Using the identities: rewrite and prove

Explain it like I am five

Proving an identity, an equation true for every allowed angle, is like showing 4 quarters and 100 pennies are the same dollar: trade for equal pieces until one side matches the other. LHS is the left-hand side of the equals sign, RHS the right-hand side. A one-side chain rewrites only one side, LHS = step = step = RHS, with a reason at each step.

Prove (1 − cos θ)(1 + cos θ) = $\sin^{2}$ θ. Start on the LHS, the busier side. Multiply out, each part times each part: 1 + cos θ − cos θ − $\cos^{2}$ θ = 1 − $\cos^{2}$ θ, since the middle terms cancel (the difference of squares: (a − b)(a + b) = $a^{2}$ − $b^{2}$ ). Subtracting $\cos^{2}$ θ from both sides of $\sin^{2}$ θ + $\cos^{2}$ θ = 1 gives 1 − $\cos^{2}$ θ = $\sin^{2}$ θ. So LHS = $\sin^{2}$ θ = RHS.

Check at 30°: (1 − $square root of 3 over 2$ )(1 + $square root of 3 over 2$ ) = 1 − $3 over 4$ = $1 over 4$ , and $\sin^{2}$ 30° = ( $1 over 2$ $) squared$ = $1 over 4$ . The check catches slips; the chain is the proof, since each step holds for every angle.

In plain words

You can pay the same amount with a $5 bill or five $1 bills. An identity makes a similar trade: you replace an expression with an equal expression wherever both are defined. A domain is the collection of allowed inputs. A common domain consists of inputs allowed by both expressions. Rewriting asks you to express one function using only another. Proving an identity, also called verifying it, asks you to show why two expressions agree for every allowed angle. You start with a side you understand, rewrite it through equal expressions and arrive at the other side. The left side, or LHS, is before the equals sign. The right side, or RHS, is after it.

θ + θ = 1

1 − θ = θ

Three forms of one identity; spotting one of them is usually the key move in a proof.

Reminder

Common denominators. $2 over 3$ + $1 over 4$ = $8 over 12$ + $3 over 12$ = $11 over 12$ ; multiply top and bottom to make equal bottoms.
Difference of squares. (1 + u)(1 − u) = 1 − $u^{2}$ because the +u and −u terms cancel.
Nonzero cancellation. $\frac{a \times b}{a}$ = b requires a ≠ 0. $7 × 3 over 7$ = 3, while division by 0 has no value.
Root signs. $x^{2}$ = $21 over 25$ gives x = ± $square root of 21 over 5$ ; an acute angle makes cosine positive.

Why it works. A one-side chain displays the reason for every equality, so a reader can follow it without assuming the desired conclusion. Working on both sides can also be valid if every step is reversible on the stated domain. But dividing by something that might be zero can discard cases, and reducing a claimed equation to a true statement does not prove it unless you justify reversing every step. Sine and cosine give all six functions a shared language. Then the Pythagorean identity replaces complicated squares with 1, or replaces 1 minus one square with the other.

RuleFor an acute angle, cos θ = and tan θ = . An identity holds on its common domain. Use sin θ ≠ 0 when dividing by sin θ and cos θ ≠ 0 when dividing by cos θ.

The same idea, five ways

Say it

Say prove the identity: show that the two expressions agree for every allowed input.

Write it

A proof rewrites equal expressions on the original common domain.

In math

LHS = expression = expression = RHS
cos θ ≠ 0 for tan θ and sec θ
sin θ ≠ 0 for cot θ and csc θ

Exchange one payment for another of equal value, keeping track of each exchange.

See it

Start with LHS

Rewrite with known equalities

Arrive at RHS

An algebra chain provides the proof; a sample input provides a check.

The same idea, other ways

As exchanging money

Five dollar bills and one five-dollar bill look different but have the same value. Rewrite an expression without changing its value, and give the reason for each exchange.

θ + θ = 1

1 − θ = θ

These are two ways to write the same Pythagorean fact.

As a chain of evidence

LHS = expression = expression = RHS is a path whose links all have reasons. A matching final expression proves equality for every input on the stated domain.

sec θ cos θ

= cos θ

= 1, with cos θ ≠ 0

Each equality must remain valid for the original allowed inputs.

As a common language

Replace sec with 1 over cosine, csc with 1 over sine, tangent with sine over cosine and cotangent with cosine over sine. Now all fractions use only two ingredients.

tan θ =

cot θ =

sec θ =

csc θ =

Check each denominator before using this shared language.

.1Rewrite using only one function

The words in terms of mean use this ingredient and no other trig function. If sine is the ingredient, use the Pythagorean identity to replace cosine. An acute angle selects the positive root.

Rule: for 0° < θ < 90°, cos θ = $square root of 1 − sin squared θ$ .
Rule: for that same acute angle, tan θ = $sin θ over square root of 1 − sin squared θ$ .
Rule: for a general angle, solving a square gives both root signs until angle information selects one.

θ = 1 − θ

cos θ = , acute θ

tan θ =

The acute-angle condition selects the positive root.

Worked exampleUse a fresh acute sine value

θ is acute and sin θ = $2 over 5$ . Find cos θ and tan θ. In words, rebuild the missing cosine from the identity, then divide sine by cosine.

The acute triangle gives a second route to the two ratios.

$\cos^{2}$ θ = 1 − $4 over 25$ = $21 over 25$ .Subtract the squared sine from 1, written as $25 over 25$ .
cos θ = $square root of 21 over 5$ .An acute cosine is positive; the principal root gives this value.
tan θ = $2 over 5$ ÷ $square root of 21 over 5$ = $2 over square root of 21$ = $2 square root of 21 over 21$ .Use the quotient identity, cancel the shared bottom and rationalize.

Answer

cos θ = $square root of 21 over 5$
tan θ = $2 square root of 21 over 21$

Check The squared sine and cosine are and , whose sum is 1.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: Use only the positive square root for every angle.

The identity fixes the square, not the sign; general angles need sign information.

✓ Instead: Use the positive root for an acute angle, and otherwise keep ± until the sign is established.

Tips and tricks

Write the acute-angle condition next to the positive root.

.2Prove (verify) an identity

Prove means show why the statement holds for every input where its original expressions have values. A chain from one side to the other makes each reason visible. You do not need to guess the angle.

Rule: rewrite using already established identities.
Rule: every cancelled denominator factor must be nonzero.
Strategy: a one-side chain is convenient; reversible two-side reasoning can also be valid when justified.

sec θ cos θ

= cos θ

= 1, cos θ ≠ 0

The proof keeps its original cosine restriction.

Worked exampleA one-flip proof

Prove sec θ cos θ = 1 where cos θ ≠ 0. In words, show that cosine multiplied by its reciprocal gives 1.

cos θ ≠ 0

× cos θ = 1

A reciprocal product is 1 only when the original number is nonzero.

sec θ cos θ = $1 over cos θ$ × cos θ.The reciprocal identity is valid for cos θ ≠ 0.
$\frac{\cos θ}{\cos θ}$ = 1.The same nonzero number divided by itself equals 1.

Answer

sec θ cos θ = 1 for cos θ ≠ 0.

Check At 405°, sec θ = and cos θ = , so their product is 1.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: An undefined original side becomes defined after simplification.

Simplification preserves values on the original domain, not outside it.

✓ Instead: Retain the original nonzero conditions.

Tips and tricks

Write LHS at the start and RHS at the end of the chain.

Strategy: step by step

State the allowed inputs before rewriting: tan and sec need cos θ ≠ 0; cot and csc need sin θ ≠ 0. A displayed fraction also needs its denominator nonzero.
Pick the messier side, usually the one with more terms or fractions.
Rewrite every tan, cot, sec and csc using sin θ and cos θ.
Combine fractions over a common denominator, or multiply out brackets.
Replace 1 − $\sin^{2}$ θ with $\cos^{2}$ θ, 1 − $\cos^{2}$ θ with $\sin^{2}$ θ, and $\sin^{2}$ θ + $\cos^{2}$ θ with 1.
Cancel common factors only when those factors are nonzero. Stop at the other side; preserve the original allowed inputs.

Strategy

Rewrite or prove without changing the domain

Does the question ask for one function in terms of another?

YesIsolate that function with the identities; choose the root sign from the stated angle information.

NoBuild a one-side proof chain.

↓

Does a denominator contain sine or cosine?

YesRecord its nonzero requirement before proceeding.

NoKeep checking any later introduced denominator.

↓

Do the fractions have different bottoms?

YesUse their product as a common denominator when both are nonzero.

NoCombine the tops.

↓

Do you see θ + θ, 1 − θ or 1 − θ?

YesReplace with 1, θ or θ respectively.

NoTry factoring or rewriting the remaining functions.

Translate the question: rewriting changes the ingredients; proving establishes equality.
Record all original nonzero denominators.
Choose one side and rewrite all functions with sine and cosine.
Combine fractions or expand brackets with a reason.
Use the Pythagorean identity and cancel only nonzero factors.
End at the requested expression and retain the domain restrictions.

Worked exampleProve sec θ − cos θ = sin θ tan θ

Prove sec θ − cos θ = sin θ tan θ for cos θ ≠ 0. In words, rewrite the left expression until it becomes the right expression.

cos θ ≠ 0

= sin θ tan θ

Keep the nonzero condition while trading one equal expression for another.

Work where cos θ ≠ 0.Both sec θ and tan θ divide by cosine; this records the original domain before any cancellation.
LHS = sec θ − cos θ = $1 over cos θ$ − cos θ.The left side has more pieces, so start there. Reciprocal identity: sec θ = $1 over cos θ$ .
= $1 over cos θ$ − $\frac{\cos^{2} θ}{\cos θ}$ = $1 − cos squared θ over cos θ$ .cos θ ≠ 0 lets you multiply cosine by $\frac{\cos θ}{\cos θ}$ , which equals 1. Match the bottoms, then subtract the tops.
= $\frac{\sin^{2} θ}{\cos θ}$ .Pythagorean identity: 1 − $\cos^{2}$ θ = $\sin^{2}$ θ.
= sin θ × $\frac{\sin θ}{\cos θ}$ = sin θ tan θ = RHS.Split $\sin^{2}$ θ into sin θ × sin θ, and $\frac{\sin θ}{\cos θ}$ = tan θ.

Answer

The left side turns into sin θ tan θ, the right side, so the identity holds wherever both sides are defined.

Check At θ = 420°, the special-triangle work gives cos θ = , sec θ = 2, sin θ = and tan θ = . Each side is . This numerical check catches errors; the algebra chain proves the identity.

Ladder: from easy to exam-hard. Press Try it first on any rung to hide its steps and use them as hints.

Rung 1Rung 1: rearrange the identity

Prove 1 − $\cos^{2}$ θ = $\sin^{2}$ θ. In words, isolate the sine square in the known identity.

θ + θ = 1

1 − θ = θ

Rearrange a known identity rather than assuming the proposed one.

$\sin^{2}$ θ + $\cos^{2}$ θ = 1.This is the established Pythagorean identity.
$\sin^{2}$ θ = 1 − $\cos^{2}$ θ.Subtract $\cos^{2}$ θ from both sides of a known true equation.
1 − $\cos^{2}$ θ = $\sin^{2}$ θ.Equality can be read in either order.

Answer

The identity holds for every real θ.

Check At 405°, the two sides both equal .

Rung 2Rung 2: one reciprocal and a cancellation

Prove sec θ cos θ = 1 for cos θ ≠ 0. In words, replace secant with its definition.

cos θ ≠ 0

= 1

Record the condition before cancelling.

LHS = $1 over cos θ$ × cos θ.Secant is reciprocal cosine on this domain.
= $\frac{\cos θ}{\cos θ}$ = 1 = RHS.The denominator and cancelled factor are nonzero.

Answer

The identity holds when cos θ ≠ 0.

Check At 420°, 2 × = 1.

Rung 3Rung 3: Proving a difference of two tan fractions equals 2 csc θ

Prove the identity $tan θ over 1 − cos θ$ − $tan θ over 1 + cos θ$ = 2 csc θ. Begin by stating the values of θ for which the identity is claimed, and give a reason for every cancellation you make.

State the allowed inputs. tan θ needs cos θ ≠ 0. The denominators 1 − cos θ and 1 + cos θ must be nonzero, so cos θ ≠ 1 and cos θ ≠ −1. Since $\sin^{2}$ θ = 1 − $\cos^{2}$ θ, cos θ = ±1 exactly when sin θ = 0, so this condition is sin θ ≠ 0. csc θ also needs sin θ ≠ 0. The identity is therefore claimed for all θ with sin θ ≠ 0 and cos θ ≠ 0, that is, θ is not a multiple of 90°.An identity holds only on its common domain. Every tan, every csc and every displayed denominator adds a condition, and these conditions are what make the later cancellation legal.
Work on the left side, $tan θ over 1 − cos θ$ − $tan θ over 1 + cos θ$ , and aim for 2 csc θ.The left side has two fractions and two tan terms, while the right side is a single term, so the left side is the messier side.
Rewrite tan θ as $\frac{\sin θ}{\cos θ}$ in both terms: $tan θ over 1 − cos θ$ = $\frac{\sin θ}{\cos θ}$ · $1 over 1 − cos θ$ = $sin θ over cos θ(1 − cos θ)$ , and in the same way $tan θ over 1 + cos θ$ = $sin θ over cos θ(1 + cos θ)$ .Once the expression contains only sin θ and cos θ, the Pythagorean identity can act on it. Dividing $\frac{\sin θ}{\cos θ}$ by 1 − cos θ multiplies the two denominators together.
Use the common denominator cos θ(1 − cos θ)(1 + cos θ). Multiply the top and bottom of the first fraction by 1 + cos θ, and the top and bottom of the second by 1 − cos θ. The numerator becomes sin θ(1 + cos θ) − sin θ(1 − cos θ) = sin θ + sin θ cos θ − sin θ + sin θ cos θ = 2 sin θ cos θ.Two fractions can be subtracted only over a shared denominator. Both denominators already contain cos θ and differ only in the factors 1 − cos θ and 1 + cos θ, which are nonzero on the domain.
Multiply out the denominator: (1 − cos θ)(1 + cos θ) = 1 − $\cos^{2}$ θ, so the left side is $2 sin θ cos θ over cos θ(1 − cos squared θ)$ . Replace 1 − $\cos^{2}$ θ with $\sin^{2}$ θ to get $2 sin θ cos θ over cos θ sin squared θ$ .(1 − cos θ)(1 + cos θ) is a difference of squares, and $\sin^{2}$ θ + $\cos^{2}$ θ = 1 rearranges to 1 − $\cos^{2}$ θ = $\sin^{2}$ θ.
Write the denominator as sin θ · sin θ cos θ and cancel the common factor sin θ cos θ: $2 sin θ cos θ over sin θ · sin θ cos θ$ = $2 over sin θ$ .On the stated domain sin θ ≠ 0 and cos θ ≠ 0, so sin θ cos θ ≠ 0, and dividing the top and bottom by it is allowed.
Rewrite $2 over sin θ$ = 2 · $1 over sin θ$ = 2 csc θ. This is the right side, so stop. Conclude that the identity holds for all θ with sin θ ≠ 0 and cos θ ≠ 0.csc θ = $1 over sin θ$ . Keep the original allowed inputs: 2 csc θ on its own is also defined where cos θ = 0, but the left side is not defined there.

Answer

− = = = 2 csc θ. The identity is true for every θ with sin θ ≠ 0 and cos θ ≠ 0, that is, θ not a multiple of 90°.

Check At θ = 60°, sin 60° = , cos 60° = and tan 60° = . Left side: ÷ − ÷ = 2 − = . Right side: 2 csc 60° = 2 · = = . The two sides match. At θ = 120°, cos 120° = − and tan 120° = −. Left side: − ÷ − (−) ÷ = − + 2 = , which equals 2 csc 120° = , so the identity also holds beyond acute angles. At θ = 90°, the right side is 2 csc 90° = 2, but tan 90° is undefined, so the left side is undefined. This confirms that cos θ ≠ 0 must stay in the domain.

Work to write

Domain: cos θ ≠ 0 (for tan θ) and sin θ ≠ 0 (for 1 − cos θ ≠ 0, 1 + cos θ ≠ 0 and csc θ), so θ is not a multiple of 90°
LHS = $sin θ over cos θ(1 − cos θ)$ − $sin θ over cos θ(1 + cos θ)$
= $sin θ(1 + cos θ) − sin θ(1 − cos θ) over cos θ(1 − cos θ)(1 + cos θ)$ = $2 sin θ cos θ over cos θ(1 − cos squared θ)$
= $2 sin θ cos θ over cos θ sin squared θ$ , using 1 − $\cos^{2}$ θ = $\sin^{2}$ θ
= $2 over sin θ$ , cancelling sin θ cos θ ≠ 0
= 2 csc θ = RHS, for all θ with sin θ ≠ 0 and cos θ ≠ 0

$tan θ over 1 − cos θ$ − $tan θ over 1 + cos θ$ = $2 sin θ cos θ over cos θ sin squared θ$ = $2 over sin θ$ = 2 csc θ. The identity is true for every θ with sin θ ≠ 0 and cos θ ≠ 0, that is, θ not a multiple of 90°.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: Both sides agree at one angle, so the identity is proved.

Different expressions can agree at one input: tan 45° = 1, while tan 30° = ≠ 1. Agreement at 45° cannot prove tan θ = 1 for every allowed θ.

✓ Instead: Use an algebra chain for all allowed inputs; use one angle only to check.

✗ Not this: Cancelling cos θ makes sec θ cos θ = 1 valid even when cos θ = 0.

The original secant is undefined when its denominator is zero. Cancellation cannot create new allowed inputs.

✓ Instead: State cos θ ≠ 0 and retain that restriction.

✗ Not this: Cancel sin θ from to get 1.

The numerator is a sum, not a product with a common sine factor.

✓ Instead: For sin θ ≠ 0, split it into + 1 = csc θ + 1.

Tips and tricks

Begin with the domain and a one-side chain. Put a reason beside every equality.
A product can lose a common factor; a sum must first be split or factored.
Try a permitted special angle after proving the identity. It can expose an arithmetic error.

Trap. Testing one angle and calling the identity proved. One angle can agree by coincidence; a proof needs algebra that works for every θ. Use a test angle only to check your algebra afterward.

Keep in mind

LHS is everything left of the equals sign and RHS everything right of it: in tan θ · cos θ = sin θ, the LHS is tan θ · cos θ.
Cancel only factors (things multiplied), never terms (things added): $\frac{\sin θ \cos θ}{\cos θ}$ = sin θ, but $sin θ + cos θ over cos θ$ must be split into tan θ + 1.
"Respectively" means in the same order: $\sin^{2}$ θ + $\cos^{2}$ θ, 1 − $\sin^{2}$ θ and 1 − $\cos^{2}$ θ become 1, $\cos^{2}$ θ and $\sin^{2}$ θ, respectively.
One matching angle proves nothing: sin θ = cos θ holds at 45° but fails at 30°, so it is not an identity.

Memory hookMSS: start on the Messy side, rewrite everything in Sine and cosine, Swap the squares (1 − θ = θ), and stop when it matches the other side.

Flash cards: say the answer out loud, then flip

What do LHS and RHS mean?

Left-hand side and right-hand side of the equals sign.

What is a one-side proof chain?

LHS = step = step = RHS: you rewrite only one side, with a reason for each step.

What does "express cos θ in terms of sin θ" mean for an acute θ?

Write cos θ using only sin θ: cos θ = .

Simplify , where cos θ ≠ 0.

= cos θ

Rewrite cot θ × sin θ in sine and cosine, then simplify (sin θ ≠ 0).

× sin θ = cos θ

Both sides of an equation give the same number at 45°. Is it proved?

No. One angle can match by luck. A proof needs an algebra chain that works for every allowed angle.