Contents / Precalculus / Analytic Trigonometry

Chapter 5

Analytic Trigonometry

Angle formulas, laws of sines and cosines, inverse trig, equations, and substitution preview.

Introduction

Angle formulas, laws of sines and cosines, inverse trig, equations, and substitution preview.

5.1Double-angle formula

Double angle lets you express trig of twice the angle using only the original angle. Extremely useful for integration and solving equations.

Double-Angle Formulas

Theorem 5.1 (Double-angle formulas). For every angle $θ$ ,

sin 2 θ = 2 sin θ cos θ,

cos 2 θ = cos^{2} θ - sin^{2} θ = 2 cos^{2} θ - 1 = 1 - 2 sin^{2} θ,

tan 2 θ = \frac{2 tan θ}{1 - tan ^{2} θ},

the last wherever $tan θ$ and $tan 2 θ$ are defined.

Proof. These come from the angle addition formulas with $α = β = θ$ : for instance $sin (θ + θ) = sin θ cos θ + cos θ sin θ = 2 sin θ cos θ$ .∎

They are identities, so they hold for every $θ$ — extremely useful for integration and solving trig equations. The three forms of $cos 2 θ$ are all equal; pick whichever leaves you with only the quantity you already know: use $2 cos^{2} θ - 1$ if you know $cos θ$ , and $1 - 2 sin^{2} θ$ if you know $sin θ$ .

Intuition. Double-angle is a "zoom" button: you know the trig of one angle and want the trig of the angle twice as wide. Instead of measuring the wide angle from scratch, the formula rebuilds it from the pieces of the small one — like predicting how far a door swings open at $2 \times$ the push from how far it swings at $1 \times$ .

Example 5.2 (Knowing the cosine, in quadrant II). Find $sin 2 θ$ given $cos θ = - \frac{3}{5}$ , with $θ$ in quadrant II.

Solution. We need $sin θ$ . From the Pythagorean identity $sin^{2} θ = 1 - cos^{2} θ = 1 - \frac{9}{25} = \frac{16}{25}$ , so $sin θ = \pm \frac{4}{5}$ . In quadrant II sine is positive, so $sin θ = + \frac{4}{5}$ . Apply the formula: $sin 2 θ = 2 sin θ cos θ = 2 \cdot \frac{4}{5} \cdot (- \frac{3}{5}) = - \frac{24}{25}$ . Sanity check: $θ$ in QII means $θ \in (9 0^{\circ}, 18 0^{\circ})$ , so $2 θ \in (18 0^{\circ}, 36 0^{\circ})$ — the lower half of the circle, where sine is negative. Our answer $- \frac{24}{25}$ is negative. ✓□

Example 5.3 (Choosing the convenient form of $cos 2 θ$ ). Given $sin θ = \frac{4}{5}$ with $θ$ in quadrant II, find $cos 2 θ$ .

Solution. Because we know $sin θ$ directly, pick the form $cos 2 θ = 1 - 2 sin^{2} θ$ so no extra work is needed. $cos 2 θ = 1 - 2 (\frac{4}{5})^{2} = 1 - 2 \cdot \frac{16}{25} = 1 - \frac{32}{25} = - \frac{7}{25}$ . Sanity check via a different form: $cos θ = - \frac{3}{5}$ (QII), so $cos^{2} θ - sin^{2} θ = \frac{9}{25} - \frac{16}{25} = - \frac{7}{25}$ . Same answer. ✓□

Common Mistakes

Double-angle misapplication: Using sin(2θ)=2sinθ instead of 2sin cos; or confusing which double-angle form of cos 2θ to pick for simplification.
Sign when using 1-2sin²: Choose the form that matches known value and quadrant.
For tan 2θ denominator 1-tan²: Zero when tan θ = ±1, corresponding to 2θ = π/2 +kπ where tan undefined.
Forgetting these are identities not equations: True for all θ (where defined).
Using for half angle backwards: Half angle solves for half from double; different.
Calculator in degree when using π: Formulas assume radian or consistent, but identities hold either.

Exercises for §5.1 →

5.2Half-angle formula

Half-angle formulas let you "split" an angle and compute exact trig values when you know the full angle (useful for 15°, 22.5° etc).

Half-Angle Formulas

Theorem 5.4 (Half-angle formulas). For every angle $θ$ ,

sin (θ /2) = \pm \frac{1 - cos θ}{2}, cos (θ /2) = \pm \frac{1 + cos θ}{2},

tan (θ /2) = \frac{sin θ}{1 + cos θ} = \frac{1 - cos θ}{sin θ} .

Proof. These are derived from the $cos 2 α$ double-angle forms by setting $α = θ /2$ and solving. For example, from $cos θ = 1 - 2 sin^{2} (θ /2)$ we get $2 sin^{2} (θ /2) = 1 - cos θ$ , hence $sin (θ /2) = \pm \frac{1 - c o s θ}{2}$ .∎

The $\pm$ sign is not free: it is fixed by the quadrant of $θ /2$ (not of $θ$ ). The square root itself always returns a nonnegative number, so you attach the sign that sine or cosine actually has in the quadrant where $θ /2$ lands. The two $tan (θ /2)$ formulas need $θ$ to avoid the points where their denominators vanish ( $1 + cos θ \neq = 0$ , i.e. $θ \neq = π + 2 kπ$ , for the first; $sin θ \neq = 0$ for the second), but at least one of the two always works.

Intuition. If double-angle is the "zoom in" button, half-angle is "zoom out": you know a full angle and want the trig of the angle half as wide. That is how you get exact values for oddball angles like $1 5^{\circ}$ or $22. 5^{\circ}$ — you halve a friendly angle you already know ( $3 0^{\circ}$ , $4 5^{\circ}$ ) instead of memorizing a new one.

Example 5.5 (Find the exact value of $sin 22. 5^{\circ}$ .). Notice $22. 5^{\circ} = \frac{4 5 ^{\circ}}{2}$ , so take $θ = 4 5^{\circ}$ , where $cos 4 5^{\circ} = \frac{2}{2}$ .

Solution.

Choose the sign: $22. 5^{\circ}$ is in quadrant I, where sine is positive, so use $+$ .
Substitute: $sin 22. 5^{\circ} = \frac{1 - \frac{2}{2}}{2}$ .
Clear the inner fraction by writing $1 = \frac{2}{2}$ : the numerator is $\frac{2 - 2}{2}$ , so the expression is $\frac{2 - 2}{4} = \frac{2 - 2}{2}$ .
Sanity check: numerically $2 - 2 \approx 0.586 \approx 0.765$ , so the result is $\approx 0.383$ . A calculator gives $sin 22. 5^{\circ} \approx 0.383$ . ✓

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Common Mistakes

Half-angle sign choice: Always determine the quadrant of the half-angle before picking + or - in front of the square root.
Forgetting the 2 in denominator: (1 - cos)/2 inside the sqrt, not (1-cos)/sqrt2.
Using for double instead: These solve for half; double angle goes the other way.
tan half two forms: Both equal; choose the one avoiding division by small number.
Domain: θ/2 must be in domain where original defined, and sqrt arg ≥0 which it is by identities.
Calculator for 22.5 without exact: Loses the nice radical form.

Exercises for §5.2 →

5.3Sum-to-product identity

These identities collapse sums of sines or cosines into products. Magic for solving equations and simplifying waves (beats).

Sum-to-Product and Product-to-Sum Identities

Theorem 5.6 (Sum-to-product identities). For all angles $A$ and $B$ ,

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}) .

These turn sums into products, which matters because a product equals zero exactly when one of its factors is zero — so an equation like $sin A + sin B = 0$ becomes two easy equations. They hold for all $A$ and $B$ . (Setting $A = B$ recovers the double-angle formula, a useful consistency check: $sin A + sin A = 2 sin A$ , and the identity gives $2 sin A cos 0 = 2 sin A$ . ✓)

Intuition. Adding two sine waves of nearly equal frequency is like two singers holding almost the same note: instead of hearing two tones you hear one tone that slowly swells and fades. The identity is the math of that swelling — it rewrites the sum as one wave (the average frequency) times a slow envelope (the half-difference frequency). That throb is what musicians call "beats."

Example 5.7 (Evaluate $sin 7 5^{\circ} + sin 1 5^{\circ}$ exactly.). Use $sin A + sin B = 2 sin (\frac{A + B}{2}) cos (\frac{A - B}{2})$ with $A = 7 5^{\circ}$ , $B = 1 5^{\circ}$ .

Solution.

Compute the half-sum and half-difference: $\frac{A + B}{2} = \frac{9 0 ^{\circ}}{2} = 4 5^{\circ}$ and $\frac{A - B}{2} = \frac{6 0 ^{\circ}}{2} = 3 0^{\circ}$ .
Substitute the known values: $2 sin 4 5^{\circ} cos 3 0^{\circ} = 2 \cdot \frac{2}{2} \cdot \frac{3}{2} = \frac{2 \cdot 3}{2} = \frac{6}{2}$ .
Sanity check: $\frac{6}{2} \approx \frac{2.449}{2} \approx 1.225$ , while directly $sin 7 5^{\circ} + sin 1 5^{\circ} \approx 0.966 + 0.259 = 1.225$ . ✓

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Common Mistakes

Sum-to-product arithmetic: (A+B)/2 and (A-B)/2 must be computed carefully; mixing up which is which ruins the identity application.
Sign in cos A - cos B: Has the minus in front of the sin sin term.
Forgetting product-to-sum the other way: Sometimes needed too.
Using when A or B make arg undefined: Check domains.
Calculator for exact: Loses the simplified radical.
Confusing with double angle: Similar looking but different (A=B case recovers double).

Exercises for §5.3 →

5.4Sum and difference formula

The addition formulas let us compute trig for sums and differences of known angles, giving exact values for 15°, 75°, etc., and are the basis for many identities.

Sum and Difference Formulas

Theorem 5.8 (Angle sum and difference formulas). For all angles $A$ and $B$ ,

sin (A \pm B) = sin A cos B \pm cos A sin B,

cos (A \pm B) = cos A cos B \mp sin A sin B,

tan (A \pm B) = \frac{tan A \pm tan B}{1 \mp tan A tan B} .

These are proved using the distance formula on unit-circle points or using Euler's formula with complex exponentials. Read the "top/bottom" signs carefully: in $cos (A \pm B)$ the sign in front of the $sin A sin B$ term is opposite to the sign in the argument (that is what $\mp$ means), so $cos (A - B) = cos A cos B + sin A sin B$ . In the $tan$ formula the denominator can vanish — that happens precisely when $A \pm B = \frac{π}{2} + kπ$ , exactly where $tan (A \pm B)$ is undefined, so the formula is faithful to the function.

They let us find exact values for angles like $1 5^{\circ} = 4 5^{\circ} - 3 0^{\circ}$ and expand expressions.

Intuition. Think of angles as amounts of turning. If you know exactly how "sideways" and "up" a $4 5^{\circ}$ turn leaves you and the same for a $3 0^{\circ}$ turn, these formulas tell you where a $4 5^{\circ} - 3 0^{\circ} = 1 5^{\circ}$ turn leaves you — no new measurement needed. Master three friendly angles ( $3 0^{\circ}, 4 5^{\circ}, 6 0^{\circ}$ ) and you can build many others for free.

Example 5.9 (Find the exact value of $sin 1 5^{\circ}$ .). Write $1 5^{\circ}$ as a difference of known angles: $1 5^{\circ} = 4 5^{\circ} - 3 0^{\circ}$ .

Solution.

Apply $sin (A - B) = sin A cos B - cos A sin B$ with $A = 4 5^{\circ}$ , $B = 3 0^{\circ}$ : $sin 1 5^{\circ} = sin 4 5^{\circ} cos 3 0^{\circ} - cos 4 5^{\circ} sin 3 0^{\circ}$ .
Substitute known values: $(\frac{2}{2}) (\frac{3}{2}) - (\frac{2}{2}) (\frac{1}{2}) = \frac{6}{4} - \frac{2}{4} = \frac{6 - 2}{4}$ .
Sanity check: $\frac{6 - 2}{4} \approx \frac{2.449 - 1.414}{4} \approx \frac{1.035}{4} \approx 0.259$ , and $sin 1 5^{\circ} \approx 0.259$ . ✓

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Common Mistakes

Sum/difference sign errors: The minus in cos(A-B) = cosA cosB + sinA sinB is frequently flipped.
For tan(A+B) the denominator sign: 1 - tanA tanB for + ; 1 + for - .
Forgetting mp signs: In cos(A±B) the sin term flips sign with the ± .
Using for half/double instead: These are for arbitrary A B; double is special case.
Domain where tan defined in tan formula: If 1 - tanA tanB =0 the sum hits asymptote.
Calculator in wrong mode for exact problems: Use exact values of 30 45 60, not decimal approx.

Exercises for §5.4 →

5.5Law of sines

The law of sines generalizes the right-triangle ratios to any triangle, allowing us to solve for missing sides or angles when we have AAS, ASA or the tricky SSA.

Law of Sines

Theorem 5.10 (Law of Sines). In any triangle $A BC$ with sides $a, b, c$ opposite angles $A, B, C$ respectively,

\frac{a}{sin A} = \frac{b}{sin B} = \frac{c}{sin C} = 2 R,

where $R$ is the circumradius.

This is derived by dropping an altitude and using area formulas or by using the extended unit-circle definition. The law applies to any triangle (not just right triangles), and the shared ratio equals $2 R$ , the diameter of the triangle's circumscribed circle.

Method 5.11 (Solving a triangle).

Label the sides $a, b, c$ so that each is opposite the angle of the matching letter.
Look for a known side paired with its opposite angle. AAS and ASA give one (find the third angle from the $18 0^{\circ}$ sum first), and so does SSA.
With such a pair, use the Law of Sines as a conversion rate to find the remaining parts.
With no such pair — SAS and SSS — start with the Law of Cosines instead, then finish with the Law of Sines.

Intuition. The Law of Sines says a triangle's sides and their opposite angles rise and fall together: the biggest side faces the biggest angle, and the ratio of each side to the sine of its facing angle is the same throughout the triangle. Know one matched side–angle pair and it acts as a conversion rate for finding any other side or angle.

Example 5.12 (Solve the triangle with $A = 4 0^{\circ}$ , $a = 8$ , $B = 6 0^{\circ}$ : find side $b$ and angle $C$ .). This is AAS. Find the third angle first: $C = 18 0^{\circ} - 4 0^{\circ} - 6 0^{\circ} = 8 0^{\circ}$ .

Solution.

Use the matched pair $\frac{a}{s i n A}$ as the conversion rate: $\frac{a}{s i n A} = \frac{b}{s i n B}$ , i.e. $\frac{8}{s i n 4 0 ^{\circ}} = \frac{b}{s i n 6 0 ^{\circ}}$ .
Solve for $b$ : $b = \frac{8 s i n 6 0 ^{\circ}}{s i n 4 0 ^{\circ}} \approx \frac{8 \cdot 0.8660}{0.6428} \approx \frac{6.928}{0.6428} \approx 10.8$ . (Keep full precision until the end; round once.)
Sanity check: the angles from smallest to largest are $A = 4 0^{\circ} < B = 6 0^{\circ} < C = 8 0^{\circ}$ , so the sides should order the same way: $a = 8 < b \approx 10.8 < c$ . Since $b > a$ matches $B > A$ , the result is consistent. ✓

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Common Mistakes

Ambiguous case (SSA): When using Law of Sines with two sides and angle opposite one of them, there may be zero, one, or two possible triangles. Always check the height h = b sin A against the given side.
Calculator mode: Degrees or radians must match the problem (usually degrees in triangle problems).
Assuming unique triangle: SSA can be ambiguous; draw the picture.
Forgetting to find the third angle first: In AAS/ASA you get third angle from 180 sum before applying law.
Rounding too early: Keep precision until final answer.
Using law of sines for SAS or SSS: Those are for law of cosines.

Exercises for §5.5 →

5.6Law of cosines

The law of cosines is Pythagoras with a correction term for the included angle. It lets us solve SAS and SSS triangles.

Law of Cosines

Theorem 5.13 (Law of Cosines). For any triangle,

c^{2} = a^{2} + b^{2} - 2 ab cos C,

and cyclically for the other two sides. Here $C$ must be the angle opposite side $c$ — equivalently, the angle included between sides $a$ and $b$ .

When $C = 9 0^{\circ}$ , $cos C = 0$ and the correction term vanishes, recovering Pythagoras $c^{2} = a^{2} + b^{2}$ . The law is what you reach for in the SAS case (two sides plus the included angle) to find the third side; rearranged, it also settles the SSS case.

Corollary 5.14 (An angle from three sides).

cos C = \frac{a ^{2} + b ^{2} - c ^{2}}{2 ab} .

This SSS form is unambiguous: $arccos$ returns a single angle in $[0^{\circ}, 18 0^{\circ}]$ , and a negative right-hand side correctly yields an obtuse angle — so the Law of Cosines never has the ambiguity that plagues SSA.

Intuition. The Law of Cosines is Pythagoras wearing a correction collar. The term $- 2 ab cos C$ measures how far the corner angle strays from a right angle. If $C$ is acute, $cos C > 0$ , the correction is subtracted, and the opposite side $c$ comes out shorter than Pythagoras would predict; if $C$ is obtuse, $cos C < 0$ , the correction adds length and $c$ comes out longer.

Example 5.15 (SAS: finding the third side). Two sides measure $a = 5$ and $b = 7$ with included angle $C = 6 0^{\circ}$ . Find side $c$ .

Solution. Apply the law with $C$ between $a$ and $b$ : $c^{2} = a^{2} + b^{2} - 2 ab cos C = 5^{2} + 7^{2} - 2 \cdot 5 \cdot 7 cos 6 0^{\circ}$ . Since $cos 6 0^{\circ} = \frac{1}{2}$ : $c^{2} = 25 + 49 - 70 \cdot \frac{1}{2} = 74 - 35 = 39$ , so $c = 39 \approx 6.24$ . Sanity check: $c$ should sit between $∣ b - a ∣ = 2$ and $b + a = 12$ (triangle inequality), and it does. ✓□

Example 5.16 (SSS: finding an angle). Find angle $C$ given sides $a = 3$ , $b = 4$ , $c = 6$ .

Solution. Rearrange for the cosine of the angle opposite $c$ : $cos C = \frac{a ^{2} + b ^{2} - c ^{2}}{2 ab} = \frac{9 + 16 - 36}{2 \cdot 3 \cdot 4} = \frac{- 11}{24} \approx - 0.4583$ . Take the inverse cosine: $C = arccos (- 0.4583) \approx 117. 3^{\circ}$ . Sanity check: the cosine came out negative, so $C$ is obtuse — and indeed $c^{2} = 36 > a^{2} + b^{2} = 25$ , which is exactly the condition for the angle opposite $c$ to exceed $9 0^{\circ}$ . ✓□

Common Mistakes

Law of Cosines for angle: When solving for angle, the numerator a²+b²-c² can be negative (obtuse); arccos handles it correctly if you keep the sign.
Using wrong angle: The cos C is for the angle opposite side c (included between a and b).
Forgetting the 2ab: The formula has the 2.
Mixing with law of sines: Use cos for SAS/SSS, sines for AAS/ASA/SSA.
Calculator in rad vs deg for arccos: arccos returns radians or degrees per mode; convert if needed for answer.
Rounding intermediate: Keep exact or high precision for arccos input.

Exercises for §5.6 →

5.7Inverse trigonometric functions

Inverse trig recover the angle from a known sine, cosine or tangent value. Their ranges are restricted to make them functions (one-to-one).

Inverse Trigonometric Functions

Definition 5.17 (Inverse trigonometric functions). The inverse trig functions recover an angle from a known trig value within a restricted principal range where the function is one-to-one.

$arcsin x$ : domain $[- 1, 1]$ , range $[- \frac{π}{2}, \frac{π}{2}]$
$arccos x$ : domain $[- 1, 1]$ , range $[0, π]$
$arctan x$ : domain all reals, range $(- \frac{π}{2}, \frac{π}{2})$
$arccsc, arcsec, arccot$ defined similarly with their own conventions.

The range restriction is the whole point. Sine (say) is not one-to-one on all of $R$ — infinitely many angles share the value $\frac{1}{2}$ — so "the angle whose sine is $x$ " has no single answer until we pin it to one slice. We choose the slice where the function is one-to-one and still hits every output value: $[- \frac{π}{2}, \frac{π}{2}]$ for $arcsin$ , $[0, π]$ for $arccos$ , $(- \frac{π}{2}, \frac{π}{2})$ for $arctan$ . These are essential for solving trig equations ("what angle has this sine?") and in calculus for integrals and derivatives.

Intuition. Sine is like a revolving door: many different pushes (angles) land you at the same spot. $arcsin$ has to undo that, but "which push?" is ambiguous — so it always reports the answer from one agreed-upon quarter-turn of the door. That is the "principal value": one consistent representative, not the only true angle.

Example 5.18 (Evaluating a sum of inverse trig values). Compute $arctan 1 + arcsin 0.5$ .

Solution. $arctan 1 = \frac{π}{4}$ because $tan \frac{π}{4} = 1$ and $\frac{π}{4} \in (- \frac{π}{2}, \frac{π}{2})$ . $arcsin 0.5 = \frac{π}{6}$ because $sin \frac{π}{6} = \frac{1}{2}$ and $\frac{π}{6} \in [- \frac{π}{2}, \frac{π}{2}]$ . Add over a common denominator: $\frac{π}{4} + \frac{π}{6} = \frac{3 π}{12} + \frac{2 π}{12} = \frac{5 π}{12}$ . Sanity check: $\frac{5 π}{12} = 7 5^{\circ}$ , which is $4 5^{\circ} + 3 0^{\circ}$ — exactly the two angles we found. ✓□

Example 5.19 (The composition trap). Evaluate $arcsin (sin \frac{5 π}{6})$ .

Solution. It is tempting to answer $\frac{5 π}{6}$ , but $\frac{5 π}{6}$ is not in $arcsin$ 's range $[- \frac{π}{2}, \frac{π}{2}]$ , so that cannot be the output. First evaluate the inside: $sin \frac{5 π}{6} = \frac{1}{2}$ . Now $arcsin \frac{1}{2} = \frac{π}{6}$ , the angle in range with that sine. So $arcsin (sin \frac{5 π}{6}) = \frac{π}{6} \neq = \frac{5 π}{6}$ . The identity $arcsin (sin x) = x$ holds only when $x$ already lies in the principal range. ✓□

Common Mistakes

Range of inverse functions: arcsin range is [-π/2,π/2]; arccos [0,π]. Using the wrong branch or forgetting the range produces answers outside the principal value.
Calculator mode: arccos etc return in current angle mode; check.
arcsin(sin x) ≠ x for all x: Only true inside the principal range; outside it gives the co-terminal in range.
Domain restrictions: arcsin and arccos only defined for [-1,1]; arctan all reals.
Confusing arctan with 1/tan: arctan is inverse function, not reciprocal (which is cot).
Quadrant when solving: After arcsin you may need to add π - result for other quadrant solutions in equations.

Exercises for §5.7 →

5.8Trigonometric equation

Solving trig equations requires identities to reduce to basic forms, then inverse trig plus the periodic general solution, followed by checking for extraneous roots.

Trigonometric Equations

Method 5.20 (Solving a trigonometric equation).

Use identities to rewrite in terms of a single function.
Solve the basic equation $sin x = k$ or similar (use the inverse function plus the general solution below).
Find all solutions in the desired interval by adding the period(s).
Check for extraneous solutions introduced by squaring or other operations.

Proposition 5.21 (General solutions of the basic equations). For integer $m$ , and with $∣ k ∣ \leq 1$ required for sine and cosine:

$sin x = k$ : two families, $x = arcsin k + 2 mπ$ and $x = π - arcsin k + 2 mπ$ (the QI/QII pair with the same sine).
$cos x = k$ : $x = \pm arccos k + 2 mπ$ (cosine is even, so a value and its negative share the cosine).
$tan x = k$ : $x = arctan k + mπ$ (period $π$ , so one family suffices).

Intuition. Solving a trig equation is like asking "at what times does the Ferris wheel put me exactly 3 meters up?" Because the wheel keeps turning, there is not one answer but a repeating schedule. So you find the answers in one full turn, then tack on " $+$ every period" to capture all the rest.

Example 5.22 (Solve $2 sin^{2} x - sin x - 1 = 0$ for $x \in [0, 2 π)$ .). The equation is quadratic in $sin x$ . Substitute $u = sin x$ : $2 u^{2} - u - 1 = 0$ .

Solution.

Factor: $(2 u + 1) (u - 1) = 0$ , so $u = 1$ or $u = - \frac{1}{2}$ .
Back-substitute the first root: $sin x = 1 \Rightarrow x = \frac{π}{2}$ (the only solution in $[0, 2 π)$ ).
Back-substitute the second: $sin x = - \frac{1}{2}$ . Sine is negative in QIII and QIV; the reference angle is $\frac{π}{6}$ , giving $x = π + \frac{π}{6} = \frac{7 π}{6}$ and $x = 2 π - \frac{π}{6} = \frac{11 π}{6}$ .
Collect: $x = \frac{π}{2}, \frac{7 π}{6}, \frac{11 π}{6}$ — three solutions.
Sanity check: substitute $x = \frac{7 π}{6}$ : $sin = - \frac{1}{2}$ , so $2 \cdot \frac{1}{4} - (- \frac{1}{2}) - 1 = \frac{1}{2} + \frac{1}{2} - 1 = 0$ . ✓ (No squaring was used here, so no extraneous roots can appear; when you do square, always verify each candidate this way.)

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Common Mistakes

General solution omissions: Forgetting the ± or the period when writing the general solution from inverse trig leads to missing families of solutions.
Quadrant when solving equations: After taking arcsin or arccos you must add the other angle in the correct quadrant where the function has the same value.
Calculator mode: Solving trig equations numerically without ensuring degree/radian mode matches the problem statement.
Extraneous solutions: Squaring both sides of a trig equation (to get sin² or similar) can introduce solutions that do not satisfy the original.
Not using identities first: Trying to solve without reducing to single function leads to harder equations.
Forgetting to check solutions in original: Especially after squaring or multiplying by expressions that could be zero.

Exercises for §5.8 →

5.9Trigonometric substitution

Trig substitution is a change of variable that turns an awkward square root of a quadratic into a clean trig function, using the Pythagorean identities. It is the payoff of everything above: inverse trig, identities, and reference triangles all combine here. This is a preview of a technique developed fully in integral calculus.

Trigonometric Substitution (Preview)

Although more fully developed in calculus, trig substitution is the technique of writing $x = a tan θ$ , $x = a sin θ$ , or $x = a sec θ$ to simplify expressions involving $a^{2} - x^{2}$ , $x^{2} + a^{2}$ , or $x^{2} - a^{2}$ . Each substitution matches one Pythagorean identity, so the radical collapses to a single trig function.

Method 5.23 (Trigonometric substitution). Match the radical to the substitution whose identity collapses it:

$a^{2} - x^{2}$ with $x = a sin θ$ uses $1 - sin^{2} θ = cos^{2} θ$ .
$x^{2} + a^{2}$ with $x = a tan θ$ uses $1 + tan^{2} θ = sec^{2} θ$ .
$x^{2} - a^{2}$ with $x = a sec θ$ uses $sec^{2} θ - 1 = tan^{2} θ$ .

Choose $θ$ in the standard range for the corresponding inverse function so the new trig factor is nonnegative and the square root simplifies without an absolute value; e.g. for the sine substitution take $θ \in [- \frac{π}{2}, \frac{π}{2}]$ , where $cos θ \geq 0$ . After substituting, the integral becomes easier, and back-substitution using inverse trig or a reference triangle recovers the answer in $x$ .

Intuition. When something under a square root looks like a Pythagorean triple missing one leg, invent an angle whose trig ratios "name" the sides of that right triangle. The radical is just one side of the triangle in disguise, so replacing $x$ with a trig expression makes it collapse to a clean single function — and now you are integrating an ordinary sine or cosine instead of a stubborn root.

Example 5.24 (Evaluate $\int \frac{d x}{9 - x ^{2}}$ .). The form $a^{2} - x^{2}$ with $a = 3$ calls for $x = 3 sin θ$ , with $θ \in [- \frac{π}{2}, \frac{π}{2}]$ . Then $d x = 3 cos θ d θ$ .

Solution.

Simplify the radical: $9 - 9 sin^{2} θ = 9 cos^{2} θ = 3 cos θ$ (nonnegative because $cos θ \geq 0$ on the chosen range, so no absolute value is needed).
Substitute everything: $\int \frac{3 cos θ d θ}{3 cos θ} = \int d θ = θ + C$ .
Back-substitute: from $x = 3 sin θ$ we get $sin θ = \frac{x}{3}$ , so $θ = arcsin \frac{x}{3}$ . Thus the integral is $arcsin \frac{x}{3} + C$ .
Sanity check by differentiating: $\frac{d}{d x} arcsin \frac{x}{3} = \frac{1}{1 - ( x /3 ) ^{2}} \cdot \frac{1}{3} = \frac{1}{3 1 - x ^{2} /9} = \frac{1}{9 - x ^{2}}$ , the original integrand. ✓

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Common Mistakes

Choosing the wrong sub for the radical form: a²-x² -> sin; x²+a² -> tan; x²-a² -> sec.
Forgetting the dx in terms of dθ: dx = a sec² dθ for tan sub etc; must substitute it too.
Not drawing the reference triangle for back-sub: The triangle with opp=x, hyp=3 etc lets you express sinθ = x/3 etc without memorizing.
Assuming cos or tan always positive: Choose the range of θ so the sub functions are positive in the interval, or use abs.
Leaving answer in θ: Must back-sub to x (or arcsin(arctan etc)).
Forgetting +C in indefinite integral: Always.

Exercises for §5.9 →

About the practice questions. They check that you can carry out this chapter's computations correctly, and each one is graded on a single answer. They are not proof exercises: working through them confirms the mechanics, not that you could prove the results yourself. For that, re-read the statements above and try to reconstruct their proofs with the page closed.

Practice this chapter