Grade by Grade
D. Trigonometric Functions · Trigonometric Ratios, Identities and Laws

Proving Trigonometric Identities: Strategies

D1.5About 70 minutes
Learning goals
  • I can choose a strategy to start a proof: work on the more complicated side and rewrite in sine and cosine.
  • I can use common denominators and factoring (including difference of squares) in a proof.
  • I can prove identities that need two or more identities and several algebra steps.
  • I can test whether an equation might be an identity before trying to prove it.

Words to know

common denominator
A shared denominator that lets you add or subtract fractions, for example a/b + c/d = (ad + bc)/(bd).
difference of squares
The factoring pattern a² − b² = (a − b)(a + b).
conjugate
For an expression like 1 − sin θ, its conjugate is 1 + sin θ. Multiplying them gives 1 − sin²θ.
restriction
A value that must be excluded because it makes a denominator zero, such as cos θ ≠ 0.

A puzzle with the answer on the box

Proving an identity is like a jigsaw puzzle: you already know what the finished picture (the other side) looks like. The challenge is finding a path to it. In lesson 6 the proofs took one or two steps. This lesson gives you a toolkit for longer ones, using algebra you learned in the unit Functions and Equivalent Expressions: common denominators, factoring and simplifying rational expressions.

1. Start with themore complicated side2. Rewrite insin θ and cos θ3. Combine fractions(common denominator)4. Factor, or usesin²θ + cos²θ = 15. Simplify. Does itmatch the other side?Yes: L.S. = R.S.Write the conclusion.Not yet? Try the other side, or another tool.
A reliable plan for proving identities. You may not need every step, and sometimes it is easier to simplify both sides to the same expression.

Your toolkit

  • Identities: csc θ = 1/sin θ, sec θ = 1/cos θ, cot θ = 1/tan θ, tan θ = sin θ/cos θ, cot θ = cos θ/sin θ, sin²θ + cos²θ = 1.
  • Algebra: common denominators, expanding, common factoring, difference of squares a² − b² = (a − b)(a + b), and cancelling common factors.
  • Look for 1 − sin²θ (= cos²θ) or 1 − cos²θ (= sin²θ). They appear all the time.
Worked example

Example 1: Reciprocals, then Pythagoras

Prove sin θ/csc θ + cos θ/sec θ = 1.

  1. L.S. = sin θ ÷ (1/sin θ) + cos θ ÷ (1/cos θ).
  2. Dividing by a fraction means multiplying by its reciprocal: L.S. = sin θ × sin θ + cos θ × cos θ = sin²θ + cos²θ.
  3. Pythagorean identity: L.S. = 1.
  4. R.S. = 1.

L.S. = R.S. ✓

Worked example

Example 2: Common denominator

Prove tan θ + 1/tan θ = 1/(sin θ cos θ).

  1. L.S. = sin θ/cos θ + cos θ/sin θ (quotient identity, and 1/tan θ = cot θ = cos θ/sin θ).
  2. Common denominator sin θ cos θ: L.S. = (sin²θ + cos²θ)/(sin θ cos θ).
  3. Pythagorean identity: L.S. = 1/(sin θ cos θ).
  4. R.S. = 1/(sin θ cos θ).

L.S. = R.S. ✓ (Restrictions: sin θ ≠ 0 and cos θ ≠ 0.)

Worked example

Example 3: Difference of squares

Prove sin⁴θ − cos⁴θ = sin²θ − cos²θ.

  1. The L.S. is a difference of squares: (sin²θ)² − (cos²θ)².
  2. Factor: L.S. = (sin²θ − cos²θ)(sin²θ + cos²θ).
  3. The second bracket equals 1: L.S. = (sin²θ − cos²θ)(1) = sin²θ − cos²θ.
  4. R.S. = sin²θ − cos²θ.

L.S. = R.S. ✓

Worked example

Example 4: Two identities at once

Prove (1 − sin²θ)(1 + tan²θ) = 1.

  1. L.S.: replace 1 − sin²θ with cos²θ, and write tan²θ = sin²θ/cos²θ.
  2. L.S. = cos²θ (1 + sin²θ/cos²θ) = cos²θ + sin²θ (expand: cos²θ × sin²θ/cos²θ = sin²θ).
  3. L.S. = 1 = R.S.

L.S. = R.S. ✓

A family of Pythagorean identities

Divide every term of sin²θ + cos²θ = 1 by cos²θ and you get tan²θ + 1 = sec²θ. Divide by sin²θ instead and you get 1 + cot²θ = csc²θ. You can also see the first one in a triangle: shrink or stretch the unit-circle triangle by dividing every side by cos θ.

θsin θ1cos θθtan θsec θ1÷ cos θEach side ÷ cos θ:sin θ ÷ cos θ = tan θcos θ ÷ cos θ = 11 ÷ cos θ = sec θPythagoras:1 + tan²θ = sec²θ
Dividing every side of the unit-circle triangle by cos θ gives a similar triangle with sides 1, tan θ and sec θ. The Pythagorean theorem in it gives 1 + tan²θ = sec²θ.
Worked example

Example 5: Multiplying by a conjugate

Prove cos θ/(1 − sin θ) = (1 + sin θ)/cos θ.

  1. L.S. has 1 − sin θ in the denominator. Multiply top and bottom by its conjugate 1 + sin θ (this is multiplying by 1, so the value does not change).
  2. L.S. = cos θ (1 + sin θ) / ((1 − sin θ)(1 + sin θ)) = cos θ (1 + sin θ)/(1 − sin²θ).
  3. Pythagorean identity: L.S. = cos θ (1 + sin θ)/cos²θ.
  4. Cancel one factor of cos θ: L.S. = (1 + sin θ)/cos θ = R.S.

L.S. = R.S. ✓ (Restrictions: cos θ ≠ 0, sin θ ≠ 1.)

Before you start: is it even true?

Substitute an angle such as 30° into both sides with a calculator. If the sides differ, it is not an identity, so stop. If they agree, it might be an identity, and now you prove it. You can also graph both sides with graphing technology: an identity gives the same graph twice. You will learn to graph y = sin x and y = cos x in the next unit, Sinusoidal Functions.

Just for fun

Why did the identity bring two sheets of paper to the test?

One for the left side and one for the right side. They aren't allowed to mix until the very last line.

Hands-on activity

Identity matching cards

You need: 16 small pieces of paper (or index cards), a pen, and a calculator.

  1. Write each of these on its own card: sin θ/cos θ, 1 − sin²θ, 1/cos θ, tan θ cos θ, 1 + tan²θ, csc θ sin θ, cos θ/sin θ, 1 − cos²θ.
  2. On eight more cards write: tan θ, cos²θ, sec θ, sin θ, sec²θ, 1, cot θ, sin²θ.
  3. Shuffle all 16 cards and lay them face up. Match each card from the first set with the simpler expression it equals.
  4. Pick any pair and check it with θ = 37° on the calculator.
  5. Bonus: use two or more cards to build a new identity, for example tan θ cos θ × csc θ = 1, and prove it.

What to look for: the matches are sin θ/cos θ = tan θ, 1 − sin²θ = cos²θ, 1/cos θ = sec θ, tan θ cos θ = sin θ, 1 + tan²θ = sec²θ, csc θ sin θ = 1, cos θ/sin θ = cot θ, and 1 − cos²θ = sin²θ.

Your turn

1

Prove cos θ sec θ − sin²θ = cos²θ.

2

Prove sin²θ/(1 − cos θ) = 1 + cos θ.

3

Prove sec θ − cos θ = sin θ tan θ.

4

Prove (1 + cos θ)(1 − cos θ) = sin²θ.

5

Prove cot θ + tan θ = csc θ sec θ.

6

Prove cos⁴θ − sin⁴θ = 1 − 2sin²θ.

More practice: warm-up

1

Write sec θ tan θ using only sin θ and cos θ.

2

Factor 1 − sin²θ as a difference of squares.

3

Write 1/sin θ + 1/cos θ as a single fraction.

4

Simplify sin²θ + cos²θ + tan²θ.

5

Simplify sin θ csc θ cos θ.

More practice: on your own

1

Prove (1 − sin²θ)/(sin θ cos θ) = cot θ.

2

Prove (sec²θ − 1)cos²θ = sin²θ.

3

Prove (sin θ + cos θ)² = 1 + 2 sin θ cos θ.

4

Prove csc θ − sin θ = cos θ cot θ.

5

Is (sin θ + cos θ)² = 1 an identity? Test before proving.

Challenge

1

Prove (1 − cos θ)/sin θ = sin θ/(1 + cos θ).

2

Prove tan²θ/(1 + tan²θ) = sin²θ.

3

Prove 1/(1 − sin θ) + 1/(1 + sin θ) = 2 sec²θ.

Check your understanding

Quiz

1.

When proving an identity, you should:

2.

sin⁴θ − cos⁴θ factors as:

3.

tan θ + cot θ simplifies to:

4.

1 + tan²θ equals:

5.

Why multiply cos θ/(1 − sin θ) by (1 + sin θ)/(1 + sin θ)?

6.

(1 + cos θ)(1 − cos θ) equals:

For parents and teachers

Big idea (D1.5): longer identities are proved with the same basic identities plus algebra: common denominators, factoring and cancelling. A good default plan is "start with the messier side and write everything in sin and cos".

How to help: If your teen is stuck, ask: "Is there a fraction you can combine? Is there a 1 − sin² or 1 − cos² you can replace? Can you factor?" Encourage testing with one angle first, so they do not spend time trying to prove something false.

Common difficulties:

  • Algebra slips when adding fractions or expanding (sin θ + cos θ)².
  • Cancelling terms instead of factors, for example cancelling cos θ from (1 + cos θ)/cos θ. Only common factors cancel.
  • Writing steps on both sides at once, as if solving an equation.