01 · Derive sine
Use t = tan(θ/2) to derive sin θ = 2t/(1 + t²).
Hint
Express sin(θ/2) as t cos(θ/2).
Worked solution
sin θ = 2t cos²(θ/2). Since cos²(θ/2) = 1/(1 + t²), the result follows wherever the substitution exists.
Understand · explore · practise
Tangent half angle substitution
Use t = tan(θ/2) to turn trigonometric equations into rational algebra. Derive the formulae, recover angle branches and check missing odd multiples of π.
Before you startHalf angle formulae, quadratic equations and inverse tangent
01 / One variable for sine and cosine
A line slope can describe nearly the whole unit circle.
Putting t = tan(θ/2) lets you express sine and cosine using fractions in t. This is useful when an equation contains both functions and neither ordinary factorisation nor a single identity makes it simple.
sin θ = 2t/(1 + t²)
cos θ = (1 − t²)/(1 + t²)
These formulae describe every point on the unit circle except (−1, 0). That missing point matters when solving equations. All angles in this lesson are in radians.
A rational point on the circleExplore
t = 2; cos θ = −3/5; sin θ = 4/5.
θ = 2 arctan(2) ≈ 2.2143 radians.
The blue line has equation y = t(x + 1). The hollow point (−1, 0) is missing for every finite t. Change the angle representative: the circle point stays fixed.
02 / Derive the rational formulae
Start with the half-angle tangent, not a memorised substitution.
Let u = θ/2 and t = tan u. This requires cos u ≠ 0, so θ cannot be an odd multiple of π. From 1 + tan² u = sec² u:
cos² u = 1/(1 + t²)
sin u cos u = t/(1 + t²)
Now sin θ = 2sin u cos u, and cos θ = 2cos² u − 1. Substitution gives the two rational formulae. Since 1 + t² is positive for real t, clearing this denominator cannot remove a real t or reverse an inequality sign.
tan θ = 2t/(1 − t²)
Use this only when t ≠ ±1; tangent is undefined at the corresponding angles. The initial substitution still omits odd multiples of π.
03 / See why one point is missing
Intersect y = t(x + 1) with x² + y² = 1.
Find the second circle intersectionWorked example
x² + t²(x + 1)² = 1
Substitute the line equation.
(x + 1)((1 + t²)x + t² − 1) = 0
One intersection is always (−1, 0).
x = (1 − t²)/(1 + t²), y = 2t/(1 + t²)
The other intersection gives cosine and sine.
x + 1 = 2/(1 + t²) > 0
No finite t gives the missing point itself.
As the magnitude of t grows, the second intersection approaches (−1, 0). Approaching a point does not make it part of the finite-parameter formula.
Watch the line generate rational circle points
Pause, replay or seek freely. The notes explain the same idea and stay in view.
04 / Recover all angles in the requested interval
Inverse tangent gives one branch; the original angle needs full turns.
θ = 2 arctan(t) + 2kπ, where k is any integer
The principal value 2 arctan(t) lies strictly between −π and π. For a negative t, add 2π to put its representative in [0, 2π). For a different interval, retain every full-turn shift that lies inside it.
t = −1 on 0 ≤ θ ≤ 4πWorked example
2 arctan(−1) = −π/2
This principal angle is outside the requested interval.
θ = −π/2 + 2kπ
Add full turns before testing the bounds.
θ = 3π/2, 7π/2
These are both valid representatives.
After recovering these angles, separately test omitted odd multiples of π in the original equation and its original domain.
05 / Solve an equation containing both sine and cosine
Turn the equation into a quadratic and preserve both roots.
Solve 3sin θ + 4cos θ = 2 for 0 ≤ θ < 2πWorked example
6t + 4(1 − t²) = 2(1 + t²)
Multiply by the positive denominator 1 + t².
3t² − 3t − 1 = 0
Collect terms and divide by 2.
t = (3 ± √21)/6
Both real roots are admissible.
θ = 2 arctan((3 + √21)/6)
This root of t is positive.
θ = 2 arctan((3 − √21)/6) + 2π
The other root is negative; select its positive angle representative.
At θ = π, the original left side is −4
The omitted angle does not solve this equation.
Exact inverse-trigonometric answers are legitimate. If a decimal is required, round only after recovering the angle and checking the interval.
06 / Keep interval endpoints distinct
One t can represent more than one angle in a closed interval.
Solve 2sin θ + cos θ = 1 for 0 ≤ θ ≤ 2πWorked example
4t + 1 − t² = 1 + t²
Substitute and clear the common denominator.
2t(t − 2) = 0
Hence t = 0 or t = 2.
t = 0 gives θ = 0 and θ = 2π
Both endpoints are included.
t = 2 gives θ = 2 arctan 2
It lies between 0 and π.
θ = π gives −1 ≠ 1
The missing circle point is not a solution.
07 / Restore a genuinely lost angle
An omitted parameter value may still solve the original equation.
Solve sin θ = 1 + cos θ for 0 ≤ θ < 2πWorked example
2t = (1 + t²) + (1 − t²) = 2
The substitution reduces the equation to t = 1.
θ = π/2
Recover the finite-parameter solution.
Test θ = π directly: 0 = 1 − 1
This missing angle also solves the original equation.
The complete solution is θ = π/2, π
A substitution alone would have lost π.
08 / Original denominators take priority
Do not restore an angle that never belonged to the equation.
Solve sin θ/(1 + cos θ) = 1 for 0 ≤ θ < 2πWorked example
The original denominator excludes θ = π
Record this before any algebra.
The quotient equals t where it is defined
So t = 1.
θ = π/2 only
The excluded angle π cannot be restored, even though it solves the cross-multiplied equation.
Solve cos θ/(1 + sin θ) = 2 on the same intervalWorked example
The original equation excludes θ = 3π/2
Its denominator is zero there.
(1 − t²)/(1 + 2t + t²) = 2, with t ≠ −1
The transformed denominator is (1 + t)².
(1 − t)/(1 + t) = 2
Cancellation retains t ≠ −1.
t = −1/3, so θ = 2 arctan(−1/3) + 2π
The resulting sine is −3/5 and cosine 4/5, giving quotient 2.
At omitted θ = π the quotient is −1
No extra solution is restored.
09 / Generate exact rational circle coordinates
Integer slope data produces a Pythagorean identity.
For t = m/n, n ≠ 0:
cos θ = (n² − m²)/(n² + m²)
sin θ = 2mn/(n² + m²)
Choose t = 2/3Worked example
cos θ = (9 − 4)/(9 + 4) = 5/13
The horizontal coordinate is positive.
sin θ = 12/13
The vertical coordinate is positive.
5² + 12² = 13²
The unit-circle identity checks the coordinates.
θ = 2 arctan(2/3) in quadrant one
The signs and inverse branch agree.
This generates rational coordinates; it does not imply that the angle itself is a rational number of radians or a familiar special angle.
10 / Your turn
Keep a separate checklist for angle branches and original restrictions.
02 · Coordinates
Find sine and cosine when t = 3.
Hint
The common denominator is 10.
Worked solution
sin θ = 3/5 and cos θ = −4/5. The representative 2 arctan 3 lies in quadrant two.
03 · Negative slope
Find sine and cosine when t = −1/2, and give θ in [0, 2π).
Hint
The principal doubled inverse tangent is negative.
Worked solution
sin θ = −4/5, cos θ = 3/5; θ = 2 arctan(−1/2) + 2π.
04 · Missing point
Why can a finite real t never give cos θ = −1?
Hint
Examine cos θ + 1.
Worked solution
cos θ + 1 = 2/(1 + t²) > 0 for every finite real t. Hence cos θ cannot equal −1.
05 · Complete turns
If t = 0, list θ in −2π ≤ θ ≤ 2π.
Hint
Use θ = 2kπ.
Worked solution
θ = −2π, 0, 2π.
06 · Two branches
If t = −√3, find θ in [0, 4π).
Hint
The principal arctangent is −π/3.
Worked solution
θ = −2π/3 + 2kπ, giving 4π/3 and 10π/3.
07 · Linear result, extra angle
Solve sin θ = 2(1 + cos θ) for 0 ≤ θ < 2π.
Hint
The substitution gives t = 2. Test π separately.
Worked solution
2t = 4, so θ = 2 arctan 2. At π both sides are zero. Thus θ = 2 arctan 2, π.
08 · Change to a quotient
Solve sin θ/(1 + cos θ) = 2 on [0, 2π).
Hint
The denominator now excludes the omitted point.
Worked solution
Only θ = 2 arctan 2. The angle π is outside the original domain.
09 · Quadratic equation
Solve sin θ + cos θ = 1 on [0, 2π].
Hint
Substitution gives 2t(1 − t) = 0.
Worked solution
t = 0 or 1 gives θ = 0, π/2, 2π. At π the left side is −1, so there is no additional solution.
10 · Tangent denominator
What happens at t = 1 in the formula tan θ = 2t/(1 − t²)?
Hint
Recover the corresponding circle point.
Worked solution
The point is (0, 1), giving θ = π/2 modulo 2π. Tangent is undefined; a zero denominator is not an infinite real answer.
11 · Original hole
Solve cos θ/(1 + sin θ) = 3 for 0 ≤ θ < 2π.
Hint
Use (1 − t)/(1 + t) with t ≠ −1.
Worked solution
1 − t = 3 + 3t, so t = −1/2. Thus θ = 2 arctan(−1/2) + 2π. Sine −4/5 and cosine 3/5 give quotient 3. The excluded angle 3π/2 is invalid, and π gives −1.
12 · Rational point
Find the circle point generated by t = 3/4.
Hint
Use m = 3 and n = 4.
Worked solution
(cos θ, sin θ) = (7/25, 24/25). Check 7² + 24² = 25².
13 · Omitted but allowed
Solve tan θ = 0 on [0, 2π) using the substitution, accounting for all original restrictions.
Hint
The rational expression gives t = 0; the substitution misses π.
Worked solution
t = 0 gives θ = 0. The original tangent is defined and zero at π, so restore it. The complete answer is 0, π. Values t = ±1 are excluded by the tangent denominator.
14 · Safe denominator
Why is multiplying by 1 + t² safe for real t, while multiplying by 1 + t requires a restriction?
Hint
Ask whether each expression can be zero.
Worked solution
1 + t² ≥ 1 for real t. But 1 + t = 0 at t = −1; cancellation or clearing an original denominator with this factor must retain that exclusion.
11 / Recap
Solve algebra, recover branches, then check omitted angles.
- Derive sine and cosine as rational expressions in t = tan(θ/2).
- Keep original denominator exclusions.
- Recover θ = 2 arctan(t) + 2kπ across the full requested interval.
- Check odd multiples of π directly in the original equation.
- Distinguish an omitted valid solution from an excluded original input.
Section 1 of 11 · One variable for sine and cosine