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Precalculus: An Investigation of Functions, by David Lippman and Melonie Rasmussen, licensed under CC BY-SA 4.0. Read the original free This shelf’s study material is shared under the same license (CC BY-SA 4.0).

Chapter 11 · pp. 463–506

Chapter 7: Trigonometric Equations and Identities

This chapter builds a larger toolkit for working with trigonometric expressions. It starts with solving harder trig equations by factoring and by rewriting with identities. It then moves to sum and difference identities, double angle, power reduction and half-angle identities, and finally to models whose amplitude or midline changes over time, such as damped springs and AM radio.

Solving trig equations that look like quadratics

You already know how to solve $f(x)=2x^2+x=0$ by factoring: $x(2x+1)=0$ , so $x=0$ or $x=-\tfrac12$ . You also know how to solve $\sin(t)=0$ from the unit circle. Putting the two together gives $2\sin^2(t)+\sin(t)$ , a polynomial whose "variable" is a trig function. Such an equation is called "quadratic in sine." (p. 463)

Method: treat the trig function like a variable, factor (or use the quadratic formula), set each factor equal to zero, and solve each case separately with the unit circle.

Example. Solve $2\sin^2(t)+\sin(t)=0$ on $[0,2\pi)$ .

Factor out the common $\sin(t)$ : $\sin(t)\,(2\sin(t)+1)=0$ .
By the zero product rule, $\sin(t)=0$ or $\sin(t)=-\tfrac12$ .
The first case gives $t=0,\pi$ . The second gives $t=\tfrac{7\pi}{6},\tfrac{11\pi}{6}$ .

There are four solutions: $0,\ \pi,\ \tfrac{7\pi}{6},\ \tfrac{11\pi}{6}$ . Graphing the function and counting its zeros is a good way to confirm you found them all. (pp. 463–464)

Factors that give no solutions

If factoring is easier for you without the trig function in the way, substitute a letter for it, factor, and then undo the substitution. Sometimes one factor leads to a value the trig function can never reach. That factor contributes no solutions.

Example. Solve $3\sec^2(t)-5\sec(t)-2=0$ on $[0,2\pi)$ .

Let $u=\sec(t)$ . Then $3u^2-5u-2=(3u+1)(u-2)=0$ .
Case $\sec(t)=-\tfrac13$ : taking the reciprocal gives $\cos(t)=-3$ . Cosine only takes values in $[-1,1]$ , so this case has no solutions.
Case $\sec(t)=2$ : this means $\cos(t)=\tfrac12$ , so $t=\tfrac{\pi}{3}$ or $\tfrac{5\pi}{3}$ .

Only the second case works, so there are exactly two solutions. (p. 465)

Using the Pythagorean identity to get one function

Mixed sine and cosine terms are hard to factor. The Pythagorean identity $\sin^2(\theta)+\cos^2(\theta)=1$ can be rearranged as

$\sin^2(\theta)=1-\cos^2(\theta)\qquad\text{or}\qquad \cos^2(\theta)=1-\sin^2(\theta),$

so a squared term can be swapped to leave only one function. (p. 466)

Example. Solve $2\sin^2(t)-\cos(t)=1$ on $[0,2\pi)$ .

Replace $\sin^2(t)$ with $1-\cos^2(t)$ : $2(1-\cos^2(t))-\cos(t)=1$ .
Distribute and move everything to one side: $-2\cos^2(t)-\cos(t)+1=0$ . Multiply by $-1$ to get $2\cos^2(t)+\cos(t)-1=0$ .
Factor: $(2\cos(t)-1)(\cos(t)+1)=0$ .
So $\cos(t)=\tfrac12$ gives $t=\tfrac{\pi}{3},\tfrac{5\pi}{3}$ , and $\cos(t)=-1$ gives $t=\pi$ .

The solutions are $\tfrac{\pi}{3},\ \pi,\ \tfrac{5\pi}{3}$ . (pp. 466–467)

Rewriting tangent, secant, cosecant and cotangent

When an equation contains $\tan$ , $\sec$ , $\csc$ or $\cot$ , use the reciprocal and quotient identities to rewrite them with sine and cosine. For example, $\tan(x)=\tfrac{\sin(x)}{\cos(x)}$ and $\sec(x)=\tfrac{1}{\cos(x)}$ . Then simplify. Often the equation becomes a quadratic in one function. (pp. 467–468)

Do not divide by a trig function that can be zero

A common mistake is to divide both sides by $\sin(x)$ to "cancel" it. Dividing assumes the quantity is never zero. Any $x$ with $\sin(x)=0$ would be a solution that you silently throw away. The safe move is to get zero on one side and factor. (p. 467)

Example. Solve $\tan(x)=3\sin(x)$ on $[0,2\pi)$ .

Rewrite: $\tfrac{\sin(x)}{\cos(x)}=3\sin(x)$ . Multiply by $\cos(x)$ : $\sin(x)=3\sin(x)\cos(x)$ .
Do not divide by $\sin(x)$ . Instead write $\sin(x)-3\sin(x)\cos(x)=0$ and factor: $\sin(x)\,(1-3\cos(x))=0$ .
Case $\sin(x)=0$ : $x=0$ or $x=\pi$ .
Case $\cos(x)=\tfrac13$ : this is not a special value, so use the technique below.

(pp. 467–468)

Equations like $\cos(\theta)=c$ with a non-special value

When $c$ is not a unit circle value, use the inverse cosine on a calculator for the first angle. Cosine is symmetric about the $x$ -axis, so the second angle in $[0,2\pi)$ is $2\pi-\theta$ .

For $\cos(x)=\tfrac13$ , the calculator gives $x=\cos^{-1}\!\left(\tfrac13\right)\approx 1.231$ . The second solution is $2\pi-1.231\approx 5.052$ . Combined with the earlier case, the full solution set for $\tan(x)=3\sin(x)$ is $x=0,\ 1.231,\ \pi,\ 5.052$ . (p. 468)

Using the quadratic formula. Suppose an equation simplifies to $4\cos^2\theta+3\cos\theta-2=0$ . It does not factor nicely, so apply the quadratic formula with $\cos\theta$ as the unknown:

$\cos\theta=\frac{-3\pm\sqrt{3^2-4(4)(-2)}}{2(4)}=\frac{-3\pm\sqrt{41}}{8}.$

The negative root gives $\cos\theta\approx-1.175$ . This is below $-1$ , so it is rejected.
The positive root gives $\cos\theta\approx 0.425$ . Then $\theta\approx\cos^{-1}(0.425)\approx 1.131$ .
By symmetry, the second solution is $2\pi-1.131\approx 5.152$ .

The solutions are $\theta\approx 1.131$ and $\theta\approx 5.152$ . (pp. 468–469)

The sum and difference identities

The four identities below let you expand the trig function of a sum or difference of two angles:

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

Memory tip: for cosine, the sign in the expansion is opposite to the sign between the angles. For sine, the sign is the same. For example, $\sin(\alpha+\beta)$ has a plus in its expansion. (p. 471)

Why the cosine difference identity is true. Put two points on the unit circle: $P=(\cos\alpha,\sin\alpha)$ and $Q=(\cos\beta,\sin\beta)$ . The angle $POQ$ measures $\alpha-\beta$ . Now add $C=(\cos(\alpha-\beta),\sin(\alpha-\beta))$ and $D=(1,0)$ . Triangle $COD$ is a rotated copy of triangle $POQ$ , so chord $CD$ and chord $PQ$ have equal length. (pp. 471–472)

Find that length two ways with the distance formula:

For $PQ$ , expand the squares and use $\sin^2+\cos^2=1$ . You get $2-2\cos\alpha\cos\beta-2\sin\alpha\sin\beta$ .
For $CD$ , the same steps give $2-2\cos(\alpha-\beta)$ .

Set them equal, cancel the 2 and the $-2$ , and the identity follows. The other three identities can be derived from this one. For instance, write $\cos(\alpha+\beta)$ as $\cos(\alpha-(-\beta))$ . (p. 472)

Using the identities: exact values, proofs and equations

Exact values. Write an unfamiliar angle as a sum or difference of special angles. Since $75^\circ=30^\circ+45^\circ$ :

$\cos75^\circ=\cos30^\circ\cos45^\circ-\sin30^\circ\sin45^\circ=\tfrac{\sqrt3}{2}\cdot\tfrac{\sqrt2}{2}-\tfrac12\cdot\tfrac{\sqrt2}{2}=\tfrac{\sqrt6-\sqrt2}{4}$

Similarly, $\tfrac{\pi}{12}=\tfrac{\pi}{3}-\tfrac{\pi}{4}$ gives $\sin\tfrac{\pi}{12}=\tfrac{\sqrt6-\sqrt2}{4}$ . The same expansion can also rewrite an expression: $\sin\!\left(x-\tfrac{\pi}{4}\right)=\tfrac{\sqrt2}{2}\sin x-\tfrac{\sqrt2}{2}\cos x$ . (pp. 472–473)

Proving identities. Start with the more complicated side and rewrite it with known identities until it matches the other side. To prove $\tfrac{\sin(a+b)}{\sin(a-b)}=\tfrac{\tan a+\tan b}{\tan a-\tan b}$ , the left side expands but does not simplify further. So work on the right side instead:

Replace each tangent with sine over cosine.
Multiply the top and bottom by $\cos a\cos b$ .
The result is $\tfrac{\sin a\cos b+\cos a\sin b}{\sin a\cos b-\cos a\sin b}$ , which is the expanded left side.

If one side stalls, try the other. (pp. 473–474)

Solving equations. Sometimes one side of an equation is already an expanded identity. In $\sin x\sin 2x+\cos x\cos 2x=\tfrac{\sqrt3}{2}$ , the left side is $\cos(x-2x)=\cos(-x)=\cos x$ . So $\cos x=\tfrac{\sqrt3}{2}$ , and $x=\tfrac{\pi}{6}+2k\pi$ or $x=\tfrac{11\pi}{6}+2k\pi$ for any integer $k$ . (p. 474)

Combining a sine and a cosine of equal period

A sinusoid $A\sin(Bx+C)$ can be expanded with the sum identity:

$A\sin(Bx+C)=A\cos C\,\sin(Bx)+A\sin C\,\cos(Bx)$

For example, $4\sin\!\left(3x+\tfrac{\pi}{3}\right)=2\sin(3x)+2\sqrt3\cos(3x)$ . The sine and cosine are stretched by different amounts, but both have the same $B$ , so they share the same period. (p. 475)

The process also runs in reverse. To write $m\sin(Bx)+n\cos(Bx)$ as $A\sin(Bx+C)$ , match $m=A\cos C$ and $n=A\sin C$ . Squaring and adding gives $m^2+n^2=A^2(\cos^2C+\sin^2C)=A^2$ , so:

$A=\sqrt{m^2+n^2},\qquad \cos C=\tfrac{m}{A},\qquad \sin C=\tfrac{n}{A}$

For $3\sin(5x)+4\cos(5x)$ , $A=\sqrt{9+16}=5$ . (pp. 475–476)

Choosing the right $C$ . One ratio usually gives two angles, so check the other ratio to pick the right quadrant. For $4\sqrt3\sin(2x)-4\cos(2x)$ :

$A=\sqrt{48+16}=8$ .
$\cos C=\tfrac{\sqrt3}{2}$ , so $C=\tfrac{\pi}{6}$ or $\tfrac{11\pi}{6}$ .
$\sin C=-\tfrac12$ is negative, so $C=\tfrac{11\pi}{6}$ .

The result is $8\sin\!\left(2x+\tfrac{11\pi}{6}\right)$ . (p. 476)

Solving $m\sin(Bx)+n\cos(Bx)=k$ . Rewrite the left side as one sinusoid, substitute $u=Bx+C$ , solve for $u$ , then undo the substitution. For $3\sin(2x)+4\cos(2x)=1$ :

$A=5$ and $C=\cos^{-1}\tfrac35\approx0.927$ . Since $\sin C=\tfrac45$ is positive, this first-quadrant angle is correct.
The equation becomes $\sin(u)=\tfrac15$ with $u=2x+0.927$ .
The solutions are $u\approx0.201$ , then $\pi-0.201\approx2.940$ , then $0.201+2\pi\approx6.485$ .
Subtract $0.927$ and divide by 2: $x\approx-0.363,\ 1.007,\ 2.779$ . The negative one is dropped if you want positive solutions. (p. 477)

Product-to-sum and sum-to-product identities

The product-to-sum identities turn a product into a sum:

$\sin\alpha\cos\beta=\tfrac12[\sin(\alpha+\beta)+\sin(\alpha-\beta)]$ $\sin\alpha\sin\beta=\tfrac12[\cos(\alpha-\beta)-\cos(\alpha+\beta)]$ $\cos\alpha\cos\beta=\tfrac12[\cos(\alpha-\beta)+\cos(\alpha+\beta)]$

To see the first one, add the sine sum and sine difference identities. The cosine terms combine into $2\sin\alpha\cos\beta$ ; then divide by 2. Example: $\sin(2t)\sin(4t)=\tfrac12[\cos(2t)-\cos(6t)]$ . A numeric product such as $\cos\tfrac{11\pi}{12}\cos\tfrac{\pi}{12}$ works the same way, giving $\tfrac12[\cos\tfrac{5\pi}{6}+\cos\pi]$ . (pp. 478–479)

The sum-to-product identities go the other direction:

$\sin u+\sin v=2\sin\tfrac{u+v}{2}\cos\tfrac{u-v}{2}$ $\sin u-\sin v=2\sin\tfrac{u-v}{2}\cos\tfrac{u+v}{2}$ $\cos u+\cos v=2\cos\tfrac{u+v}{2}\cos\tfrac{u-v}{2}$ $\cos u-\cos v=-2\sin\tfrac{u+v}{2}\sin\tfrac{u-v}{2}$

They come from the first product-to-sum identity. Let $u=\alpha+\beta$ and $v=\alpha-\beta$ , so $\alpha=\tfrac{u+v}{2}$ and $\beta=\tfrac{u-v}{2}$ . For the difference of sines, write $\sin u-\sin v=\sin u+\sin(-v)$ and apply the sum rule. (pp. 479–480)

Uses. These identities help with exact values and proofs.

Evaluating: $\cos15^\circ-\cos75^\circ=-2\sin45^\circ\sin(-30^\circ)=\tfrac{\sqrt2}{2}$ .
Proving: to show $\tfrac{\cos4t-\cos2t}{\sin4t+\sin2t}=-\tan t$ , convert the top to $-2\sin3t\sin t$ and the bottom to $2\sin3t\cos t$ . The common $2\sin3t$ cancels, leaving $-\tfrac{\sin t}{\cos t}$ . (p. 480)

Counting every cycle when solving

When you factor an equation, each factor may have a different period. Count all the full cycles of that factor inside the interval. Take $\sin(\pi t)+\sin(3\pi t)=\cos(\pi t)$ on $0\le t<2$ :

Sum-to-product on the left gives $2\sin(2\pi t)\cos(\pi t)=\cos(\pi t)$ , so $\cos(\pi t)\,[2\sin(2\pi t)-1]=0$ .
$\cos(\pi t)$ has period 2, so the interval holds one cycle. Then $\pi t=\tfrac{\pi}{2}$ or $\tfrac{3\pi}{2}$ , so $t=\tfrac12,\tfrac32$ .
$\sin(2\pi t)=\tfrac12$ has period 1, so the interval holds two cycles. Then $2\pi t=\tfrac{\pi}{6},\tfrac{5\pi}{6}$ on the first cycle and $\tfrac{13\pi}{6},\tfrac{17\pi}{6}$ on the second. This gives $t=\tfrac1{12},\tfrac5{12},\tfrac{13}{12},\tfrac{17}{12}$ .

Altogether there are six solutions. Stopping after the first cycle of the second factor would miss two of them. (pp. 481–482)

Double angle identities

The double angle identities are the sum identities with both angles equal. For sine, $\sin(\theta+\theta)=\sin\theta\cos\theta+\cos\theta\sin\theta$ , so $\sin(2\theta)=2\sin\theta\cos\theta$ . Cosine works the same way: $\cos(2\theta)=\cos^2\theta-\sin^2\theta$ (p. 487).

The cosine identity has three equivalent forms. Replace $\cos^2\theta$ with $1-\sin^2\theta$ to get $\cos(2\theta)=1-2\sin^2\theta$ . Replace $\sin^2\theta$ with $1-\cos^2\theta$ to get $\cos(2\theta)=2\cos^2\theta-1$ . Choose the form that leaves only the function you already know (p. 487).

Finding values. Suppose $\sin\theta=\tfrac35$ and $\theta$ is in quadrant II.

Use the sine-only form: $\cos2\theta=1-2\cdot\tfrac{9}{25}=\tfrac{7}{25}$ .
For $\sin 2\theta$ you also need $\cos\theta$ . The Pythagorean identity gives $\cos^2\theta=\tfrac{16}{25}$ . Cosine is negative in quadrant II, so $\cos\theta=-\tfrac45$ .
Then $\sin2\theta=2\cdot\tfrac35\cdot\left(-\tfrac45\right)=-\tfrac{24}{25}$ (p. 488).

Simplifying. Look for a pattern that matches an identity.

$2\cos^2(12^\circ)-1$ matches $\cos 2\theta$ , so it equals $\cos24^\circ$ .
$8\sin3x\cos3x=4\cdot(2\sin3x\cos3x)=4\sin6x$ (p. 488).
For $\dfrac{\cos2t}{\cos t-\sin t}$ , pick the form $\cos^2t-\sin^2t$ . It factors into $(\cos t-\sin t)(\cos t+\sin t)$ , so the fraction reduces to $\cos t+\sin t$ (p. 489).

Proving. To show $\sec2\theta=\dfrac{\sec^2\theta}{2-\sec^2\theta}$ , start with the more complicated right side. Rewrite it in cosines as $\dfrac{1/\cos^2\theta}{2-1/\cos^2\theta}$ . Multiply the top and bottom by $\cos^2\theta$ to get $\dfrac{1}{2\cos^2\theta-1}$ . That is $\dfrac{1}{\cos2\theta}=\sec2\theta$ (pp. 489–490).

Solving. An equation with $\cos2t$ and $\cos t$ mixes different periods. Use $\cos2t=2\cos^2t-1$ to put everything in terms of $\cos t$ . For $\cos2t=\cos t$ : $2\cos^2t-\cos t-1=0\quad\Rightarrow\quad(2\cos t+1)(\cos t-1)=0.$ So $\cos t=-\tfrac12$ or $\cos t=1$ . On $[0,2\pi)$ the solutions are $t=0,\tfrac{2\pi}{3},\tfrac{4\pi}{3}$ (p. 490).

Projectile aiming. A cannonball is fired at 100 m/s at angle $\theta$ , with height $h(t)=-4.9t^2+100\sin(\theta)t$ . Factor $h(t)=0$ to find the landing time $t=\tfrac{100\sin\theta}{4.9}$ . Put this into $x(t)=100\cos(\theta)t=900$ to get $\sin\theta\cos\theta=\tfrac{900\cdot4.9}{100^2}$ . Multiplying by 2 creates a double angle: $\sin2\theta=0.882$ . So $2\theta\approx1.080$ and $\theta\approx0.540$ rad, about $30.94^\circ$ (pp. 491–492).

Power reduction and half-angle identities

The double angle forms can be solved for the squared term. From $\cos2\theta=2\cos^2\theta-1$ we get $\cos^2\theta=\frac{1+\cos2\theta}{2},$ and from $\cos2\theta=1-2\sin^2\theta$ we get $\sin^2\theta=\frac{1-\cos2\theta}{2}.$ These power reduction identities swap a square for a first power of a double angle (pp. 492–493).

Replace $\theta$ with $\tfrac{\theta}{2}$ and take a square root to get the half-angle identities: $\cos\tfrac{\theta}{2}=\pm\sqrt{\tfrac{1+\cos\theta}{2}},\qquad \sin\tfrac{\theta}{2}=\pm\sqrt{\tfrac{1-\cos\theta}{2}}.$ The sign depends on the quadrant of $\tfrac{\theta}{2}$ , not of $\theta$ . For $\cos15^\circ$ , use $\theta=30^\circ$ . Since $15^\circ$ is in quadrant I, take the positive root: $\cos15^\circ=\sqrt{\tfrac{1+\sqrt3/2}{2}}=\sqrt{\tfrac12+\tfrac{\sqrt3}{4}}$ (pp. 493–494).

Removing powers. To write $\cos^4x$ with no exponents, apply power reduction twice: $\cos^4x=\left(\tfrac{1+\cos2x}{2}\right)^2=\tfrac{1+2\cos2x+\cos^22x}{4}.$ Then replace $\cos^22x$ with $\tfrac{1+\cos4x}{2}$ . The result is $\tfrac38+\tfrac12\cos2x+\tfrac18\cos4x$ (pp. 493–494).

Models with a changing midline

A changing midline model is $f(t)=A\sin(Bt)+g(t)$ . The wave oscillates above and below the moving average $g(t)$ , which may be linear, exponential or quadratic (pp. 498–499).

Build the model in four steps:

Get the amplitude from the size of the swing.
Get $B=\tfrac{2\pi}{\text{period}}$ .
Choose the starting shape: cosine for a peak, negative cosine for a minimum, sine for a start on the midline.
Use the trend function as the midline.

For elk averaging 2000, growing 4% a year, swinging $\pm50$ , and starting in winter (the minimum), the model is $P(t)=-50\cos(2\pi t)+2000(1.04)^t$ (p. 498). For a resort averaging 4000 tourists, gaining 200 a year, swinging $\pm1000$ , and starting at peak season, the model is $N(t)=1000\cos(2\pi t)+4000+200t$ (pp. 500–501).

Fitting data. Suppose $f(t)=A\sin(\tfrac{\pi}{2}t)+mt+b$ . Then the period is 4, so the graph sits on the midline at $t=0$ and $t=2$ . With points $(0,5)$ and $(2,9)$ the slope is 2, so the midline is $2t+5$ . Use another point, $(1,10)$ : $10=A+7$ , so $A=3$ (pp. 499–500).

Models with a changing amplitude

Here the amplitude is a function: $f(t)=A(t)\sin(Bt)+k$ . It can be linear, exponential or quadratic. A constant $A(t)$ gives an ordinary sinusoid (p. 501).

A common mistake. A friction-damped spring should not use a linear amplitude like $A(t)=10-t$ . It hits 0 at $t=10$ , then becomes negative, and the swings grow again. Real springs lose a fixed percentage of amplitude each cycle, which is exponential decay (p. 502).

Damped harmonic motion is $f(t)=ab^t\sin(Bt)+k$ or $f(t)=ae^{rt}\sin(Bt)+k$ . Use cosine if the motion starts at a maximum. If the amplitude drops 20% a second, then $b=1-0.20=0.80$ . A spring of natural length 20 ft, pulled back 6 ft, with a 2-second period, gives $f(t)=6(0.80)^t\cos(\pi t)+20$ (p. 502).

If you are given two amplitudes instead of a percentage, substitute them into $A(t)=ab^t$ . A start of 10 cm and 5 cm after 2 s gives $a=10$ and $5=10b^2$ , so $b=\sqrt{0.5}\approx0.707$ . With 4 cycles per second, $B=8\pi$ , so $f(t)=10(0.707)^t\cos(8\pi t)+30$ (p. 503).

AM radio. The signal wave becomes the amplitude of the carrier. A 110 Hz note of amplitude 3 is $3\sin(220\pi t)$ , because $B=2\pi\cdot110$ . A 2000 Hz carrier is $\sin(4000\pi t)$ . The broadcast is the product $f(t)=3\sin(220\pi t)\sin(4000\pi t)$ (pp. 503–504).

Check yourself

Solve $2\cos^2x-\cos x-1=0$ on $[0,2\pi)$ , and explain what happens when a factor gives a value outside the range of cosine.
Why is dividing both sides of $\sin x\cos x=\sin x$ by $\sin x$ risky? How should you solve it?
Given $\cos\theta=-\tfrac{5}{13}$ with $\theta$ in quadrant III, find $\sin2\theta$ and $\cos2\theta$ .
Use a sum or difference identity to find the exact value of $\sin75^\circ$ , then rewrite $\sin x+\cos x$ in the form $A\sin(x+C)$ .
A spring starts at its maximum, 8 cm from rest, oscillates 3 times per second, and loses 10% of its amplitude each second. Write a model for its position.