Precalculus · Chapter 6

Chapter 6
Trigonometric identities

Lectures 22–23  |  many names for the same quantity

In this chapter
  1. Pythagorean identities and the symmetry of the circle
  2. Addition formulas, double angles, and substitution

An identity is an equation that is true for every value of the variable. \(x^{2}-1=(x-1)(x+1)\) is one; \(x^{2}-1=0\) is not, since it only holds at two particular values. The distinction matters: an equation is something you solve, an identity is something you use.

spot check
Which of these is an identity?

Trigonometry has an unusual number of identities, and there is a reason. Sine and cosine are the coordinates of a point on a circle, and a circle is an extremely symmetric object — you can reflect it in either axis, rotate it, or turn it inside out, and it lands on itself. Every one of those symmetries becomes an equation. That is all an identity is: a symmetry of the circle, written in algebra.

Lecture 22 — Pythagorean identities and the symmetry of the circle

Cartoon: the same quantity written several waysOne stick figure points at a board showing the same quantity written two different ways. Another asks which one is the answer. Which one isthe answer?sin² = 1 − cos²= (1−cos)(1+cos)
“All of them. That is the point.”

The identity you already have

The point \((\cos t,\sin t)\) lies on the unit circle, whose equation is \(x^{2}+y^{2}=1\). Substituting immediately gives the most-used identity in mathematics.

the Pythagorean identities $$\cos^{2}t+\sin^{2}t=1$$

Dividing through by \(\cos^{2}t\) gives \(1+\tan^{2}t=\sec^{2}t\); dividing instead by \(\sin^{2}t\) gives \(1+\cot^{2}t=\csc^{2}t\). The three are one fact in three costumes, and the second two are not worth memorising separately — derive them in the margin when you need them.

spot check

If \(\sin t=0.6\) and \(\cos t\) is positive, what is \(\cos t\)?

decimals, fractions, pi and sqrt( ) are all fine

The notation \(\cos^{2}t\) means \((\cos t)^{2}\), not \(\cos(t^{2})\). It is a bad convention and it is universal.

the relations that come free

Reciprocal: \(\csc t=\dfrac{1}{\sin t}\), \(\sec t=\dfrac{1}{\cos t}\), \(\cot t=\dfrac{1}{\tan t}\).

Quotient: \(\tan t=\dfrac{\sin t}{\cos t}\), \(\cot t=\dfrac{\cos t}{\sin t}\).

Even and odd: \(\cos(-t)=\cos t\) and \(\sin(-t)=-\sin t\), so \(\tan(-t)=-\tan t\).

Cofunction: \(\sin\!\left(\tfrac{\pi}{2}-t\right)=\cos t\) and \(\cos\!\left(\tfrac{\pi}{2}-t\right)=\sin t\).

spot check

\(\sin(-0.3)\) equals , and \(\cos\!\left(\tfrac{\pi}{2}-t\right)\) equals .

The cofunction relations are the two acute angles of a right triangle talking to each other: they add to \(\tfrac{\pi}{2}\), and what is opposite one is adjacent to the other. That is the whole proof, and it is why "co" appears in cosine, cotangent and cosecant — each is the function of the complementary angle.

Where the rest come from

The four points the circle’s symmetry relates A unit circle with a point at 40 degrees and its three mirror images: reflected in the vertical axis at 140 degrees, through the origin at 220 degrees, and in the horizontal axis at 320 degrees. Each reflection changes the sign of one or both coordinates, which is where the identities come from. t π − t π + t −t (cos t, sin t) (−cos t, sin t) (−cos t, −sin t) (cos t, −sin t) one picture, four identities: sin(π − t) = sin t, cos(π − t) = −cos t, sin(−t) = −sin t, and so on
Every identity in this lecture is a statement about one of these reflections. The circle is symmetric in both axes and about the origin, so each point has three companions, and each companion differs only in the signs of its coordinates — which is to say, in the signs of a cosine and a sine.
spot check

Reflecting the point \((\cos t,\sin t)\) across the \(y\)-axis gives the terminal point of . So \(\cos(\pi-t)=\) .

Reflect the point, read the identity

Move the point, then choose a reflection. The identity it produces is written underneath — and it is nothing more than a statement about which coordinate changed sign.

Simplifying, and proving

Two different jobs use these identities, and it helps to know which one you are doing.

example 6.1 — simplifying

Simplify \(\dfrac{\sin t}{\csc t}+\dfrac{\cos t}{\sec t}\) and \(\dfrac{\sec t-\cos t}{\tan t}\).

For the first, the reciprocals collapse: \(\dfrac{\sin t}{1/\sin t}=\sin^{2}t\), and likewise the second term is \(\cos^{2}t\). So the whole thing is \(\sin^{2}t+\cos^{2}t=1\).

For the second, put the numerator over a common denominator: $$\frac{\frac{1}{\cos t}-\cos t}{\tan t}=\frac{\frac{1-\cos^{2}t}{\cos t}}{\frac{\sin t}{\cos t}}=\frac{\sin^{2}t}{\cos t}\cdot\frac{\cos t}{\sin t}=\sin t.$$

The working rule when an expression looks hopeless: write everything in terms of sine and cosine. Almost every simplification in this lecture yields to that one move followed by ordinary fraction algebra from Chapter 0.

example 6.2 — proving an identity

Prove that \(\dfrac{\cos t}{1-\sin t}=\dfrac{1+\sin t}{\cos t}\).

Start on the left and multiply top and bottom by \(1+\sin t\) — the conjugate trick from Chapter 0, chosen because it will create a difference of squares: $$\frac{\cos t}{1-\sin t}\cdot\frac{1+\sin t}{1+\sin t}=\frac{\cos t(1+\sin t)}{1-\sin^{2}t}=\frac{\cos t(1+\sin t)}{\cos^{2}t}=\frac{1+\sin t}{\cos t}.$$

Two rules of etiquette for proofs like this. Work on one side only and transform it until it becomes the other; do not treat the identity as an equation and operate on both sides, since that assumes what you are trying to show. And when stuck, look for the move that manufactures \(1-\sin^{2}t\) or \(1-\cos^{2}t\), because that is the doorway to the Pythagorean identity.

Solving trigonometric equations

Now the other job. Here the equation is not an identity, and the question is which values of \(t\) make it true. Two things go wrong for students, and both are avoidable.

the two-step method
  1. Use identities and algebra to reduce to something of the form \(\sin t=k\), \(\cos t=k\) or \(\tan t=k\) — factoring if necessary.
  2. Find all solutions in one full turn, using the symmetry of the circle, then add multiples of the period.

Your calculator's \(\sin^{-1}\) returns one answer. The circle has two points at most heights, and the second one is the one everybody forgets.

spot check

How many solutions does \(\sin t=0.3\) have in \([0,2\pi)\)?

decimals, fractions, pi and sqrt( ) are all fine

example 6.3 — find every solution

Solve \(2\sin^{2}t-\sin t-1=0\) for \(0\le t<2\pi\).

It is a quadratic in \(\sin t\). Writing \(u=\sin t\): \(2u^{2}-u-1=(2u+1)(u-1)=0\), so \(\sin t=-\tfrac12\) or \(\sin t=1\).

\(\sin t=1\) at \(t=\tfrac{\pi}{2}\) only — the top of the circle is a single point.

\(\sin t=-\tfrac12\) means a height of \(-\tfrac12\), which happens at two points, in the third and fourth quadrants: \(t=\tfrac{7\pi}{6}\) and \(t=\tfrac{11\pi}{6}\).

So there are three solutions: \(\tfrac{\pi}{2},\ \tfrac{7\pi}{6},\ \tfrac{11\pi}{6}\). Drawing the horizontal line \(y=-\tfrac12\) across the circle and counting crossings is the reliable way to be sure you have them all.

Reading the solutions off the unit circleA unit circle with a horizontal dashed line at height minus one half, cutting the circle at two points in the lower half, and a radius drawn straight up to the single point at the top where the height is one. y = −1/27π/611π/6π/2, where sin t = 1sin t = 1 touches the circleat one point onlysin t = −1/2 cuts it at two,in quadrants III and IVthree solutions in one turn
Drawing the horizontal line at the required height and counting crossings is the reliable way to be sure you have every solution. A height strictly between \(-1\) and \(1\) always cuts twice; a height of exactly \(\pm1\) touches once; anything beyond misses entirely.
example 6.4 — a modelling equation

The depth of water in a harbour is \(h(t)=2.4\cos\!\left(\tfrac{\pi}{6}t\right)+5.6\) metres, \(t\) hours after midnight. A ship needs 4.5 m of water. Between what times is it safe to enter, in the first twelve hours?

Safe when \(2.4\cos\!\left(\tfrac{\pi}{6}t\right)+5.6\ge4.5\), that is \(\cos\!\left(\tfrac{\pi}{6}t\right)\ge-0.4583\).

The cosine drops below \(-0.4583\) between the two angles whose cosine equals it. \(\cos^{-1}(-0.4583)=2.0466\) rad, and by symmetry the other is \(2\pi-2.0466=4.2366\) rad.

So the unsafe stretch is \(2.0466\le\tfrac{\pi}{6}t\le4.2366\), giving \(3.91\le t\le8.09\) hours.

The ship can enter before about 3:55 a.m. or after about 8:05 a.m. Notice that the answer came out as an interval, and that the symmetry of the circle — not the calculator — supplied the second endpoint.

Harbour depth against time, with the unsafe window shadedA wave falling from 8 metres at midnight to a minimum of 3.2 metres at six in the morning and back. A dashed line at 4.5 metres cuts it at about 3.9 and 8.1 hours, and the stretch between those two times is shaded as too shallow. the ship needs 4.5 m3:558:05too shallow to enter246810468hours after midnight
The inequality becomes a picture: the curve is below the line exactly between the two crossing times. The second crossing came from the symmetry of the circle, not from the calculator, which only ever reports the first.
exercises 6.1
  1. Simplify: (a) \(\cos t\tan t\); (b) \(\dfrac{1-\cos^{2}t}{\sin t}\); (c) \(\sec t-\sec t\sin^{2}t\); (d) \(\dfrac{\cot t}{\csc t}\).
  2. If \(\sin t=\tfrac35\) and \(t\) is in the second quadrant, find \(\cos t\), \(\tan t\) and \(\sec t\) exactly.
  3. Prove \(\tan t+\cot t=\sec t\csc t\).
  4. Prove \(\dfrac{1+\sin t}{\cos t}+\dfrac{\cos t}{1+\sin t}=2\sec t\).
  5. Solve for \(0\le t<2\pi\): (a) \(2\cos t+\sqrt3=0\); (b) \(\tan^{2}t=3\); (c) \(2\sin^{2}t=\sin t\).
  6. Explain why \(\sin t=1.4\) has no solutions, without a calculator.
  7. A wheel-mounted reflector has height \(h(t)=0.4\cos(3.9t)+0.5\) metres. Find the first two times it is exactly 0.7 m above the ground.
  8. Use the circle to explain why \(\cos t=k\) has two solutions in one turn when \(-1<k<1\), one solution when \(k=\pm1\), and none otherwise.
  1. (a) \(\sin t\); (b) \(\sin t\); (c) \(\sec t(1-\sin^{2}t)=\sec t\cos^{2}t=\cos t\); (d) \(\dfrac{\cos t/\sin t}{1/\sin t}=\cos t\).
  2. \(\cos^{2}t=1-\tfrac{9}{25}=\tfrac{16}{25}\), and cosine is negative in the second quadrant, so \(\cos t=-\tfrac45\). Then \(\tan t=-\tfrac34\) and \(\sec t=-\tfrac54\).
  3. \(\dfrac{\sin t}{\cos t}+\dfrac{\cos t}{\sin t}=\dfrac{\sin^{2}t+\cos^{2}t}{\sin t\cos t}=\dfrac{1}{\sin t\cos t}=\sec t\csc t\).
  4. Common denominator \(\cos t(1+\sin t)\): the numerator is \((1+\sin t)^{2}+\cos^{2}t=1+2\sin t+\sin^{2}t+\cos^{2}t=2+2\sin t\). So the sum is \(\dfrac{2(1+\sin t)}{\cos t(1+\sin t)}=\dfrac{2}{\cos t}\).
  5. (a) \(\cos t=-\tfrac{\sqrt3}{2}\) at \(t=\tfrac{5\pi}{6},\tfrac{7\pi}{6}\). (b) \(\tan t=\pm\sqrt3\) at \(t=\tfrac{\pi}{3},\tfrac{2\pi}{3},\tfrac{4\pi}{3},\tfrac{5\pi}{3}\). (c) \(\sin t(2\sin t-1)=0\), giving \(t=0,\pi,\tfrac{\pi}{6},\tfrac{5\pi}{6}\). Do not divide by \(\sin t\) — that would throw away two of the four answers.
  6. \(\sin t\) is the height of a point on a circle of radius 1, so it can never exceed 1. There is no point at height 1.4 to find.
  7. \(\cos(3.9t)=0.5\), so \(3.9t=\tfrac{\pi}{3}\) or \(3.9t=2\pi-\tfrac{\pi}{3}=\tfrac{5\pi}{3}\), giving \(t=0.268\) s and \(t=1.342\) s.
  8. A vertical line at \(x=k\) cuts the circle twice when \(-1<k<1\), because the circle has two points at every such horizontal position; it touches once at \(k=\pm1\), the extreme left and right points; and misses entirely when \(|k|>1\), since no point of the circle is that far from the vertical axis.

Lecture 23 — Addition formulas, double angles, and substitution

Cartoon: two speakers and a quiet spot in the roomTwo speakers face a room. A stick figure stands between them at a spot where the sound cancels, arms raised. I can’t hearanything here.
the two waves arrive exactly out of step

Everything so far has related one angle to itself. The formulas in this lecture relate two angles, and they are what makes the whole subject usable: they are why a mixture of waves can be rewritten as a single wave, why two speakers can cancel each other in a particular seat of a room, and why the Fourier method of Lecture 18 works at all.

The first thing to be clear about is what is not true. \(\sin(\alpha+\beta)\) is not \(\sin\alpha+\sin\beta\). Test it with \(\alpha=\beta=\tfrac{\pi}{2}\): the left side is \(\sin\pi=0\) and the right side is \(2\). Sine is not a function that distributes over addition, and neither is any other interesting function.

the addition and subtraction formulas $$\sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta$$ $$\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta$$

Replacing \(\beta\) by \(-\beta\) and using the even-odd relations flips the middle signs:

$$\sin(\alpha-\beta)=\sin\alpha\cos\beta-\cos\alpha\sin\beta,\qquad \cos(\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta$$

Note the cosine formulas carry the opposite sign to the one in the bracket. That trap is worth a moment's care every single time.

spot check

Suppose \(\cos\alpha=0.8\), \(\sin\alpha=0.6\), \(\cos\beta=0.6\), \(\sin\beta=0.8\). What is \(\cos(\alpha+\beta)\)?

decimals, fractions, pi and sqrt( ) are all fine

Where the addition formula comes from Two right triangles stacked: one with angle alpha resting on the horizontal axis, and a second with angle beta built on the hypotenuse of the first. Dropping heights from the far corner splits the total height into two pieces, which are sin alpha cos beta and cos alpha sin beta. α β cos α sin β sin α cos β 1 the total height above the axis is sin(α + β) and the dashed line cuts it into exactly two pieces sin α cos β + cos α sin β
The addition formula is not an algebraic accident; it is a picture. Build a second right triangle on the hypotenuse of the first, drop a vertical from the far corner, and the height splits into two pieces whose lengths are exactly the two terms of the formula.
example 6.5 — exact values for new angles

Find the exact value of \(\cos15^{\circ}\) and of \(\sin75^{\circ}\).

\(15^{\circ}=45^{\circ}-30^{\circ}\), and both of those come from the special triangles: $$\cos15^{\circ}=\cos45^{\circ}\cos30^{\circ}+\sin45^{\circ}\sin30^{\circ} =\frac{\sqrt2}{2}\cdot\frac{\sqrt3}{2}+\frac{\sqrt2}{2}\cdot\frac12=\frac{\sqrt6+\sqrt2}{4}.$$ That is \(0.9659\), which a calculator confirms.

For \(75^{\circ}=45^{\circ}+30^{\circ}\): \(\sin75^{\circ}=\dfrac{\sqrt2}{2}\cdot\dfrac{\sqrt3}{2}+\dfrac{\sqrt2}{2}\cdot\dfrac12=\dfrac{\sqrt6+\sqrt2}{4}\) as well — the same number, as the cofunction relation promised, since \(75^{\circ}\) and \(15^{\circ}\) are complementary.

Double angles, from nothing new

Set \(\beta=\alpha\) in the addition formulas and three more identities fall out for free.

double-angle formulas $$\sin2\alpha=2\sin\alpha\cos\alpha$$ $$\cos2\alpha=\cos^{2}\alpha-\sin^{2}\alpha=2\cos^{2}\alpha-1=1-2\sin^{2}\alpha$$

The three versions of the cosine formula come from substituting \(\sin^{2}=1-\cos^{2}\) or the other way round. Which one to use depends on what you want left standing at the end.

spot check

If \(\sin\alpha=0.6\) and \(\cos\alpha=0.8\), what is \(\sin2\alpha\)?

decimals, fractions, pi and sqrt( ) are all fine

example 6.6 — the best angle to throw at

A projectile launched at speed \(v\) and angle \(\theta\) travels a horizontal distance \(R=\dfrac{v^{2}}{g}\,2\sin\theta\cos\theta\). Rewrite this with a double angle and find the angle that throws farthest.

\(2\sin\theta\cos\theta\) is exactly \(\sin2\theta\), so $$R=\frac{v^{2}}{g}\sin2\theta.$$

Now the answer is visible without any calculation at all. \(\sin\) is largest when its input is \(90^{\circ}\), so \(2\theta=90^{\circ}\) and \(\theta=45^{\circ}\).

It also shows something less obvious: since \(\sin2\theta=\sin(180^{\circ}-2\theta)\), the angles \(\theta\) and \(90^{\circ}-\theta\) give the same range. A throw at \(30^{\circ}\) and a throw at \(60^{\circ}\) land in the same place, one flat and fast, the other high and slow. Rewriting the formula did the work that optimisation would otherwise have had to do.

Sums into products, and why rooms have quiet spots

sum to product $$\sin A+\sin B=2\sin\!\left(\frac{A+B}{2}\right)\cos\!\left(\frac{A-B}{2}\right)$$ $$\cos A+\cos B=2\cos\!\left(\frac{A+B}{2}\right)\cos\!\left(\frac{A-B}{2}\right)$$

Each follows by writing \(A=\tfrac{A+B}{2}+\tfrac{A-B}{2}\) and \(B=\tfrac{A+B}{2}-\tfrac{A-B}{2}\), applying the addition formulas, and adding.

idea — this is what a beat is

Lecture 16 showed two nearby frequencies producing a throb and left the explanation for later. Here it is. Adding two waves of frequencies \(f_1\) and \(f_2\) gives, by the first formula above, a wave at the average frequency \(\tfrac{f_1+f_2}{2}\) multiplied by a slow cosine at half the difference.

So the ear hears a single tone at the average pitch, whose loudness is being swelled and squashed by the slow factor. When the difference is small, that factor takes seconds to complete a cycle — and the beat is audible. The guitar tuner of Lecture 16 is listening to an identity.

The same formula explains the cartoon at the top of this lecture. Two speakers playing the same note reach a given seat with different delays, which is to say different phases. Where the phases differ by half a cycle, the sum-to-product factor is zero, and the seat is silent. Concert halls are designed to avoid this and noise-cancelling headphones are designed to cause it.

spot check

Two tones of 300 Hz and 304 Hz sound together. At what pitch (in Hz) do you hear the single tone?

decimals, fractions, pi and sqrt( ) are all fine

example 6.7 — one wave instead of two

Write \(3\sin x+4\cos x\) as a single sine wave \(A\sin(x+\varphi)\).

Expanding the target with the addition formula gives \(A\sin x\cos\varphi+A\cos x\sin\varphi\). Matching coefficients: $$A\cos\varphi=3,\qquad A\sin\varphi=4.$$

Squaring and adding kills \(\varphi\): \(A^{2}(\cos^{2}\varphi+\sin^{2}\varphi)=9+16=25\), so \(A=5\). Dividing gives \(\tan\varphi=\tfrac43\), so \(\varphi=0.927\) rad.

$$3\sin x+4\cos x=5\sin(x+0.927).$$

This is worth more than it looks. The left side is a sum of two things and its maximum is not obvious; the right side is a single wave, so its amplitude is 5, its maximum is 5, and you can see the whole of its behaviour at a glance. Note also that \(A\) came from \(\sqrt{3^{2}+4^{2}}\) — the same Pythagorean combination that gave the amplitude of a Fourier component in Lecture 18, for exactly the same reason.

Substitution: making a root disappear

One last use, and it is a preview. Expressions containing \(\sqrt{1-x^{2}}\) are awkward; the root refuses to simplify and the algebra goes nowhere. But the Pythagorean identity says \(1-\sin^{2}\theta=\cos^{2}\theta\), which is a perfect square. So substitute \(x=\sin\theta\) and the root evaporates: $$\sqrt{1-x^{2}}=\sqrt{1-\sin^{2}\theta}=\cos\theta.$$

spot check

With \(x=\sin\theta\) and \(\theta\) between \(-\tfrac{\pi}{2}\) and \(\tfrac{\pi}{2}\): if \(x=0.6\), what is \(\sqrt{1-x^{2}}\)?

decimals, fractions, pi and sqrt( ) are all fine

example 6.8 — the root evaporates

Simplify \(\dfrac{x}{\sqrt{1-x^{2}}}\) using the substitution \(x=\sin\theta\), and say what geometric object the substitution corresponds to.

With \(x=\sin\theta\) the expression becomes \(\dfrac{\sin\theta}{\cos\theta}=\tan\theta\).

Geometrically the substitution puts \(x\) on the unit circle: it says "let \(x\) be the height of a point at angle \(\theta\)", and then \(\sqrt{1-x^{2}}\) is automatically the width. The awkward root was only awkward because we were insisting on coordinates when the natural description was an angle.

In calculus this move is called trigonometric substitution, and it turns a whole family of otherwise impossible problems into routine ones. You will recognise it when it arrives: it is this, with one extra step.

exercises 6.2
  1. Find exact values for \(\sin105^{\circ}\) and \(\tan75^{\circ}\), using \(105=60+45\) and \(75=45+30\).
  2. Given \(\sin\alpha=\tfrac35\) with \(\alpha\) acute and \(\cos\beta=-\tfrac{5}{13}\) with \(\beta\) in the second quadrant, find \(\sin(\alpha+\beta)\) and \(\cos(\alpha-\beta)\).
  3. Prove \(\sin(\pi-t)=\sin t\) from the addition formula, and check it against the reflection picture of Lecture 22.
  4. If \(\cos\theta=\tfrac13\), find \(\cos2\theta\) and \(\sin2\theta\) (taking \(\theta\) acute). Which version of the double-angle formula was least work?
  5. Write \(\sin x-\sqrt3\cos x\) as a single sine wave, and state its amplitude and phase shift.
  6. Two tuning forks at 440 Hz and 446 Hz sound together. Use the sum-to-product formula to state the frequency you hear and the rate at which the loudness swells.
  7. Simplify \(\sqrt{9-x^{2}}\) using \(x=3\sin\theta\). What restriction on \(\theta\) keeps the answer positive?
  8. Show that \(\cos3\theta=4\cos^{3}\theta-3\cos\theta\), by writing \(3\theta=2\theta+\theta\).
  1. \(\sin105^{\circ}=\sin60\cos45+\cos60\sin45=\dfrac{\sqrt6+\sqrt2}{4}=0.9659\). \(\tan75^{\circ}=\dfrac{\sin75}{\cos75}=2+\sqrt3=3.732\).
  2. \(\cos\alpha=\tfrac45\), \(\sin\beta=\tfrac{12}{13}\). Then \(\sin(\alpha+\beta)=\tfrac35\cdot\left(-\tfrac{5}{13}\right)+\tfrac45\cdot\tfrac{12}{13}=\tfrac{33}{65}\), and \(\cos(\alpha-\beta)=\tfrac45\left(-\tfrac{5}{13}\right)+\tfrac35\cdot\tfrac{12}{13}=\tfrac{16}{65}\).
  3. \(\sin(\pi-t)=\sin\pi\cos t-\cos\pi\sin t=0-(-1)\sin t=\sin t\). The picture says the same thing: reflecting across the vertical axis leaves the height alone.
  4. \(\cos2\theta=2\left(\tfrac19\right)-1=-\tfrac79\), using the version written in cosines only, which needed nothing else. For the sine, \(\sin\theta=\tfrac{2\sqrt2}{3}\), so \(\sin2\theta=2\cdot\tfrac{2\sqrt2}{3}\cdot\tfrac13=\tfrac{4\sqrt2}{9}\).
  5. \(A=\sqrt{1+3}=2\) and \(\tan\varphi=-\sqrt3\) with the cosine part positive, so \(\varphi=-\tfrac{\pi}{3}\): the expression is \(2\sin\!\left(x-\tfrac{\pi}{3}\right)\), amplitude 2, shifted \(\tfrac{\pi}{3}\) right.
  6. The tone heard is the average, 443 Hz. The slow factor has frequency \(\tfrac{446-440}{2}=3\) Hz, and since loudness peaks twice per cycle of that factor, the beat is heard 6 times a second — the difference of the two frequencies.
  7. \(\sqrt{9-9\sin^{2}\theta}=3\sqrt{\cos^{2}\theta}=3|\cos\theta|\), which is \(3\cos\theta\) provided \(-\tfrac{\pi}{2}\le\theta\le\tfrac{\pi}{2}\), where the cosine is not negative.
  8. \(\cos(2\theta+\theta)=\cos2\theta\cos\theta-\sin2\theta\sin\theta=(2\cos^{2}\theta-1)\cos\theta-2\sin^{2}\theta\cos\theta\). Replacing \(\sin^{2}\theta\) with \(1-\cos^{2}\theta\) gives \(2\cos^{3}\theta-\cos\theta-2\cos\theta+2\cos^{3}\theta=4\cos^{3}\theta-3\cos\theta\).

Where this chapter leaves you

Identities are the circle's symmetry written down. Reflecting a point in an axis changes the sign of one coordinate, and that is a whole family of them; the equation of the circle itself is the Pythagorean identity; and building one triangle on another gives the addition formulas, from which double angles, sums-to-products and the single-wave form all follow without further ideas.

What they buy you is freedom of expression: the same quantity can be written in whichever form makes the next step easy. That is why a beat has an explanation, why the best throwing angle is obvious once the formula is rewritten, and why a root that blocked all progress can be made to vanish by choosing to describe a number as an angle.

Chapter 7 leaves the circle for the plane itself, where a vector is a journey, a matrix is a function that moves every point at once, and the rotation you have been doing by hand all chapter becomes a single object you can multiply.

Next: Chapter 7 — Plane vectors and matrix algebra.