where x>0, negative where x<0), and \cos(-\theta) = \cos\theta — swinging the same angle below the axis instead of above it reaches a point with the same x-coordinate, so cosine is an even function. That [-1,1] ceiling is exactly why the ratio \frac{\mathbf u\cdot\mathbf v}{|\mathbf u||\mathbf v|} in the angle-between-vectors formula is always safe to feed into \cos^{-1} — whatever \mathbf u and \mathbf v are, that ratio collapses to a cosine, and a cosine is always in range.
Cosine as an x-coordinate
Drag \theta around the circle. Watch \cos\theta track the point's x-coordinate on the reference triangle — and check that it never leaves [-1, 1].
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Check your understanding (counts toward your progress)
Solve c^2 = a^2+b^2-2ab\cos C for \cos C.
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Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.
Why
Isolate the cosine term: 2ab\cos C = a^2+b^2-c^2, then divide both sides by 2ab to get \cos C = \frac{a^2+b^2-c^2}{2ab}.
Correct answer
Also accepted:
Why
Isolate \cos C by moving every other term to the other side of the equation, then divide by what's left multiplying it.
A triangle has sides a=6, b=8, c=9, with C opposite side c. Find C in degrees.
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Why
\cos C = \frac{a^2+b^2-c^2}{2ab} = \frac{36+64-81}{96} = \frac{19}{96} \approx 0.198, so C = \cos^{-1}(0.198) \approx 78.6°.
Correct answer
Also accepted:
Why
Since you know all three sides and want the angle opposite c, the Law of Cosines solved for \cos C is the tool to reach for.
Since cosine is an even function, \cos(-\theta) equals
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-\cos\theta is what an ODD function would give — that's the property sine has, not cosine.
\sin\theta is a completely different function's value; negating the angle doesn't turn a cosine into a sine.
Same issue — negating \theta never swaps cosine for a (negated) sine value.
Why
Reflecting \theta below the x-axis to -\theta lands on a point with the same x-coordinate, so \cos(-\theta) = \cos\theta for every angle.
Evenness means \cos(-\theta)=\cos(\theta) -- picture the cosine graph and what reflecting \theta across 0 does to it.
Practice: Cosine as a coordinate, drilled (5 problems · doesn’t affect your progress)
On the unit circle, \cos\theta equals the ___ of the point where \theta's terminal side meets the circle. (For reference, the choices given are: x-coordinate; y-coordinate; radius; slope.)
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That's the definition of \sin\theta, not \cos\theta — the y-coordinate is sine's job; cosine reads the horizontal coordinate.
The radius is fixed at 1 for every angle on the unit circle — it never changes with \theta, so it can't be what cosine reads off.
The slope of the terminal side is \sin\theta / \cos\theta (that's \tan\theta), not \cos\theta by itself.
Why
The unit-circle definition sets \cos\theta = x and \sin\theta = y for the point (x,y) where the terminal side crosses the circle.
x-coordinate
Convert \frac{5\pi}{6} radians to degrees. (Work through it specifically for \frac{5\pi}{6}.)
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Why
Multiply by \frac{180°}{\pi}: \frac{5\pi}{6}\cdot\frac{180°}{\pi} = \frac{5(180°)}{6} = 150°.
Correct answer
Also accepted:
Why
In which quadrant is \cos\theta negative while \sin\theta is positive? (For reference, the choices given are: I; II; III; IV.)
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In Quadrant I both coordinates of the terminal point are positive, so both cosine and sine are positive there.
In Quadrant III both coordinates are negative — sine would be negative too, not positive.
In Quadrant IV the x-coordinate is positive, so cosine is positive there, not negative.
Why
Quadrant II has x<0 and y>0 for every terminal point that lands in it, so \cos\theta<0 while \sin\theta>0.
Recall the sign pattern of cosine and sine in each of the four quadrants, going counterclockwise from quadrant I.
II
Find \cos\!\left(\frac{4\pi}{3}\right) exactly, as a fraction (no decimal). (Work through it specifically for \cos\!\left(\frac{4\pi}{3}\right).)
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Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.
Why
\frac{4\pi}{3} = 240°, which is 60° into Quadrant III, so its reference angle is 60° and cosine is negative there: \cos 240° = -\cos 60° = -\frac{1}{2}.
Correct answer
Also accepted:
Why
Find the reference angle for \frac{4\pi}{3} and determine which quadrant it lands in to get the correct sign.
Since cosine is an even function, \cos(-\theta) equals (For reference, the choices given are: -\cos\theta; \cos\theta; \sin\theta; -\sin\theta.)
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-\cos\theta is what an ODD function would give — that's the property sine has, not cosine.
\sin\theta is a completely different function's value; negating the angle doesn't turn a cosine into a sine.
Same issue — negating \theta never swaps cosine for a (negated) sine value.
Why
Reflecting \theta below the x-axis to -\theta lands on a point with the same x-coordinate, so \cos(-\theta) = \cos\theta for every angle.
Evenness means \cos(-\theta)=\cos(\theta) -- picture the cosine graph and what reflecting \theta across 0 does to it.
\cos\theta
The Law of Cosines: three sides in, one angle out
Think of two roads leaving the same town at some angle to each other, one of length a and one of length b. A third road is built straight across to connect their far ends, and that connecting road is side c. If the two roads left town at a perfect right angle, the Pythagorean theorem alone would tell you the length of the connector. Any other angle makes the connector a little shorter or longer than that right-angle answer. The -2ab\cos C term is precisely that correction. It shrinks c when the roads lean toward each other, since C is acute and the correction term subtracts. It stretches c when the roads splay apart, since C is obtuse and \cos C itself goes negative, so the correction term adds.
For a triangle with sides a, b, c opposite angles A, B, C:
c^2 = a^2 + b^2 - 2ab\cos C.Solved for the angle instead of the side, this becomes a direct way to recover C from three known side lengths alone:
\cos C = \frac{a^2+b^2-c^2}{2ab}.Set C = 90° and the correction term -2ab\cos C vanishes entirely (since \cos 90° = 0), leaving the ordinary Pythagorean theorem — the Law of Cosines is the Pythagorean theorem's general-triangle upgrade, with a cosine supplying exactly the correction a non-right angle needs.
Theorem
The Law of Cosines
Recalled from College Algebra · Law of Cosines: SAS and SSS triangles — first taught there.
For any triangle with sides a, b, c opposite angles A, B, C: c^2 = a^2 + b^2 - 2ab\cos C
Place triangle ABC with angle C at vertex C, flanked by sides CA = b and CB = a; the side opposite C is AB = c. Drop an altitude from vertex B down to line CA, meeting it at point D, with height h = BD.
Triangle BDC is a right triangle with hypotenuse CB = a and angle C at vertex C, so CD = a\cos C, \qquad h = a\sin C. (If C is obtuse, \cos C is negative and D falls past C, outside segment CA — the signed length CD = a\cos C still comes out correct, so nothing below needs to change for that case.)
That leaves AD = CA - CD = b - a\cos C. Triangle ABD has its right angle at D and hypotenuse AB = c, so the Pythagorean theorem gives c^2 = AD^2 + h^2 = (b - a\cos C)^2 + (a\sin C)^2. Expand and group the a^2 terms together: c^2 = b^2 - 2ab\cos C + a^2\cos^2 C + a^2\sin^2 C = b^2 - 2ab\cos C + a^2\left(\cos^2 C + \sin^2 C\right). The Pythagorean identity collapses \cos^2 C + \sin^2 C to 1: c^2 = a^2 + b^2 - 2ab\cos C. Relabeling the same argument around the other two vertices gives the A- and B-rotations.
Sanity check: set C = 90°. Then \cos C = 0, the whole correction term vanishes, and what's left is c^2 = a^2 + b^2 — the Pythagorean theorem, exactly as it must be, since a right triangle is just the C = 90° special case of every triangle.
Recipe Find a triangle's angle from its three side lengths Use when: you know all three side lengths of a triangle and need the angle opposite one of them, with no angle given directly Goal: the measure of the angle opposite the side you named, without ever drawing the triangle to scale
Quick recipe
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1
Name the angle you want C, and call the side directly opposite it c. The other two sides get the letters a and b, in either order — it doesn't matter which is which as long as neither is c.
Example
A triangle has sides 6, 8, and 9, and you want the angle opposite the side of length 9. Call that angle C, so c = 9; the other two sides are a = 6 and b = 8.
Why: the Law of Cosines always subtracts the squared side opposite the angle you're solving for — mislabeling which side is c swaps in the wrong number at that step.
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2
Start from c^2 = a^2+b^2-2ab\cos C and isolate \cos C before substituting any numbers: \cos C = \frac{a^2+b^2-c^2}{2ab}
Example
With a=6, b=8, c=9, this is the formula you're about to fill in: \cos C = \frac{6^2+8^2-9^2}{2(6)(8)}
Why: solving for \cos C symbolically first, instead of substituting into c^2=a^2+b^2-2ab\cos C and rearranging afterward, avoids untangling a fraction from inside an equation.
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3
Compute the numerator and denominator separately, then divide to get a decimal.
Example
\cos C = \frac{36+64-81}{96} = \frac{19}{96} \approx 0.1979- the numerator a^2+b^2-c^2 comes out positive — \cos C>0, so C will be acute (less than 90°).
- the numerator comes out negative — \cos C<0, so C will be obtuse (more than 90°).
Why: knowing which sign to expect catches an arithmetic slip before you even take the inverse cosine, rather than after.
-
4
Apply the inverse cosine to the decimal from the last step to recover the angle itself.
Example
C = \cos^{-1}(0.1979) \approx \boxed{78.6°}Why: \cos^{-1} is the one operation that turns a cosine value back into an angle measure — the previous steps only ever produced \cos C, not C.
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5
Compare a^2+b^2 to c^2: if they're equal, the triangle is right and C should land exactly at 90°; otherwise the sign from Step 3 should match whether C came out above or below 90°. Also confirm 0°<C<180°, the only range a triangle's angle can occupy.
Example
Here a^2+b^2=100 and c^2=81, so a^2+b^2>c^2 — Step 3 predicted an acute angle, and 78.6° is indeed just under 90°, matching the prediction.
Why: this catches a mislabeled side or a sign error before the angle gets used anywhere downstream, such as in the dot-product formula this recipe feeds in Chapter 1.
Common mistakes
- Labeling the wrong side as c — the numerator's subtracted term must always be the side opposite the angle being solved for, not just "the third side."
- Forgetting the 2ab in the denominator, or building it from the wrong two sides.
- Trusting a computed \cos C without checking its sign against the acute/obtuse prediction from Step 3.
- Reporting an angle outside 0°–180°, which usually means a mode error (degrees vs. radians) on the calculator's \cos^{-1} key, not a real triangle.
Worked example
Find an angle from three side lengths
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1
A triangle has sides a = 6, b = 8, and c = 9, with C the angle opposite side c — the same recipe as above, run start to finish on these exact numbers. Find C.
-
2c is opposite C, so c² is the one that gets subtracted
Solve the Law of Cosines for \cos C and substitute the three sides: \cos C = \frac{a^2+b^2-c^2}{2ab} = \frac{6^2+8^2-9^2}{2(6)(8)}
c is opposite C, so c² is the one that gets subtracted
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3
Simplify the numerator and denominator separately: \cos C = \frac{36+64-81}{96} = \frac{19}{96} \approx 0.1979
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4
Undo the cosine with \cos^{-1}: C = \cos^{-1}(0.1979) \approx \boxed{78.6°} Check: C isn't 90°, and indeed 19/96 \ne 0 — the sides don't satisfy the Pythagorean relation, so the angle correctly comes out slightly less than a right angle rather than exactly one.
Loading interactive…
Worked example
Convert, then read an exact cosine value off the circle
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1
Find \cos\!\left(\dfrac{11\pi}{6}\right) exactly.
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2
Convert the radian measure to degrees to locate the angle: \frac{11\pi}{6}\cdot\frac{180°}{\pi} = \boxed{330°}
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3
330° sits in Quadrant IV, 30° short of a full turn — its reference angle is \theta' = 360° - 330° = 30°.
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4
Look up the exact value at the 30° reference angle, \cos 30° = \frac{\sqrt3}{2}, then apply the Quadrant IV sign (cosine is positive there, since x>0 in that quadrant): \cos\!\left(\frac{11\pi}{6}\right) = \textcolor{#4f46e5}{+}\frac{\sqrt3}{2} = \boxed{\frac{\sqrt3}{2}}
Loading interactive…
Check your understanding (counts toward your progress)
Find \cos\!\left(\frac{4\pi}{3}\right) exactly, as a fraction (no decimal).
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.
Why
\frac{4\pi}{3} = 240°, which is 60° into Quadrant III, so its reference angle is 60° and cosine is negative there: \cos 240° = -\cos 60° = -\frac{1}{2}.
Correct answer
Also accepted:
Why
Find the reference angle for \frac{4\pi}{3} and determine which quadrant it lands in to get the correct sign.
On the unit circle, \cos\theta equals the ___ of the point where \theta's terminal side meets the circle.
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
That's the definition of \sin\theta, not \cos\theta — the y-coordinate is sine's job; cosine reads the horizontal coordinate.
The radius is fixed at 1 for every angle on the unit circle — it never changes with \theta, so it can't be what cosine reads off.
The slope of the terminal side is \sin\theta / \cos\theta (that's \tan\theta), not \cos\theta by itself.
Why
The unit-circle definition sets \cos\theta = x and \sin\theta = y for the point (x,y) where the terminal side crosses the circle.
Which of these numbers could NOT be a value of \cos\theta for any angle \theta?
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0.5 is achievable — for instance \cos 60° = 0.5 exactly.
Cosine can absolutely be negative — it's negative throughout Quadrants II and III, so -0.9 is a value it actually takes.
1 is achievable at the extreme: \cos 0° = 1, when the terminal point sits right at (1,0).
Why
Every point on the unit circle has x-coordinate between -1 and 1, so \cos\theta can never exceed that range in either direction — 1.2 is outside the circle entirely.
In which quadrant is \cos\theta negative while \sin\theta is positive?
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In Quadrant I both coordinates of the terminal point are positive, so both cosine and sine are positive there.
In Quadrant III both coordinates are negative — sine would be negative too, not positive.
In Quadrant IV the x-coordinate is positive, so cosine is positive there, not negative.
Why
Quadrant II has x<0 and y>0 for every terminal point that lands in it, so \cos\theta<0 while \sin\theta>0.
Recall the sign pattern of cosine and sine in each of the four quadrants, going counterclockwise from quadrant I.
Practice: The Law of Cosines, start to finish (4 problems · doesn’t affect your progress)
The ratio \frac{\mathbf u\cdot\mathbf v}{|\mathbf u||\mathbf v|} in the angle-between-vectors formula can never exceed which value in absolute value? (For reference, the choices given are: 0; 1; 2; It depends on the vectors' components.)
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0 would mean the two vectors are always perpendicular — that's only true for special pairs, not every pair of vectors.
2 is too generous a bound — no cosine value ever reaches that high, no matter how the vectors are chosen.
The individual dot product and magnitudes do depend on the vectors, but this particular ratio always collapses to a cosine value, and every cosine obeys the same fixed bound regardless of which vectors produced it.
Why
That ratio is defined to equal \cos\theta for the angle \theta between \mathbf u and \mathbf v, and every cosine value is trapped in [-1,1] — so the ratio's absolute value can never exceed 1.
Recall the definition \cos\theta = \frac{\mathbf u\cdot\mathbf v}{|\mathbf u||\mathbf v|} and what the largest and smallest possible values of cosine are.
1
Solve c^2 = a^2+b^2-2ab\cos C for \cos C. (Work through it specifically for c^2 = a^2+b^2-2ab\cos C and \cos C.)
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.
Why
Isolate the cosine term: 2ab\cos C = a^2+b^2-c^2, then divide both sides by 2ab to get \cos C = \frac{a^2+b^2-c^2}{2ab}.
Correct answer
Also accepted:
Why
Isolate \cos C by moving every other term to the other side of the equation, then divide by what's left multiplying it.
A triangle has sides a=6, b=8, c=9, with C opposite side c. Find C in degrees. (Work through it specifically for a=6 and b=8 and c=9.)
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Why
\cos C = \frac{a^2+b^2-c^2}{2ab} = \frac{36+64-81}{96} = \frac{19}{96} \approx 0.198, so C = \cos^{-1}(0.198) \approx 78.6°.
Correct answer
Also accepted:
Why
Since you know all three sides and want the angle opposite c, the Law of Cosines solved for \cos C is the tool to reach for.
Two sides of a triangle are a=10 and b=14, with an included angle C=50° between them. Find the third side c. (Work through it specifically for a=10 and b=14 and C=50°.)
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Why
c^2 = a^2+b^2-2ab\cos C = 100+196-2(10)(14)\cos 50° = 296 - 280(0.6428) \approx 116.0, so c \approx \sqrt{116.0} \approx 10.77.
Correct answer
Also accepted:
Why
You have two sides and the included angle, so apply the Law of Cosines directly for the third side c.
Where this goes next
Every piece reviewed here reassembles almost untouched in Chapter 1. The Law of Cosines applied to the triangle formed by \mathbf u, \mathbf v, and \mathbf u - \mathbf v — with the dot product standing in for the componentwise algebra — is exactly how the formula \mathbf u\cdot\mathbf v = |\mathbf u||\mathbf v|\cos\theta gets derived from scratch, rather than handed down as a rule to memorize. The [-1,1] ceiling on cosine noted above is the same fact that later guarantees the Cauchy–Schwarz inequality, one of this course's first real theorems. None of that requires new trigonometry beyond what's above — only a new willingness to see \mathbf u, \mathbf v, and the triangle between them where, before, there was only a triangle.
What's next
With angle and cosine pinned down in the notation this course will use, the next section turns to a different piece of bookkeeping — writing a sum of many terms compactly with \Sigma-notation, the shorthand every later definition (dot products, matrix entries, determinants) leans on to avoid writing out ten terms by hand.