12.3 Trigonometric Identities and Equations

12.3 Trigonometric Identities and Equations

Quick Answer: Identities rewrite an expression without changing its value. On MAT you use the square and quotient identities, reduction formulae, and compound and double-angle formulae to simplify, then you solve. Special angles 0°, 30°, 45°, 60°, and 90° supply exact numbers. Items are unscaffolded: simplify and solve in one stem, with no hence.

What the MAT booklet asks

The booklet lists solving trigonometric equations and using identities; simplification of trigonometric expressions using identities and reduction formulae where necessary; special angles; compound and double angles. Independent OpenExamPrep practice for this skill is one multiple-choice item that does the whole chain. NSC often splits simplify and hence solve; MAT does not.

You still have no calculator. If an equation reduces to sin θ = 1/2, the solutions come from the special-angle table and the general-solution pattern, not from a device. Official venue guidance also lists calculators among prohibited items, so exact values are not optional extras.

Core identities

Venue sessions do not let you bring a personal formula booklet. Learn these.

Square identity. sin²θ + cos²θ = 1, so sin²θ = 1 − cos²θ and cos²θ = 1 − sin²θ.

Quotient identity. tan θ = sin θ / cos θ, provided cos θ ≠ 0.

Reduction formulae (degrees). These move any angle to an acute angle whose sine or cosine you know, while tracking the CAST sign.

• sin(90° − θ) = cos θ and cos(90° − θ) = sin θ
• sin(90° + θ) = cos θ and cos(90° + θ) = −sin θ
• sin(180° − θ) = sin θ and cos(180° − θ) = −cos θ
• sin(180° + θ) = −sin θ and cos(180° + θ) = −cos θ
• sin(360° − θ) = −sin θ and cos(360° − θ) = cos θ
• sin(−θ) = −sin θ and cos(−θ) = cos θ (sine odd, cosine even)

Compound angles.

• sin(A + B) = sin A cos B + cos A sin B
• sin(A − B) = sin A cos B − cos A sin B
• cos(A + B) = cos A cos B − sin A sin B
• cos(A − B) = cos A cos B + sin A sin B

Double angles (set B = A).

• sin 2A = 2 sin A cos A
• cos 2A = cos²A − sin²A = 2 cos²A − 1 = 1 − 2 sin²A

Choose the form of cos 2A that matches what you are given. If you know sin A, the form 1 − 2 sin²A is usually fastest.

Worked example — reduction and a special angle

Evaluate sin 210° and cos 150° by hand.

• 210° = 180° + 30°, so sin 210° = −sin 30° = −1/2
• 150° = 180° − 30°, so cos 150° = −cos 30° = −√3/2

A four-option set will include +1/2 and −√3/2 as traps that forgot the CAST sign.

Worked example — compound angle with special angles

Evaluate sin 75° as sin(45° + 30°):

sin 75° = sin 45° cos 30° + cos 45° sin 30°

= (√2/2)(√3/2) + (√2/2)(1/2)

= √6/4 + √2/4

= (√6 + √2)/4

The companion values are cos 75° = (√6 − √2)/4 and sin 15° = (√6 − √2)/4. Mixing the plus and minus between sine and cosine of 75° is the usual error.

Worked example — double angle from a ratio

Given cos θ = 3/5 and θ acute, find cos 2θ and sin 2θ.

First sin θ = 4/5, because √(1 − 9/25) = 4/5 and sine is positive in the first quadrant.

• cos 2θ = 2 cos²θ − 1 = 2(9/25) − 1 = 18/25 − 25/25 = −7/25
• sin 2θ = 2 sin θ cos θ = 2(4/5)(3/5) = 24/25

Cosine of the double angle is negative: when θ is the acute angle with cosine 3/5, 2θ is already greater than 90°.

General solutions, then list in an interval

After you reach a single-function equation:

• sin θ = sin α ⇒ θ = α + 360°k or θ = 180° − α + 360°k, k an integer
• cos θ = cos α ⇒ θ = ±α + 360°k
• tan θ = tan α ⇒ θ = α + 180°k

Then list every solution in the interval the stem names, usually 0° ≤ θ < 360° or 0° ≤ θ ≤ 360°.

Unscaffolded: simplify, then solve, in one item

Example A — double angle hiding in a product. Solve 2 sin θ cos θ = 1/2 for 0° ≤ θ < 360°.

The left side is sin 2θ, so sin 2θ = 1/2.

Reference angle 30°. Then 2θ = 30° or 150°, and also 30° + 360° = 390° and 150° + 360° = 510°.

Divide by 2: θ = 15°, 75°, 195°, 255°.

Stopping at 15° and 75° is an incomplete list; MAT options often include that half-list as a distractor.

Example B — factor, with no second part. Solve 2 sin θ cos θ − cos θ = 0 for 0° ≤ θ < 360°.

Factor: cos θ (2 sin θ − 1) = 0.

cos θ = 0 ⇒ θ = 90°, 270°.

sin θ = 1/2 ⇒ θ = 30°, 150°.

Solutions: 30°, 90°, 150°, 270°.

If you divide both sides by cos θ at the start you lose 90° and 270°. Cancelling a factor is a classic lost-root error.

Example C — reduction inside an equation. Solve sin(90° + θ) = 1/2 for 0° ≤ θ < 360°.

sin(90° + θ) = cos θ, so cos θ = 1/2.

θ = 60° or θ = 300°.

The stem never told you to reduce first; you had to recognise the reduction, then solve, in one question.

Example D — special angle after a double-angle rewrite. Solve 1 − 2 sin²θ = 0 for 0° ≤ θ < 360°.

That is cos 2θ = 0, so 2θ = 90° or 270°, and also 90° + 360° = 450° and 270° + 360° = 630°.

θ = 45°, 135°, 225°, 315°.

Equivalently, sin²θ = 1/2, so sin θ = ±√2/2, which yields the same four solutions from the special-angle table.

Restrictions

Whenever you use tangent, or divide by sine or cosine, exclude values that make a denominator zero. After solving, check that each candidate is in the domain. An equation that produced tan θ = 1 still cannot include 90°, where tangent is undefined.

MAT tactic

1. Rewrite using an identity until you see a single function of θ or of 2θ.
2. Insert special-angle values only after the identity has done the heavy lifting.
3. Write the general solution, then list in the given interval — incomplete lists are common wrong options.
4. Factor rather than divide, so you keep extra roots.
5. Watch CAST signs when reducing 150°, 210°, 330°, and related angles.