Trig identities: the eight you derive, the rest you look up, the double-angle formulas, and how to solve a trig equation without losing a solution

There are three kinds of trig identities. The eight basic ones you derive in a line each from the unit circle: sin²θ + cos²θ = 1 and its two divided forms, tan = sin/cos and cot = cos/sin, and the three reciprocals. The compound-angle formulas you look up or memorise once, and the double-angle formulas fall out of them by setting both angles equal. Everything else is algebra. This page derives the eight, states the rest, proves one identity line by line, and solves one equation without losing a solution.

The trig identities, grouped by where they come from

A list of thirty identities is hard to hold. Three groups with an origin each are easy.

Group 1: the Pythagorean identities

On the unit circle, the point at angle θ is (cos θ, sin θ), and every point on the circle satisfies x² + y² = 1. Substitute and you have the first identity:

• sin²θ + cos²θ = 1

Divide every term by cos²θ:

• tan²θ + 1 = sec²θ

Divide every term by sin²θ instead:

• 1 + cot²θ = csc²θ

Three identities, one picture, two divisions. The unit circle these come from is worth rebuilding first if the picture is shaky.

Group 2: the definitions

These are not facts to learn but names to know.

• tan θ = sin θ/cos θ and cot θ = cos θ/sin θ
• sec θ = 1/cos θ, csc θ = 1/sin θ and cot θ = 1/tan θ

That makes eight identities in groups 1 and 2, every one derived in a line.

Group 3: the compound-angle formulas

These do not drop out of the circle in one line, so you memorise them once, with their sign pattern:

• sin(A ± B) = sin A cos B ± cos A sin B
• cos(A ± B) = cos A cos B ∓ sin A sin B
• tan(A ± B) = (tan A ± tan B)/(1 ∓ tan A tan B)

The pattern to remember: sine keeps the sign, cosine flips it. The half-angle and product-to-sum formulas also exist, and they are look-up formulas. A formula sheet should carry them; your head does not need to.

In MHF4U these are ideas 19, "Fundamental trigonometric identities", 20, "Compound-angle and double-angle formulas", and 21, "Solving trigonometric equations". They come after idea 18, "Reciprocal trigonometric functions", and the course then moves to its last strand with idea 22, "Average and instantaneous rates of change". The MHF4U course map shows where they sit.

Double-angle formulas, derived not memorised

Set B = A in the group 3 formulas and the double angles appear.

• sin 2A = sin A cos A + cos A sin A = 2 sin A cos A
• cos 2A = cos A cos A − sin A sin A = cos²A − sin²A
• tan 2A = 2 tan A/(1 − tan²A)

Cosine has two more forms, from the Pythagorean identity:

• Replace sin²A with 1 − cos²A: cos 2A = 2cos²A − 1.
• Replace cos²A with 1 − sin²A: cos 2A = 1 − 2sin²A.

Which form do you use? The one that leaves a single trig function in the equation you are working on. If the rest of the equation is in sin x, use 1 − 2sin²x.

Worked example 1: proving an identity, line by line

Prove that tan θ + cot θ = sec θ csc θ.

1. Work on the left side only. It is the more complicated side, and the right side stays untouched until the end.
2. Write everything in sin and cos. sin θ/cos θ + cos θ/sin θ.
3. Use a common denominator, sin θ cos θ. (sin²θ + cos²θ)/(sin θ cos θ).
4. Use the Pythagorean identity on the top. The numerator is 1, so the expression is 1/(sin θ cos θ).
5. Split into reciprocals. (1/sin θ)(1/cos θ) = csc θ sec θ, which is the right side.

The left side has become the right side, so the identity is proved.

The common wrong line: "subtract cot θ from both sides" in line 2, as if the identity were an equation to solve. An identity is proved by changing one side into the other. Moving terms across assumes the two sides are already equal, which is the thing you were asked to show.

A second wrong line: writing sin²θ − cos²θ on the top in line 3, from a sign slip. That expression is −cos 2θ and it goes nowhere. Booth and colleagues (2013) found that studying incorrect examples with the fault explained may be especially useful for understanding, which is why each fault is named here.

Worked example 2: how to solve trig equations on [0, 2π)

Solve 2sin²x − sin x − 1 = 0 for 0 ≤ x < 2π.

1. Spot the quadratic. Let s = sin x. The equation is 2s² − s − 1 = 0.
2. Factor. (2s + 1)(s − 1) = 0. Check: 2s² − 2s + s − 1 = 2s² − s − 1.
3. Solve for s. s = −1/2 or s = 1.
4. Solve sin x = 1. On one turn, only x = π/2.
5. Solve sin x = −1/2. The reference angle is π/6. Sine is negative in quadrants III and IV, so x = π + π/6 = 7π/6 and x = 2π − π/6 = 11π/6.
6. List every solution. x = π/2, 7π/6, 11π/6.

The common wrong line, first version: dividing both sides by a trig function to simplify. Take sin x cos x = sin x. Dividing by sin x leaves cos x = 1, so x = 0, and the solution x = π is gone, because sin π = 0 also makes both sides zero. Factor instead: sin x (cos x − 1) = 0 gives sin x = 0 or cos x = 1, so x = 0 or π. The rule is: factor out, never divide out.

The common wrong line, second version: finding one angle for sin x = −1/2 and stopping. Inverse sine gives −π/6, which is not even in [0, 2π). A sine value between −1 and 1 is taken at two angles per turn, so look for both.

Sibling to try cold

Two of them. Cover the examples above and write every line yourself.

1. Solve 2cos²x + cos x − 1 = 0 for 0 ≤ x < 2π.
2. Prove (1 − cos²θ)/sin θ = sin θ.

Check

1. The equation factors as (2cos x − 1)(cos x + 1) = 0. So cos x = 1/2, at x = π/3 and 5π/3, or cos x = −1, at x = π. Three solutions: π/3, π, 5π/3.
2. The numerator 1 − cos²θ is sin²θ by the Pythagorean identity, and sin²θ/sin θ = sin θ. One line.

If you found only two solutions to the first, you missed one of the two angles for cos x = 1/2, or the single angle for cos x = −1. Write that in your error log. Studying a worked example first and then trying a problem cold is the order worked examples, then practice problems sets out.

The three moves that solve almost every identity

Write everything in sin and cos. Tan, cot, sec and csc are disguises. Once they are gone, you can see what cancels. Use this first, almost every time.

Find a common denominator. A sum of fractions rarely simplifies until it is one fraction. In the proof above, the common denominator is what made sin²θ + cos²θ appear.

Swap 1 for sin²θ + cos²θ, or the reverse. When you see sin² + cos², replace it with 1. When you are stuck with a 1 and need squares, replace the 1 with sin² + cos². The double-angle forms of cos 2θ are the same move.

These three moves are not a guarantee. Some identities need a factoring step or a compound-angle formula. But they finish most of the ones a Grade 12 test asks, and when you are stuck, trying each in turn is a plan. Test yourself on them with the page closed; testing yourself before you look explains why.

Where it goes next

In calculus, the derivative rules for sin and cos lean on sin²x + cos²x = 1 constantly, and integrating sin²x starts by rewriting it with cos 2x = 1 − 2sin²x, before u-substitution and integration by parts take over the harder integrals. The precalculus ideas calculus assumes list trig identities for that reason.

When a sign error or a lost solution keeps costing you, chapter 4 of the free playbook is about finding the line where the mark went. The rest of the MHF4U explainers are on the math help hub.

Trig identities on one screen

• Pythagorean: sin²θ + cos²θ = 1; tan²θ + 1 = sec²θ; 1 + cot²θ = csc²θ.
• Definitions: tan = sin/cos, cot = cos/sin, sec = 1/cos, csc = 1/sin, cot = 1/tan.
• Compound angles: sin(A ± B) = sin A cos B ± cos A sin B; cos(A ± B) = cos A cos B ∓ sin A sin B. Sine keeps the sign, cosine flips it.
• Double angles: sin 2A = 2 sin A cos A; cos 2A = cos²A − sin²A = 2cos²A − 1 = 1 − 2sin²A.
• Proving: one side only; sin and cos; common denominator; swap 1 for sin² + cos².
• Solving: factor out, never divide out; two angles per value unless it is 1 or −1.