Trigonometric identities

Identities are equations that hold for every (permitted) value of the angle. With them, knowing one function you can find the others, simplify expressions, prove equalities and reduce large angles to acute ones. These skills are needed constantly in trigonometric equations and later in differentiation and integration. The lesson covers the fundamental identity and its consequences, choosing the sign by quadrant, techniques for simplifying and proving, and the reduction formulas. Tangent and cotangent are written tg and ctg here, as in Uzbek and Russian textbooks (tan and cot elsewhere).

§1. The fundamental trigonometric identity

The point (cos α; sin α) on the unit circle is at distance 1 from the centre. By Pythagoras’ theorem (or the distance formula):

This holds for every α — it is the "basic law" of trigonometry. It leads to:

To get the second, divide the fundamental identity by cos²α: sin²α/cos²α + 1 = 1/cos²α. The third comes from dividing by sin²α. These formulas let you find cos and sin when tg or ctg is known.
Identities hold only where the expressions are defined: tg α · ctg α = 1 makes no sense at α = πk/2.

§2. Finding the other functions from one

From the fundamental identity:

Important: the sign in front of the root is chosen according to the quadrant of the angle. The identity gives only the square; the quadrant decides the sign. This is where most mistakes happen.
Procedure:

1. Identify the quadrant and write down the sign of each function.
2. Find the square using the appropriate identity.
3. Take the root and attach the sign decided in step 1.
4. Find the rest via tg = sin/cos and ctg = 1/tg.

Worked example

sin α = 3/5 with α in quadrant II. Find cos α and tg α.

1. In quadrant II, cos α < 0.
2. cos α = −√(1 − 9/25) = −√(16/25) = −4/5.
3. tg α = (3/5) : (−4/5) = −3/4.

Worked example

tg α = 2 with α in quadrant III. Find sin α and cos α.

1. 1/cos²α = 1 + 4 = 5 → cos²α = 1/5.
2. In quadrant III, cos α < 0: cos α = −1/√5 = −√5/5.
3. sin α = tg α · cos α = −2/√5 = −2√5/5.

§3. Simplifying expressions

Main techniques:

• Replace 1 with sin²α + cos²α (or the other way round).
• Write tg and ctg as sin/cos and cos/sin.
• Use a common denominator and factorise.
• Apply the special-product formulas.

Examples:

• (1 − cos²α)/(1 − sin²α) = sin²α/cos²α = tg²α.
• sin⁴α + cos⁴α = (sin²α + cos²α)² − 2sin²α cos²α = 1 − 2sin²α cos²α.
• (sin α + cos α)² = 1 + 2 sin α cos α.

Homogeneous fractions with known `tg`. If tg α = 3, then for (sin α + cos α)/(sin α − cos α) divide top and bottom by cos α: (tg α + 1)/(tg α − 1) = 4/2 = 2. There is no need to find the angle itself.

Worked example

Find sin α · cos α if sin α + cos α = 1/2.

1. Square both sides: sin²α + 2 sin α cos α + cos²α = 1/4.
2. 1 + 2 sin α cos α = 1/4.
3. sin α cos α = (1/4 − 1)/2 = −3/8.

§4. Proving identities

To prove an identity is to show both sides are equal for all permitted α. Methods:

• Simplify the more complicated side until it becomes the other.
• Reduce both sides to the same expression.
• Show that left side minus right side equals 0.

"Working" on both sides at once (for example, squaring both sides) is not a proof — it can introduce extraneous cases.
For example, (1 − sin α)(1 + sin α) = cos²α: the left side is 1 − sin²α = cos²α — proved. Or tg α + ctg α = 1/(sin α cos α): the left side is (sin²α + cos²α)/(sin α cos α) = 1/(sin α cos α).

§5. Reduction formulas

These reduce functions of π/2 ± α, π ± α, 3π/2 ± α, 2π ± α to functions of α. You need not memorise them — a simple rule is enough:

Why? A turn of π/2 + α on the unit circle swaps the coordinates (x and y), so sin and cos swap. A turn of π + α moves the point to the diametrically opposite position — both coordinates simply change sign.
Examples: sin(π − α) = sin α, cos(π − α) = −cos α, sin(π/2 + α) = cos α, cos(3π/2 + α) = sin α, tg(π + α) = tg α.

Worked example

Simplify cos(3π/2 + α).

1. 3π/2 is on the vertical axis, so the name changes: cos → sin.
2. If α is acute, 3π/2 + α is in quadrant IV, where the original function cos > 0, so the sign is +.
3. Answer: sin α.

§6. Large angles and mixed calculations

First remove whole turns (multiples of 360°), then apply a reduction formula: sin 1110° = sin(1110° − 3·360°) = sin 30° = 1/2; cos 300° = cos(360° − 60°) = cos 60° = 1/2.

Worked example

Evaluate sin 150° · cos 240° + tg 225°.

1. sin 150° = sin(180° − 30°) = sin 30° = 1/2.
2. cos 240° = cos(180° + 60°) = −cos 60° = −1/2; tg 225° = tg(180° + 45°) = 1.
3. 1/2 · (−1/2) + 1 = 3/4.

Worked example

Evaluate 2(sin⁴(π/8) + cos⁴(3π/8) + sin⁴(5π/8) + cos⁴(7π/8)).

1. cos(3π/8) = sin(π/8), sin(5π/8) = cos(π/8), cos⁴(7π/8) = cos⁴(π/8).
2. The sum: 2sin⁴(π/8) + 2cos⁴(π/8) = 2(1 − ½ sin²(π/4)) = 2 · 3/4 = 3/2.
3. 2 · 3/2 = 3.

Check yourself

1. cos α = −5/13 with α in quadrant II. Find tg α.

2. Simplify sin²α · ctg²α + sin²α.

3. What is sin(π/2 + α) equal to?

4. tg α = 1/2. Find (2 sin α − cos α)/(sin α + 3 cos α).

5. Evaluate cos 1020°.

6. sin α + cos α = 1/3. Find sin α cos α.

7. Simplify (1 − sin α)(1 + sin α)/cos α.

8. Find (sin α + cos α)/(sin α − cos α) if tg α = 3.

Common mistakes

• Ignoring the sign of the quadrant when taking a square root.
• Forgetting the change of name in a reduction formula, or choosing the sign from the new function.
• Using sin²α + cos²α = 1 as if it were sin α + cos α = 1.
• Removing whole turns incorrectly from a large angle (subtracting 180° instead of 360°).
• Confusing sin²α with sin α² (the first is the square of the sine).
• Transforming both sides at once when proving an identity and calling it a proof.