01 · One identity
Simplify 7sin² θ + 7cos² θ − 3.
Hint
Factor out 7.
Worked solution
7(1) − 3 = 4
All real θ.
Understand · explore · practise
Trigonometric identities
Use sin²θ + cos²θ = 1 and tanθ = sinθ/cosθ to simplify, prove identities, find exact ratios and eliminate an angle while keeping domain restrictions.
Before you startUnit-circle signs, exact values, factorising and fractions
01 / Two core identities
Pythagoras connects the coordinates.
A point on a unit circle satisfies x² + y² = 1. Substitute x = cos θ and y = sin θ. The gradient of the radius supplies the second identity:
sin² θ + cos² θ = 1
tan θ = sin θ / cos θ, for cos θ ≠ 0
sin² θ means (sin θ)², whereas sin 2θ means the sine of twice the angle. An identity holds for every value in its stated domain. An equation such as sin θ = 1/2 is true only for particular angles.
Squares add to one; signs still matterChoose and compare
θ = 120°. cos² θ ≈ 0.25 and sin² θ ≈ 0.75; their sum is 1. cos θ ≈ -0.5, but the principal square root √(cos² θ) ≈ 0.5 is non-negative.
Watch the two squared coordinates add to one
Pause, replay or seek freely. The notes explain the same idea and stay in view.
02 / Rearrange carefully
A square does not preserve a sign.
sin² θ = 1 − cos² θ
cos² θ = 1 − sin² θ
√(1 − sin² θ) = |cos θ|
The square root symbol means the non-negative root. At θ = 120°, cos θ = −1/2 but √(1 − sin² θ) = 1/2. Therefore sin θ / √(1 − sin² θ) equals tan θ only where cos θ is positive; it equals −tan θ where cos θ is negative, and is undefined where cos θ = 0.
You may replace θ by any complete angle expression. For example, sin²(3x − 20°) + cos²(3x − 20°) = 1. Keep the entire argument the same in both terms.
03 / Simplify expressions
Turn unlike ratios into sine and cosine.
Two routes to a simpler expressionWorked example
1 − cos² θ = sin² θ
A direct rearrangement of the identity.
(1 − cos² θ)/sin θ = sin θ, for sin θ ≠ 0
Cancel a factor only where the original denominator is non-zero.
(1 + tan² θ)cos² θ = cos² θ + sin² θ = 1
The original tangent requires cos θ ≠ 0, even though the final constant is defined everywhere.
Fourth powers
cos⁴ θ − sin⁴ θ
= (cos² θ − sin² θ)(cos² θ + sin² θ)
= cos² θ − sin² θ = 1 − 2sin² θ
Factor the difference of squares before substituting. This identity holds for every real θ.
04 / Prove an identity
Rewrite one side until it becomes the other.
To prove an identity, give a chain of equivalent expressions on the common domain. Do not start by assuming the desired equality is already true.
(1 − cos θ)(1 + cos θ)/sin θ
= (1 − cos² θ)/sin θ
= sin² θ/sin θ
= sin θ, where sin θ ≠ 0
This proves the identity wherever its original left side exists. It does not define that left side at 0°, 180° or their whole-turn equivalents.
A proof using two angles
sin² A cos² B − cos² A sin² B
= sin² A(1 − sin² B) − (1 − sin² A)sin² B
= sin² A − sin² B
The cross terms cancel. Both sides exist for all real A and B. For a quotient such as tan A/tan B, additionally require cos A ≠ 0, cos B ≠ 0 and sin B ≠ 0.
05 / Recover exact ratios
Use the quadrant to choose the sign.
Suppose sin θ = −5/13 and 180° < θ < 270°. The identity gives cos² θ = 144/169. Cosine is negative in this quadrant.
cos θ = −12/13
tan θ = (−5/13)/(−12/13) = 5/12
If instead tan φ = −3/4 and 270° < φ < 360°, use a 3–4–5 reference triangle: sin φ = −3/5 and cos φ = 4/5. The reference triangle gives magnitudes; the quadrant supplies signs.
When the angle belongs to a triangle
A triangle has adjacent sides 5 and 6 and opposite side 7. The cosine rule gives cos A = (25 + 36 − 49)/60 = 1/5. Since an interior triangle angle has positive sine, sin A = 2√6/5. Its area is ½ × 5 × 6 × 2√6/5 = 6√6. No inverse angle or rounding is needed.
06 / Eliminate an angle
An equation alone may describe too many points.
If x = 3 cos θ and y = 2 sin θ, then:
x²/9 + y²/4 = 1
With all real θ, this traces the whole ellipse, so −3 ≤ x ≤ 3 and −2 ≤ y ≤ 2. With 0° ≤ θ ≤ 90°, keep only its first-quadrant arc: x ≥ 0 and y ≥ 0. The restrictions belong with the final equation.
A parabola segment
If x = sin θ and y = cos² θ for unrestricted real θ, then y = 1 − x² with −1 ≤ x ≤ 1. The equation y = 1 − x² on its own would wrongly include x = 2.
A rational relation
If x = sin θ and y = tan² θ, tangent excludes cos θ = 0. Hence:
y = x²/(1 − x²), with −1 < x < 1
Every allowed x is attained by an angle with non-zero cosine; y is then non-negative.
07 / Combine two coordinates
Add squares to cancel the cross terms.
Let x = sin θ + cos θ and y = cos θ − sin θ. Squaring and adding gives:
x² + y² = 2(sin² θ + cos² θ) = 2
With unrestricted θ this traces the whole circle of radius √2. The inverse relations sin θ = (x − y)/2 and cos θ = (x + y)/2 show that every point on x² + y² = 2 corresponds to valid unit-circle coordinates.
If θ is restricted, transform its range as well. For 0° ≤ θ ≤ 90°, sin θ and cos θ are non-negative, so x ≥ |y|. This selects the right-hand arc between (1,1) and (1,−1), passing through (√2,0).
08 / Your turn
Preserve the domain as you simplify.
State restrictions caused by every original denominator or tangent.
02 · Cancel with care
Simplify (1 − sin² θ)/cos θ.
Hint
The numerator is cos² θ.
Worked solution
cos θ, where cos θ ≠ 0
The original expression is undefined when cos θ = 0.
03 · The square root
Evaluate √(1 − sin² 210°).
Hint
This is |cos 210°|.
Worked solution
√3/2
The negative value of cos 210° is not the principal square root.
04 · Exact signs
cos θ = −8/17 and 90° < θ < 180°. Find sin θ and tan θ.
Hint
Sine is positive in quadrant II.
Worked solution
sin θ = 15/17
tan θ = −15/8
05 · Prove it
Prove sin⁴ θ − cos⁴ θ = 2sin² θ − 1.
Hint
Factor the difference of squares.
Worked solution
(sin² θ − cos² θ)(sin² θ + cos² θ)
= sin² θ − (1 − sin² θ)
= 2sin² θ − 1
06 · Lost domain
A student simplifies cos² θ(1 + tan² θ) to 1 and claims the original expression equals 1 at θ = 90°. Explain the error.
Hint
Evaluate the original tangent first.
Worked solution
tan 90° is undefined. Multiplying an undefined expression by zero does not define it. The identity is valid only for cos θ ≠ 0.
07 · Eliminate θ
x = 4 sin θ and y = 3 cos θ, with 0° ≤ θ ≤ 90°. Eliminate θ and state the required arc.
Hint
Square x/4 and y/3.
Worked solution
x²/16 + y²/9 = 1
x ≥ 0, y ≥ 0
The first-quadrant arc, including endpoints (0,3) and (4,0).
08 · A restricted curve
x = cos θ and y = sin² θ for real θ. Find the relation and the exact x-range.
Hint
Cosine lies in [−1,1].
Worked solution
y = 1 − x², −1 ≤ x ≤ 1
09 · Two angles
Simplify cos² A sin² B − sin² A cos² B.
Hint
Replace both cosine squares.
Worked solution
(1 − sin² A)sin² B − sin² A(1 − sin² B)
= sin² B − sin² A
09 / Recap
The algebra and its domain travel together.
- Use sin² θ + cos² θ = 1 for matching complete arguments.
- Convert tangent to sin/cos when that makes the structure clearer.
- √(cos² θ) = |cos θ|.
- A quadrant chooses the sign of a recovered square root.
- Cancellation does not restore excluded values.
- Eliminating θ requires the resulting range or arc as well as the equation.
Section 1 of 9 · Two core identities