9 - Trigonometric Equations and Applications

A progressive guide to trigonometric identities, equation-solving strategies, inverse functions, oblique triangles, and sinusoidal applications.

Core identities and algebraic tools

Trigonometric equations connect unknown angles with ratios, distances, and repeating behavior. Begin by distinguishing an identity from an equation. A is true for every permitted input, whereas a trigonometric equation is generally true only for selected values.

Important identities include:

  • Reciprocal identities: csc⁡θ=1sin⁡θ\csc\theta=\frac{1}{\sin\theta}, sec⁡θ=1cos⁡θ\sec\theta=\frac{1}{\cos\theta}, and cot⁡θ=1tan⁡θ\cot\theta=\frac{1}{\tan\theta}.

  • Quotient identities: tan⁡θ=sin⁡θcos⁡θ\tan\theta=\frac{\sin\theta}{\cos\theta} and cot⁡θ=cos⁡θsin⁡θ\cot\theta=\frac{\cos\theta}{\sin\theta}.

  • Pythagorean identities: sin⁡2θ+cos⁡2θ=1\sin^2\theta+\cos^2\theta=1, 1+tan⁡2θ=sec⁡2θ1+\tan^2\theta=\sec^2\theta, and 1+cot⁡2θ=csc⁡2θ1+\cot^2\theta=\csc^2\theta.

  • Even-odd identities: cos⁡(−θ)=cos⁡θ\cos(-\theta)=\cos\theta and sin⁡(−θ)=−sin⁡θ\sin(-\theta)=-\sin\theta.

  • Cofunction identities: sin⁡θ=cos⁡(π2−θ)\sin\theta=\cos\left(\frac{\pi}{2}-\theta\right) and cos⁡θ=sin⁡(π2−θ)\cos\theta=\sin\left(\frac{\pi}{2}-\theta\right).

Sum-and-difference identities expand compound angles:

sin⁡(α+β)=sin⁡αcos⁡β+cos⁡αsin⁡β\sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta
cos⁡(α+β)=cos⁡αcos⁡β−sin⁡αsin⁡β\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta

Double-angle identities can change an equation into a more useful form:

sin⁡(2θ)=2sin⁡θcos⁡θ\sin(2\theta)=2\sin\theta\cos\theta
cos⁡(2θ)=cos⁡2θ−sin⁡2θ=2cos⁡2θ−1=1−2sin⁡2θ\cos(2\theta)=\cos^2\theta-\sin^2\theta=2\cos^2\theta-1=1-2\sin^2\theta

Always preserve domain restrictions. For example, quotient identities are valid only where their denominators are nonzero.

Takeaway: Identify the expression type first, then choose an identity that creates a useful equivalent form without ignoring restrictions.

Verifying identities safely

To verify an identity, transform one side into the other through a connected sequence of equivalent steps. Usually begin with the more complicated side and simplify it until it matches the other side.

For example, consider

1−cos⁡2θsin⁡θ=sin⁡θ.\frac{1-\cos^2\theta}{\sin\theta}=\sin\theta.

Using the Pythagorean identity gives

1−cos⁡2θsin⁡θ=sin⁡2θsin⁡θ=sin⁡θ.\frac{1-\cos^2\theta}{\sin\theta}=\frac{\sin^2\theta}{\sin\theta}=\sin\theta.

The cancellation requires sin⁡θ≠0\sin\theta\neq 0, which is already required for the original left side to be defined.

A reliable method is:

  1. Choose the more complicated side.

  2. Replace parts of it with reciprocal, quotient, Pythagorean, cofunction, or angle identities.

  3. Simplify one step at a time.

  4. Stop when the other side is reached.

  5. Check the domain of every expression.

Manipulating both sides independently can hide an invalid step or assume the desired result before proving it.

Takeaway: Identity verification is a transformation process, not a procedure for solving for a variable.

Solving trigonometric equations

A trigonometric equation requires all values in the requested interval or a complete general solution. A useful procedure is:

  1. Simplify algebraically or with identities.

  2. Isolate one trigonometric function when possible.

  3. Find a reference angle or use an inverse function.

  4. Use the sign of the function to identify every relevant quadrant.

  5. Apply when writing general solutions.

  6. Check values that may have been excluded by division or cancellation.

For sin⁡θ=k\sin\theta=k, if α=sin⁡−1(k)\alpha=\sin^{-1}(k), the general solutions are

θ=α+2πn\theta=\alpha+2\pi n

or

θ=π−α+2πn,n∈Z.\theta=\pi-\alpha+2\pi n,\qquad n\in\mathbb{Z}.

For cos⁡θ=k\cos\theta=k, the general form is

θ=±cos⁡−1(k)+2πn,n∈Z.\theta=\pm\cos^{-1}(k)+2\pi n,\qquad n\in\mathbb{Z}.

For tan⁡θ=k\tan\theta=k, the general form is

θ=tan⁡−1(k)+πn,n∈Z.\theta=\tan^{-1}(k)+\pi n,\qquad n\in\mathbb{Z}.

For example, solve 2sin⁡x=32\sin x=\sqrt{3} on [0,2π)[0,2\pi). Isolating sine gives sin⁡x=32\sin x=\frac{\sqrt{3}}{2}. The reference angle is π3\frac{\pi}{3}, and sine is positive in Quadrants I and II. Therefore,

x=π3,x=2π3.x=\frac{\pi}{3},\qquad x=\frac{2\pi}{3}.

For tan⁡x=1\tan x=1 on [0,2π)[0,2\pi), tangent is positive in Quadrants I and III:

x=π4,x=5π4.x=\frac{\pi}{4},\qquad x=\frac{5\pi}{4}.

If the argument is 2x2x, as in cos⁡(2x)=12\cos(2x)=\frac12, solve for 2x2x over [0,4π)[0,4\pi) before dividing by 22. This yields

x=π6,5π6,7π6,11π6.x=\frac{\pi}{6},\frac{5\pi}{6},\frac{7\pi}{6},\frac{11\pi}{6}.

Takeaway: An inverse-function result supplies a starting angle; quadrant analysis and produce the complete solution set.

Factoring and multiple-angle equations

Factoring and substitution are useful when an equation contains powers of one trigonometric function. For

2sin⁡2x−sin⁡x−1=0,2\sin^2x-\sin x-1=0,

let u=sin⁡xu=\sin x. Then

2u2−u−1=0=(2u+1)(u−1).2u^2-u-1=0=(2u+1)(u-1).

Thus,

sin⁡x=−12orsin⁡x=1.\sin x=-\frac12\qquad\text{or}\qquad\sin x=1.

On [0,2π)[0,2\pi), the solutions are

x=7π6,11π6,π2.x=\frac{7\pi}{6},\frac{11\pi}{6},\frac{\pi}{2}.

When factoring, solve every resulting trigonometric equation and combine the solution sets. Substitute answers into the original equation when division, cancellation, or another potentially restrictive operation was used.

Double-angle identities can also change an equation into a form suited to factoring or isolation. For example,

cos⁡(2θ)=cos⁡2θ−sin⁡2θ=2cos⁡2θ−1=1−2sin⁡2θ.\cos(2\theta)=\cos^2\theta-\sin^2\theta=2\cos^2\theta-1=1-2\sin^2\theta.

Takeaway: Treat a repeated trigonometric expression like an algebraic variable, but verify that the resulting values lie in the range of the original function and satisfy the original equation.

Inverse functions and principal angles

An returns a principal angle, not every angle with a given trigonometric value. Sine, cosine, and tangent are periodic, so they are not one-to-one over their full domains. Restricted domains are therefore used to define their inverses.

The principal ranges are:

  • sin⁡−1x\sin^{-1}x: [−π2,π2]\left[-\frac{\pi}{2},\frac{\pi}{2}\right].

  • cos⁡−1x\cos^{-1}x: [0,π][0,\pi].

  • tan⁡−1x\tan^{-1}x: (−π2,π2)\left(-\frac{\pi}{2},\frac{\pi}{2}\right).

The notation sin⁡−1x\sin^{-1}x means inverse sine, whereas csc⁡x\csc x means the reciprocal of sine. For example,

sin⁡−1(12)=π6.\sin^{-1}\left(\frac12\right)=\frac{\pi}{6}.

Composition order matters. For −1≤x≤1-1\le x\le1,

sin⁡(sin⁡−1x)=x.\sin(\sin^{-1}x)=x.

However, sin⁡−1(sin⁡x)=x\sin^{-1}(\sin x)=x only when xx lies in the principal range of inverse sine.

If a right triangle has opposite side 77 and adjacent side 1010, then

tan⁡θ=710,θ=tan⁡−1(710)≈34.99∘.\tan\theta=\frac{7}{10},\qquad \theta=\tan^{-1}\left(\frac{7}{10}\right)\approx34.99^\circ.

Use degree mode when the requested answer is in degrees. If θ=cos⁡−1x\theta=\cos^{-1}x, then θ\theta lies in Quadrants I or II, so

sin⁡(cos⁡−1x)=1−x2,−1≤x≤1.\sin(\cos^{-1}x)=\sqrt{1-x^2},\qquad -1\le x\le1.

The positive square root follows from the principal range of inverse cosine.

Takeaway: Inverse functions provide principal values. Interpret them using the relevant domain, range, angle unit, and quadrant information.

Solving oblique triangles

An has no right angle. Its angles are commonly labeled AA, BB, and CC, with opposite side lengths aa, bb, and cc. The angle sum is

A+B+C=180∘A+B+C=180^\circ

or, in radians,

A+B+C=π.A+B+C=\pi.

The is

asin⁡A=bsin⁡B=csin⁡C.\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}.

Use it for ASA, AAS, and some SSA problems. For example, if A=42∘A=42^\circ, B=71∘B=71^\circ, and a=15a=15, first find

C=180∘−42∘−71∘=67∘.C=180^\circ-42^\circ-71^\circ=67^\circ.

Then

b=15sin⁡71∘sin⁡42∘≈21.30,c=15sin⁡67∘sin⁡42∘≈20.91.b=\frac{15\sin71^\circ}{\sin42^\circ}\approx21.30, \qquad c=\frac{15\sin67^\circ}{\sin42^\circ}\approx20.91.

SSA can be ambiguous because

sin⁡θ=sin⁡(180∘−θ).\sin\theta=\sin(180^\circ-\theta).

If one possible angle is B1=sin⁡−1(k)B_1=\sin^{-1}(k), also test B2=180∘−B1B_2=180^\circ-B_1. Keep a candidate only if A+B<180∘A+B<180^\circ. This may produce zero, one, or two triangles.

The provides

a2=b2+c2−2bccos⁡A.a^2=b^2+c^2-2bc\cos A.

Use it for SAS and SSS data. With b=8b=8, c=11c=11, and included angle A=52∘A=52^\circ,

a2=82+112−2(8)(11)cos⁡52∘,a^2=8^2+11^2-2(8)(11)\cos52^\circ,

so a≈8.75a\approx8.75. For SSS data, rearrange the formula:

C=cos⁡−1(a2+b2−c22ab).C=\cos^{-1}\left(\frac{a^2+b^2-c^2}{2ab}\right).

Use the included angle with the two known sides, keep the angle unit consistent, and delay rounding until the final answer.

Takeaway: Match the data pattern to the law: ASA or AAS usually suggests the , SAS or SSS suggests the , and SSA requires an ambiguity check.

Modeling periodic behavior and distances

A describes a quantity that repeats in a regular pattern:

y=Asin⁡(Bx+C)+Dy=A\sin(Bx+C)+D

or

y=Acos⁡(Bx+C)+D.y=A\cos(Bx+C)+D.

Its parameters describe the graph as follows:

  • Amplitude: ∣A∣|A|.

  • Period, when xx is measured in radians: 2π∣B∣\frac{2\pi}{|B|}.

  • Horizontal phase shift: −CB-\frac{C}{B}.

  • Midline: y=Dy=D.

To find when a modeled quantity reaches a target value, substitute the target into the model and solve the resulting trigonometric equation. List every solution in the requested interval because repeated solutions can represent successive tides, cycles, or other recurring events.

Triangle laws support non-right-triangle applications. A measured baseline and angles from its endpoints can be combined with the to find a remote distance. Two known distances and their included angle can be combined with the to find a third distance.

Before finalizing an application problem, check:

  • Whether the calculator is in degree or radian mode.

  • Whether all relevant solutions in the interval have been included.

  • Whether a denominator could be zero.

  • Whether the angle used with the is actually included.

  • Whether an SSA configuration has a second possible triangle.

  • Whether early rounding has changed the result.

Takeaway: Translate the context into a trigonometric equation or triangle model, solve completely, and interpret only the solutions that fit the stated interval and situation.