Where the fundamental identities come from, the reciprocal, quotient, Pythagorean, cofunction and negative-angle families, finding all six function values from one, simplifying trigonometric expressions, and verifying identities.
Subject: Algebra 2 · 65 slides · symbolic lesson
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Title
Algebra 2 · Chapter 14 — Trigonometric Graphs, Identities, and Equations
Verify Trigonometric Identities
Objectives
Five outcomes. Relationships that hold for every angle at once.
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.3 Verify Trigonometric Identities §14.3, pp. 923-929 — the lesson these objectives are drawn from
Warm-up
On the unit circle the point where the terminal side crosses is cosine theta, sine theta, and the radius is 1.
Discussion prompt
Write the equation of the unit circle. Now substitute those coordinates. What have you produced?
Hint: The circle of radius 1 centred at the origin.
Answer:
\[ x^2+y^2 = 1 \;\Longrightarrow\; \cos^2\theta+\sin^2\theta = 1 \]
That equation holds for every angle theta, without exception. An equation true for all values of its variable is called an identity, and this one is the most important in trigonometry.
It is not something to be solved. It is something to be used.
Concept
A trigonometric identity is an equation true for all values of the variable in its domain. The fundamental identities come in five families: reciprocal, quotient, Pythagorean, cofunction and negative-angle. They are used to evaluate, to simplify, and to verify further identities.
trigonometric identity — A trigonometric equation that is true for every value of the variable in its domain, as opposed to an equation that holds only for particular values. Identities are verified rather than solved.
\[ \sin^2\theta+\cos^2\theta = 1 \]
Every one of the fundamental identities can be read off the unit circle or off the definitions, so none of them has to be memorised without understanding.
Figure (svg): Two columns comparing an identity with an ordinary equation
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.3 Verify Trigonometric Identities §14.3, pp. 924-924
Section
Section 1
Concept
Substituting cosine and sine for x and y in the unit circle's equation gives the first Pythagorean identity. Dividing it through by cosine squared or by sine squared gives the other two.
\[ \sin^2\theta+\cos^2\theta = 1; \; 1+\tan^2\theta = \sec^2\theta; \; 1+\cot^2\theta = \csc^2\theta \]
The reciprocal and quotient identities are simply the definitions of the six functions written out, so they need no derivation at all.
Figure (svg): The Pythagorean identity read off the unit circle
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.3 Verify Trigonometric Identities §14.3, pp. 924-924 — Fundamental Trigonometric Identities
Picture it
The unit circle and the identities it produces.
Figure (svg): The Pythagorean identity read off the unit circle
Dividing by cosine squared turns sine over cosine into tangent and 1 over cosine into secant, which is exactly the second identity.
Worked example
The derivation on page 924.
\[ \text{Show that } \sin^2\theta+\cos^2\theta = 1 \text{ for every angle } \theta. \]
Write the unit circle's equation
Why: Radius 1, centred at the origin.
\[ x ^{2} + y ^{2} = 1 \]
Recall the coordinates
Why: On the unit circle x is cosine and y is sine.
\[ x = \cos \theta, y = \sin \theta \]
Substitute
Why: Replace x and y.
\[ \cos ^{2} \theta + \sin ^{2} \theta = 1 \]
Note the scope
Why: Every angle has a crossing point.
Figure (svg): The Pythagorean identity read off the unit circle
\[ \cos^2\theta+\sin^2\theta = 1 \]
Verify: test it at a special angle
Why: At 30 degrees the sine is one half and the cosine is root 3 over 2, so the squares are one quarter and three quarters, summing to exactly 1. Testing an identity at one angle cannot prove it, but a failed test would instantly disprove it.
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.3 Verify Trigonometric Identities §14.3, pp. 924-924
Fill the middle
Example 1, Step 1.
Fill in the blanks
\sin^2\theta+\cos^2\theta = 1 \;\Longrightarrow\; \cos^2\theta = 1-sin^2 theta
Why: One minus sine squared theta. This rearrangement is used more often than the identity in its original form.
Worked example
The boxed identities on page 924.
\[ \text{Derive } 1+\tan^2\theta = \sec^2\theta \text{ and } 1+\cot^2\theta = \csc^2\theta. \]
Start from the first identity
Why: Sine squared plus cosine squared is 1.
Divide every term by cosine squared
Why: Three separate quotients.
\[ \tan ^{2} + 1 = \sec ^{2} \]
Return and divide by sine squared
Why: Again three quotients.
\[ 1 + \cot ^{2} = \csc ^{2} \]
Note when each is invalid
Why: Division needs a nonzero denominator.
Figure (svg): The solution to Worked example derive the other two shown as a ladder of expressions, one row per algebraic move
\[ 1+\tan^2\theta = \sec^2\theta; \qquad 1+\cot^2\theta = \csc^2\theta \]
Verify: check the second at 45 degrees
Why: Tangent of 45 is 1 and secant is root 2, so 1 plus 1 is 2 and root 2 squared is 2. They agree. Notice the identity is silent at 90 degrees, where both tangent and secant are undefined — which is what in its domain means in the definition.
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.3 Verify Trigonometric Identities §14.3, pp. 924-924
Trap
\[ \sin^2\theta+\cos^2\theta = 1 \]
Solve for theta
Why: The equals sign invites solving.
\[ \theta = ? \quad \text{(no particular answer)} \]
There is no solution to find, because EVERY angle satisfies it. Solving asks which values work, and here the answer is all of them.
\[ \text{use it as a substitution} \]
Treat an identity as a tool, not a problem
Why: It replaces one expression by an equal one.
\[ 1-\cos^2\theta = \sin^2\theta \]
That rearrangement is how the identity actually earns its keep, and it appears in nearly every problem in this lesson.
Matching
Five families of fundamental identities.
Match the pairs
Why: The first two are just the definitions restated, the third comes from the circle, and the fourth from its symmetry about the horizontal axis. Only the Pythagorean family required any work to establish.
Sorting
Is it true for every value?
Sort into buckets
Sort each statement.
The two ordinary equations are exactly the kind Lesson 14.5 will teach you to solve. Telling the two apart decides whether you verify or solve.
Prediction
Commit before reasoning.
Predict first
Dividing sine squared plus cosine squared equals 1 by cosine squared gives 1 plus tangent squared equals secant squared. Why is that legitimate?
Correct: Because dividing both sides of a true equation by the same nonzero quantity keeps it true, and each quotient simplifies to a named function.
\[ \frac{\sin^2}{\cos^2}+\frac{\cos^2}{\cos^2} = \frac{1}{\cos^2} \]
Why: Sine squared over cosine squared is tangent squared, cosine squared over itself is 1, and 1 over cosine squared is secant squared, so the whole equation rewrites into the second identity. The only cost is a restriction: wherever cosine is zero the division was illegal, and indeed both tangent and secant are undefined at exactly those angles. The identity's domain and its derivation agree.
Section
Section 2
Concept
The cofunction identities say each function of an angle equals its co-function of the complement. The negative-angle identities say cosine is unchanged by a sign flip while sine and tangent reverse.
\[ \sin\!\left(\tfrac{\pi}{2}-\theta\right) = \cos\theta; \qquad \sin(-\theta) = -\sin\theta \]
Cosine keeps its sign because reflecting across the horizontal axis leaves the x-coordinate alone, while sine and tangent both change.
Figure (svg): The five families of fundamental identities
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.3 Verify Trigonometric Identities §14.3, pp. 924-924 — Cofunction and Negative Angle Identities
Picture it
All five families of fundamental identities.
Figure (svg): The five families of fundamental identities
Cosine is the only even function among the three, which is why only it survives a negative angle unchanged.
Worked example
The boxed identities on page 924.
\[ \text{Explain why } \sin\!\left(\tfrac{\pi}{2}-\theta\right) = \cos\theta. \]
Draw a right triangle with acute angle theta
Why: Its other acute angle is the complement.
\[ \frac{\pi}{2}\text{ minus } \theta \]
Name the sides from theta
Why: Opposite, adjacent, hypotenuse.
Name them from the other angle
Why: The two legs swap roles.
Compare
Why: Sine of one is cosine of the other.
Figure (svg): The five families of fundamental identities
\[ \sin\!\left(\tfrac{\pi}{2}-\theta\right) = \cos\theta \]
Verify: test it at 30 degrees
Why: Sine of 60 is root 3 over 2 and cosine of 30 is root 3 over 2, and 60 is the complement of 30. The names carry the fact: co-sine literally means the sine of the complement, and the same holds for co-tangent and co-secant.
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.3 Verify Trigonometric Identities §14.3, pp. 924-924
Fill the middle
Example 2, first step.
Fill in the blanks
\tan\!\left(\tfraccot___-\theta\right) = ___\,\theta
Why: Cotangent theta. Each function of an angle equals the co-named function of its complement, which is where the co in cosine, cotangent and cosecant comes from.
Worked example
The boxed identities on page 924.
\[ \text{Explain why } \cos(-\theta) = \cos\theta \text{ but } \sin(-\theta) = -\sin\theta. \]
Draw both angles on the unit circle
Why: One counterclockwise, one clockwise.
Compare the x-coordinates
Why: Reflection leaves x alone.
Compare the y-coordinates
Why: Reflection negates y.
Deduce the tangent
Why: Y over x, with y negated.
Figure (svg): The solution to Worked example explain the negative-angle identities shown as a ladder of expressions, one row per algebraic move
\[ \cos(-\theta) = \cos\theta; \qquad \sin(-\theta) = -\sin\theta \]
Verify: test at 60 degrees
Why: Cosine of negative 60 is one half, the same as cosine of 60; sine of negative 60 is negative root 3 over 2, the opposite of sine of 60. In function language cosine is EVEN and sine is odd, which is visible in their graphs: cosine is symmetric about the vertical axis and sine about the origin.
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.3 Verify Trigonometric Identities §14.3, pp. 924-924
Error analysis
A student simplifies cos(-t) + cos t.
Annotate
On: \( \cos(-t)+\cos t = -\cos t+\cos t = 0 \)
The picture settles it: reflecting a point across the horizontal axis moves it up or down, never left or right, so the x-coordinate — the cosine — cannot change.
Sorting
Does a negative angle change the sign?
Sort into buckets
Sort each function by how it behaves at a negative angle.
A function and its reciprocal always share this property, since flipping a fraction cannot change its sign. So cosine and secant are even, and the other four are odd.
Matching
One identity each.
Match the pairs
Why: The first and last both simplify to cosine t but for entirely different reasons: one is a cofunction relationship and the other is evenness. Recognising which identity applies is what the family names are for.
Prediction
Commit before reasoning.
Predict first
Cosine is an even function. What does that say about its graph?
Correct: It is symmetric about the vertical axis, since the value at -t equals the value at t.
\[ \cos(-t) = \cos t \;\Longleftrightarrow\; \text{y-axis symmetry} \]
Why: Folding the cosine graph along the y-axis makes the two halves coincide exactly, which is the graphical meaning of an even function. Sine, being odd, has the other kind of symmetry: rotating its graph half a turn about the origin reproduces it. Both symmetries are visible in the graphs of Lesson 14.1, and both come straight from what reflecting a point across the horizontal axis does to its coordinates.
Section
Section 3
Concept
Given one function value and the quadrant, use a Pythagorean identity to find a second value, choosing the sign from the quadrant. The remaining four then follow from the quotient and reciprocal identities.
\[ \cos^2\theta = 1-\sin^2\theta \]
Taking a square root always produces two signs, and only the stated interval can decide between them.
Figure (svg): One known value producing the other five
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.3 Verify Trigonometric Identities §14.3, pp. 925-925 — Find trigonometric values
Picture it
Example 1: sine theta is four fifths with theta in quadrant two.
Figure (svg): One known value producing the other five
The identity gave plus or minus three fifths, and quadrant two chose the negative. Everything else is division.
Worked example
Example 1.
\[ \text{Given } \sin\theta = \tfrac{4}{5} \text{ with } \tfrac{\pi}{2} < \theta < \pi, \text{ find the other five values.} \]
Use the Pythagorean identity
Why: One minus sixteen twenty-fifths.
\[ \cos ^{2} \theta = \frac{9}{25} \]
Take square roots
Why: Both signs appear.
\[ \cos \theta =\text{ plus or minus } \frac{3}{5} \]
Choose the sign
Why: Quadrant two has cosine negative.
\[ \cos \theta = -\frac{3}{5} \]
Divide for the rest
Why: Quotient and reciprocal identities.
\[ \tan - \frac{4}{3}, \cot - \frac{3}{4}, \csc 5 / 4, \sec - \frac{5}{3} \]
Figure (svg): One known value producing the other five
\[ \cos\theta = -\tfrac{3}{5}, \; \tan\theta = -\tfrac{4}{3} \]
Verify: check the signs against quadrant two
Why: Only sine and cosecant came out positive, and quadrant two is exactly where sine and cosecant are positive while the other four are negative. Every one of the six agrees with the sign table of Lesson 13.3.
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.3 Verify Trigonometric Identities §14.3, pp. 925-925
Fill the middle
Example 1, Step 1.
Fill in the blanks
\cos^2\theta = 1-\!\left(\tfrac9___\right)^2 = 1-\tfrac______ = \tfrac___}___
Why: Nine twenty-fifths, so the cosine is plus or minus three fifths. The quadrant then selects the negative option.
Worked example
Guided Practice 1 and 2.
\[ \text{Find the other five values when } \cos\theta = \tfrac{1}{6} \text{ in quadrant I, and when } \sin\theta = -\tfrac{3}{7} \text{ in quadrant III.} \]
First: find sine
Why: One minus one thirty-sixth is 35/36.
\[ \sin \theta = \sqrt{35}\text{ over } 6 \]
First: divide for the rest
Why: All positive in quadrant one.
\[ \tan \sqrt{35}, \sec 6, \csc 6 \sqrt{35} / 35 \]
Second: find cosine
Why: One minus nine forty-ninths is 40/49.
\[ \cos \theta = -2 \sqrt{10}\text{ over } 7 \]
Second: divide for the rest
Why: Only tangent and cotangent positive.
\[ \tan 3 \sqrt{10} / 20, \csc - \frac{7}{3} \]
Figure (svg): The solution to Worked example two more shown as a ladder of expressions, one row per algebraic move
\[ \sin\theta = \tfrac{\sqrt{35}}{6}; \qquad \cos\theta = -\tfrac{2\sqrt{10}}{7} \]
Verify: check the second against quadrant three
Why: In quadrant three both coordinates are negative, so sine and cosine are negative but their quotient is positive — and indeed tangent came out positive while sine and cosine did not. Two negatives dividing to a positive is what makes quadrant three the tangent quadrant.
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.3 Verify Trigonometric Identities §14.3, pp. 926-926
Trap
\[ \sin\theta = \tfrac{4}{5} \;\Longrightarrow\; \cos\theta = \pm\tfrac{3}{5} \]
Report both possibilities
Why: The square root genuinely has two.
\[ \tan\theta = \pm\tfrac{4}{3} \quad \text{(incomplete)} \]
The question gave an interval, and that interval is there precisely to choose between the two signs. Reporting both ignores information.
\[ \tfrac{\pi}{2} < \theta < \pi \;\Longrightarrow\; \text{quadrant II} \]
Use the interval to pick the sign
Why: Cosine is negative there.
\[ \cos\theta = -\tfrac{3}{5}, \; \tan\theta = -\tfrac{4}{3} \]
Without an interval the question would genuinely have two answers, which is why one is always supplied.
Fill the middle
Example 1, Step 2.
Fill in the blanks
\tan\theta = \frac3___ = -\frac______}
Why: Negative four thirds. The fifths cancel, leaving the ratio of the numerators with the sign carried through.
Comparison
Fill the blanks. Sine four fifths, in two places.
Comparison matrix
| Question | Quadrant I | Quadrant II |
|---|---|---|
| sine | 4/5 | 4/5 |
| cosine | 3/5 | -3/5 |
| tangent | 4/3 | -4/3 |
| secant | 5/3 | -5/3 |
The sizes are identical and four of the six signs differ, which is exactly the reference-angle structure of Lesson 13.3 seen through the identities.
Prediction
Commit before reasoning.
Predict first
Why does a problem giving sine theta always also give an interval for theta?
Correct: Because the Pythagorean identity gives cosine only up to a sign, and the quadrant decides it.
\[ \cos\theta = \pm\sqrt{1-\sin^2\theta} \]
Why: A sine of four fifths occurs at two angles per turn, one in quadrant one and one in quadrant two, and those two have opposite cosines. The identity cannot tell them apart, because squaring destroys sign information. Only the interval can, which is why omitting it would leave the question with two equally correct answers.
Section
Section 4
Concept
To simplify a trigonometric expression, replace pieces of it using the fundamental identities until nothing further cancels. Writing the name of the identity beside each line keeps the work checkable.
\[ \tan\!\left(\tfrac{\pi}{2}-\theta\right)\sin\theta = \cos\theta \]
Rewriting everything in terms of sine and cosine is the reliable default when no better step suggests itself.
Figure (svg): A trigonometric expression simplified one identity at a time
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.3 Verify Trigonometric Identities §14.3, pp. 925-925 — Simplify a trigonometric expression
Picture it
Example 3, worked in full.
Figure (svg): A trigonometric expression simplified one identity at a time
Five lines, five named steps, ending at a single term. Every line is an equality justified by a specific identity.
Worked example
Examples 2 and 3.
\[ \text{Simplify } \tan\!\left(\tfrac{\pi}{2}-\theta\right)\sin\theta \text{ and } \csc\theta\cot^2\theta+\frac{1}{\sin\theta}. \]
First: apply the cofunction identity
Why: Tangent of the complement is cotangent.
First: apply the quotient identity
Why: Cotangent is cosine over sine.
Second: replace the fraction
Why: One over sine is cosecant.
\[ \csc \cot ^{2} + \csc \]
Second: use the Pythagorean identity and distribute
Why: Cotangent squared is cosecant squared minus 1.
\[ \csc ^{3} \theta \]
Figure (svg): A trigonometric expression simplified one identity at a time
\[ \cos\theta; \qquad \csc^3\theta \]
Verify: test the second numerically at 30 degrees
Why: Cosecant of 30 is 2 and cotangent is root 3, so the original is 2 times 3 plus 2, which is 8; and cosecant cubed is 2 cubed, also 8. A numerical check at one angle will not prove a simplification but will catch almost every mistake.
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.3 Verify Trigonometric Identities §14.3, pp. 925-925
Fill the middle
Example 3.
Fill in the blanks
\cot^2\theta = \csc^2\theta-1
Why: One. Rearranging the third Pythagorean identity is what lets the expression collapse to a single cosecant term.
Worked example
Guided Practice 3, 4 and 5.
\[ \text{Simplify } \sin x\cot x\sec x, \; \frac{\tan x\csc x}{\sec x}, \; \frac{\cos\!\left(\tfrac{\pi}{2}-\theta\right)-1}{1+\sin(-\theta)}. \]
First: write everything in sine and cosine
Why: Sine times cosine over sine times 1 over cosine.
\[ 1 \]
Second: the same approach
Why: Sine over cosine, times 1 over sine, times cosine.
\[ 1 \]
Third: apply cofunction and negative angle
Why: Cosine of the complement is sine; sine of negative is negative sine.
\[ \frac{\sin \theta - 1}{1 - \sin \theta} \]
Third: factor out a negative
Why: The numerator is the negative of the denominator.
\[ -1 \]
Figure (svg): The solution to Worked example three more shown as a ladder of expressions, one row per algebraic move
\[ 1; \quad 1; \quad -1 \]
Verify: notice the third one's structure
Why: Sine theta minus 1 is exactly the negative of 1 minus sine theta, so the quotient is negative 1 for every theta where the denominator is nonzero. Spotting that a numerator is the negative of a denominator is worth more than any amount of further expansion.
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.3 Verify Trigonometric Identities §14.3, pp. 926-926
Error analysis
A student simplifies the quotient of sine theta plus 1 by sine theta.
Annotate
On: \( \frac{\sin\theta+1}{\sin\theta} = 1+1 = 2 \)
This is an algebra error rather than a trigonometry one, and it is the commonest single mistake in this lesson. Split the fraction first, then simplify each piece.
Matching
Rewrite in sine and cosine if in doubt.
Match the pairs
Why: Two of the four collapse all the way to 1, which is common: an expression built from three or four functions of the same angle very often cancels completely once everything is written in sine and cosine.
Sorting
Algebra rules still apply.
Sort into buckets
Sort each proposed step.
The illegal step is pure algebra, not trigonometry. Splitting the fraction first, as the legal version does, turns it into two honest pieces.
Prediction
Commit before reasoning.
Predict first
You are stuck simplifying an expression with tangents, secants and cotangents in it. What should you try?
Correct: Rewrite every function in terms of sine and cosine, then use ordinary fraction algebra.
\[ \tan = \tfrac{\sin}{\cos}, \; \sec = \tfrac{1}{\cos}, \; \cot = \tfrac{\cos}{\sin} \]
Why: All six functions are built from sine and cosine, so that rewriting always succeeds in putting everything on common ground, after which combining and cancelling fractions is familiar work. It is rarely the shortest route but it is the one that always works, which makes it the right thing to reach for when no clever step presents itself.
Section
Section 5
Concept
To verify an identity, begin with the expression on one side and use algebra and identities to turn it into the other side. Never operate on both sides at once, since that would assume what you are trying to prove.
\[ \frac{\sec^2\theta-1}{\sec^2\theta} = \sin^2\theta \]
Multiplying by a conjugate over itself is multiplying by 1, and it creates a difference of squares that a Pythagorean identity can then simplify.
Figure (svg): An identity verified by working on one side only
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.3 Verify Trigonometric Identities §14.3, pp. 926-926 — Verify a trigonometric identity
Picture it
Example 5, verified in full.
Figure (svg): An identity verified by working on one side only
Every line is the previous one rewritten, and the conjugate trick in the middle is what creates the difference of squares the identity needs.
Worked example
Example 4.
\[ \text{Verify } \frac{\sec^2\theta-1}{\sec^2\theta} = \sin^2\theta. \]
Split the fraction
Why: Two separate quotients.
\[ 1 - 1 / \sec ^{2} \theta \]
Apply the reciprocal identity
Why: One over secant is cosine.
\[ 1 - \cos ^{2} \theta \]
Apply the Pythagorean identity
Why: One minus cosine squared.
\[ \sin ^{2} \theta \]
Compare with the target
Why: The two sides now match.
Figure (svg): The solution to Worked example a short verification shown as a ladder of expressions, one row per algebraic move
\[ 1-\cos^2\theta = \sin^2\theta \]
Verify: notice the shortcut hiding in the first line
Why: The numerator secant squared minus 1 is tangent squared by the second Pythagorean identity, so the whole fraction is tangent squared over secant squared, which is sine squared over cosine squared times cosine squared. Both routes work; the printed one needs fewer identities.
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.3 Verify Trigonometric Identities §14.3, pp. 926-926
Fill the middle
Example 4, first step.
Fill in the blanks
\fraccos^2 theta___ = 1-\frac______ = 1-___
Why: One minus cosine squared theta, which the Pythagorean identity turns into sine squared theta.
Worked example
Example 5.
\[ \text{Verify } \sec x+\tan x = \frac{\cos x}{1-\sin x}. \]
Write the left side over cosine
Why: Reciprocal and quotient identities.
\[ \frac{1 + \sin x}{\cos x} \]
Multiply by the conjugate over itself
Why: One minus sine x, top and bottom.
\[ \text{multiplying by } 1 \]
Expand the numerator
Why: A difference of squares.
\[ \frac{1 - \sin ^{2} x}{\cos x(1 - \sin x)} \]
Apply the Pythagorean identity and cancel
Why: One minus sine squared is cosine squared.
\[ \cos x / (1 - \sin x) \]
Figure (svg): An identity verified by working on one side only
\[ \frac{\cos x}{1-\sin x} \]
Verify: explain why the conjugate was chosen
Why: The target has 1 minus sine x underneath, and the left side had nothing like it, so that factor had to be introduced deliberately. Multiplying by it over itself is legal because the quotient is 1, and choosing the conjugate is what makes the numerator collapse via the difference of squares.
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.3 Verify Trigonometric Identities §14.3, pp. 926-926
Trap
\[ \sec x+\tan x = \frac{\cos x}{1-\sin x} \]
Cross-multiply and simplify both sides
Why: The equation is treated as one to be solved.
\[ (\sec x+\tan x)(1-\sin x) = \cos x \]
This assumes the very equation being proved. A verification must not use its own conclusion as a step.
\[ \text{start with } \sec x+\tan x \text{ alone} \]
Transform one side only
Why: Each line equals the one above it, independently of the claim.
\[ \;\dots\; = \frac{\cos x}{1-\sin x} \]
Arriving at the other side then proves the identity, because a chain of equalities has been built without assuming it.
Fill the middle
Example 5.
Fill in the blanks
\text1 - sin x 1-\sin x \text___ \frac______}
Why: One minus sine x. Multiplying by a quantity over itself changes nothing, which is what makes the step legal.
Sorting
Only one side may be transformed.
Sort into buckets
Sort each move.
The distinction matters logically, not just stylistically: a proof that uses its own conclusion proves nothing, however true the conclusion happens to be.
Prediction
Commit before reasoning.
Predict first
Why is operating on both sides of an identity forbidden when verifying it?
Correct: Because doing so assumes the two sides are already equal, which is exactly what is being proved.
\[ \text{LHS} = \dots = \text{RHS} \; \text{ proves it; assuming } \text{LHS}=\text{RHS} \text{ does not} \]
Why: Every legitimate operation on both sides of an equation presumes there IS an equation to operate on. Starting from the claim and reaching something true shows only that the claim WOULD imply that truth, not that the claim holds. Transforming one side into the other builds a chain of equalities that never assumes the conclusion, which is why it is the required form. Solving an equation is different precisely because there the goal is to find which values make it true, not to prove it always is.
Comparison
Fill the blanks. Two things an equals sign can mean.
Comparison matrix
| Question | Verifying an identity | Solving an equation |
|---|---|---|
| True for | every value in the domain | particular values only |
| Goal | show the two sides are the same | find which values work |
| Method | transform one side only | operate on both sides |
| Answer looks like | a chain of equalities | a list of values |
The instruction word settles which you are doing. Verify means prove it always holds; solve means find where it holds.
Pattern
Identify, substitute, simplify.
Cancelling is legal only across a whole factor, never across one term of a sum.
Check
The quadrant chooses the sign.
Check your understanding
If sin theta = 4/5 and theta is in quadrant II, what is cos theta?
Answer: A
Why: The identity gives cosine squared equal to 9/25, and quadrant II makes cosine negative.
Check
Cosine is even; sine is odd.
Check your understanding
What does cos(-t) equal?
Answer: A
Why: Reflecting a point across the horizontal axis leaves its x-coordinate unchanged.
Check
Substitute identities one at a time.
Check your understanding
Simplify tan(pi/2 - t) sin t.
Answer: A
Why: The cofunction identity gives cot t, and cot t times sin t is cos t.
Real world
A projectile launched at angle theta with speed v travels a horizontal distance of v squared over 32, times 2 sine theta cosine theta.
Discussion prompt
Physics texts usually write this range formula with a single sine instead. Can you see why the two agree, and what it says about the best launch angle?
Hint: Lesson 13.3 used the range formula with sine of 2 theta in it.
Answer:
\[ 2\sin\theta\cos\theta = \sin 2\theta \]
\[ d = \frac{v^2}{32}\sin 2\theta \]
The two forms are the same expression, related by the double-angle identity that Lesson 14.7 will establish.
The rewritten form makes the answer obvious in a way the original does not. Sine peaks at 1 when its argument is 90 degrees, so the range is greatest when 2 theta is 90, that is at a launch angle of 45 degrees. Staring at 2 sine theta cosine theta could leave you differentiating; recognising it as sine of 2 theta needs no calculus at all. That is what identities buy: not new facts, but a form in which the facts you want are visible.
Commit first
Answer, then rate your confidence honestly.
Predict first
Is sine squared theta plus cosine squared theta equal to 1 an equation to be solved?
Correct: No — it is an identity, true for every theta, so there is nothing to solve.
\[ \sin^2 100^\circ+\cos^2 100^\circ = 1 \text{ too} \]
Why: Every angle whatsoever satisfies it, because it is just the unit circle's own equation with the coordinates renamed. Asking which values of theta make it true has the unhelpful answer all of them. That is precisely what distinguishes an identity from an equation, and it is why identities are verified and used as substitutions rather than solved. Lesson 14.5 will return to genuine equations, where solving is exactly the right thing to do.
Explain it
They have solved equations and never met an identity.
Discussion prompt
In four sentences or fewer, explain the difference between an identity and an equation.
Hint: Compare with x plus x equals 2x.
Answer:
An equation like x plus 3 equals 7 is true for one particular value, and solving it finds that value. An identity like x plus x equals 2x is true for every value you could put in.
Trigonometric identities are the second kind: sine squared plus cosine squared is 1 for every angle there is. So you never solve them, you use them to rewrite other expressions.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For signs, name the quadrant explicitly before taking any square root. For even and odd, only cosine and secant keep their sign, because only they depend on x alone. For conjugates, look at what the target expression has that yours lacks, and introduce it. For verifying, physically cover the other side of the page until you have finished.
Connect it up
Paper. Fifteen minutes.
Draw it
Build an identities page. Top left: draw the unit circle with a labelled point, write the circle's equation, and derive all three Pythagorean identities from it, showing what you divided by each time. Top right: write out the reciprocal, quotient, cofunction and negative-angle families in full, marking beside the last which functions are even and which are odd. Middle: take one given value with a quadrant and find the other five, boxing the moment where the quadrant chose a sign. Bottom left: simplify three expressions of your own choosing, naming the identity used on every single line. Bottom right: verify one identity that needs a conjugate, keeping the target side untouched and covered until the last line.
If any verification on your page operated on both sides at once, it proves nothing. Redo it starting from one side alone.
Recap
Five things, and relationships that hold everywhere at once.
| If you see | Then |
|---|---|
| One value and an interval | Use a Pythagorean identity, then the quadrant for the sign |
| A complement inside a function | Use a cofunction identity |
| A negative angle | Only cosine and secant keep their sign |
| A messy expression | Rewrite everything in sine and cosine |
| An identity to verify | Transform one side only |
| A factor the target has and you lack | Multiply by it over itself |
Lesson 14.4 puts the graphing of 14.1 and 14.2 to work on real data: writing a trigonometric model from a maximum, a minimum and a period.
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.3 Verify Trigonometric Identities §14.3, pp. 923-929 — everything on these slides traces back here
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