Difficulty: Intermediate to Advanced | Prerequisites: Functions, logarithms and exponentials study notes
This is the largest and most varied section of the precalculus final. It covers trigonometric functions (evaluating, solving equations, proving identities), angle formulas, polar coordinate conversion, systems of equations, and conic sections. You need solid comfort with the unit circle and basic trig ratios before diving in. These topics form the bridge to calculus, where trig functions are differentiated and integrated constantly.
Trig problems on the final fall into a few predictable types: evaluate a trig function given constraints (use quadrant analysis), solve a trig equation in [0, 2π) (rewrite using identities, then solve), apply angle addition or power-reducing formulas (pattern-match to known identities), convert between rectangular and polar (substitute x = r cos θ, y = r sin θ), solve systems by substitution, and put ellipses in standard form. The algebra is mechanical once you know which identity to reach for.
Unit circle
The circle of radius 1 centred at the origin. The coordinates of any point on it are (cos θ, sin θ). This is the foundation for all trig values.
Reference angle
The acute angle that a given angle makes with the x-axis. You find trig values of any angle by computing the trig values of its reference angle and then adjusting the sign for the quadrant.
CAST rule (All Students Take Calculus)
A mnemonic for which trig functions are positive in each quadrant. All in QI, Sine in QII, Tangent in QIII, Cosine in QIV.
Inverse trigonometric functions (arcsin, arccos, arctan)
Functions that return the angle whose sine, cosine, or tangent is a given value. They have restricted ranges to ensure they are well-defined.
Double-angle identities
Formulas expressing trig functions of 2u in terms of trig functions of u. For example, cos 2u = 2cos²u − 1 = 1 − 2sin²u.
Power-reducing identities
Rearrangements of the double-angle formulas that express sin²u or cos²u in terms of cos 2u. sin²u = (1 − cos 2u)/2, cos²u = (1 + cos 2u)/2.
Angle addition/subtraction formulas
sin(u ± v) = sin u cos v ± cos u sin v. cos(u ± v) = cos u cos v ∓ sin u sin v. These let you find exact values for non-standard angles.
Polar coordinates
A system where each point is described by (r, θ): distance from the origin and angle from the positive x-axis. Related to rectangular coordinates by x = r cos θ, y = r sin θ.
Ellipse (standard form)
(x − h)²/a² + (y − k)²/b² = 1, centred at (h, k) with semi-axes a and b. Think of it as a stretched circle.
When you are given one trig ratio and a sign constraint, the process is: determine the quadrant, set up a reference triangle, and use the Pythagorean theorem to find the missing side.
Example: Find csc θ if tan θ = 77/2 and sin θ < 0.
tan θ > 0 places θ in QI or QIII. sin θ < 0 places θ in QIII or QIV. The overlap is QIII.
In QIII, both x and y are negative. Since tan θ = y/x = 77/2 = (−77)/(−2), set y = −77 and x = −2.
Find r: r = √(x² + y²) = √(4 + 5929) = √5933.
csc θ = r/y = √5933/(−77) = −√5933/77.
The CAST mnemonic is essential here. Always determine the quadrant first, then assign the correct signs to x and y.
To simplify an expression like sin(arccos(1/x)), construct a reference triangle from the inner inverse trig function.
Example: Find an algebraic expression for sin(arccos(1/x)).
Let θ = arccos(1/x). Then cos θ = 1/x = adj/hyp.
Construct the triangle: adjacent = 1, hypotenuse = x, opposite = √(x² − 1) (by the Pythagorean theorem).
Therefore sin(arccos(1/x)) = sin θ = opp/hyp = √(x² − 1)/x.
This technique works for any composition of a trig function with an inverse trig function. Draw the triangle, label the known sides, find the missing side, and read off the ratio you need.
The typical approach: use an identity to rewrite everything in terms of one trig function, then solve as a polynomial.
Example: Solve cos 2x + cos x = 0 in [0, 2π).
Rewrite cos 2x using the double-angle identity:
cos 2x = 2cos²x − 1
So the equation becomes: 2cos²x − 1 + cos x = 0, which is 2cos²x + cos x − 1 = 0.
Let y = cos x. Then 2y² + y − 1 = 0.
Using the quadratic formula: y = (−1 ± 3)/4, giving y = 1/2 or y = −1.
Solving cos x = 1/2: The reference angle is π/3 (from the special triangle with adj = 1, hyp = 2). Cosine is positive in QI and QIV, so x = π/3 and x = 2π − π/3 = 5π/3.
Solving cos x = −1: x = π.
Solutions: x = π/3, π, 5π/3.
Always state the interval you are working in and account for all quadrants where the trig function takes the required sign.
These formulas let you find exact trig values for angles that are sums or differences of standard angles (π/6, π/4, π/3).
sin(u − v) = sin u cos v − cos u sin v
sin(u + v) = sin u cos v + cos u sin v
cos(u − v) = cos u cos v + sin u sin v
cos(u + v) = cos u cos v − sin u sin v
Example: Find sin(π/12) exactly.
Express π/12 as a difference of standard angles: π/12 = π/4 − π/6.
sin(π/4 − π/6) = sin(π/4) cos(π/6) − cos(π/4) sin(π/6)
= (1/√2)(√3/2) − (1/√2)(1/2)
= √3/(2√2) − 1/(2√2)
= (√3 − 1)/(2√2)
The special triangles (45-45-90 and 30-60-90) provide the values you need. Memorise sin and cos for π/6, π/4, and π/3.
Deriving sin(u + v) from cos(u − v):
This is a common proof question. Use the co-function identity sin(θ) = cos(π/2 − θ):
sin(u + v) = cos(π/2 − (u + v)) = cos((π/2 − u) − v)
Apply the cos difference formula and simplify using cos(π/2 − u) = sin u and sin(π/2 − u) = cos u.
These identities are rearrangements of each other. The double-angle identity for cosine has three equivalent forms:
cos 2u = cos²u − sin²u
cos 2u = 2cos²u − 1
cos 2u = 1 − 2sin²u
Rearranging the last two gives the power-reducing identities:
sin²u = (1 − cos 2u)/2
cos²u = (1 + cos 2u)/2
Example: Prove that sin⁴x = (1/8)(3 − 4 cos 2x + cos 4x).
Start by writing sin⁴x = (sin²x)².
Apply the power-reducing identity: sin²x = (1 − cos 2x)/2.
So sin⁴x = [(1 − cos 2x)/2]² = (1/4)(1 − cos 2x)²
= (1/4)(1 − 2 cos 2x + cos²2x)
Now apply the power-reducing identity again to cos²2x:
cos²2x = (1 + cos 4x)/2
Substitute back: (1/4)(1 − 2 cos 2x + (1 + cos 4x)/2)
= (1/4)(2/2 − 2 cos 2x + (1 + cos 4x)/2)
= (1/8)(2 + 1 + cos 4x − 4 cos 2x)
= (1/8)(3 − 4 cos 2x + cos 4x). ∎
Proving cos²u = (1 + cos 2u)/2 from the angle addition formula:
Start with cos(u + v) = cos u cos v − sin u sin v. Set v = u:
cos 2u = cos²u − sin²u = cos²u − (1 − cos²u) = 2cos²u − 1
Rearranging: cos²u = (1 + cos 2u)/2.
Use the substitutions x = r cos θ, y = r sin θ, and the identity x² + y² = r².
Example: Convert (x + 3)² + (y + 3)² = 18 to a polar equation.
Expand:
(r cos θ + 3)² + (r sin θ + 3)² = 18
r²cos²θ + 6r cos θ + 9 + r²sin²θ + 6r sin θ + 9 = 18
Use cos²θ + sin²θ = 1:
r² + 6r cos θ + 6r sin θ + 18 = 18
r² + 6r cos θ + 6r sin θ = 0
Factor out r:
r(r + 6 cos θ + 6 sin θ) = 0
Since r ≠ 0 for a non-trivial solution:
r = −6 cos θ − 6 sin θ
The Pythagorean identity cos²θ + sin²θ = 1 is the tool that collapses the r² terms. This step appears in virtually every rectangular-to-polar conversion of a circle equation.
For nonlinear systems, substitution is the standard approach: solve one equation for one variable, then substitute into the other.
Example 1: Solve y² = −x + 9 and y = −x.
Substitute y = −x into the first equation: (−x)² = −x + 9, so x² + x − 9 = 0.
By the quadratic formula: x = (−1 ± √37)/2.
Since y = −x, the two solutions are:
(x, y) = ((−1 + √37)/2, (1 − √37)/2) and (x, y) = ((−1 − √37)/2, (1 + √37)/2).
Example 2: Solve y² = x and x² = −8y.
From the first equation, x = y². Substitute into the second: (y²)² = −8y, so y⁴ + 8y = 0.
Factor: y(y³ + 8) = 0, giving y = 0 or y = −2.
For y = 0: x = 0. For y = −2: x = (−2)² = 4.
Solutions: (0, 0) and (4, −2).
Always substitute back to find the corresponding values of the other variable. Check that each pair satisfies both original equations.
To sketch or analyse an ellipse, get the equation into standard form: (x − h)²/a² + (y − k)²/b² = 1.
Example: Sketch (x + 3)² + 16(y − 2)² = 4.
Divide through by 4:
(x + 3)²/4 + 4(y − 2)² = 1, which is (x + 3)²/2² + (y − 2)²/(1/2)² = 1.
Centre: (−3, 2). Semi-axis in x: a = 2. Semi-axis in y: b = 1/2.
To find the four key points:
Set x = −3: y = 2 ± 1/2, giving (−3, 1.5) and (−3, 2.5)
Set y = 2: x = −3 ± 2, giving (−1, 2) and (−5, 2)
These four points define the bounding box of the ellipse. Sketch the ellipse through them.
The larger denominator under the squared terms tells you which axis is the major axis. Here a > b, so the major axis is horizontal.
The foci lie on the major axis at distance c = √(a² − b²) from the centre. Here c = √(4 − 1/4) = √(15/4) = √15/2. The foci are at (−3 ± √15/2, 2). (Note: foci were flagged as not on this particular final exam, but they may appear on other exams.)
Students often mix up which double-angle form to use. Use the version that eliminates the trig function you do not want. If you want everything in terms of cosine, use cos 2u = 2cos²u − 1.
When solving trig equations, students sometimes find only the reference angle solution and forget the second solution in the other valid quadrant. Cosine is positive in two quadrants, sine is positive in two quadrants. Check both.
Students confuse polar and rectangular substitution direction. To go from rectangular to polar, replace x with r cos θ and y with r sin θ. To go the other way, use x² + y² = r², x = r cos θ, y = r sin θ.
In ellipse problems, students sometimes forget to divide through to get 1 on the right-hand side before reading off a and b. The equation must equal 1 in standard form.
⚠️ Expect at least one "evaluate a trig function given constraints" problem. The quadrant analysis method is always the same.
⚠️ Solving trig equations in [0, 2π) is a near-certain exam question. Know the unit circle values for sin and cos cold.
⚠️ Angle addition/subtraction problems typically ask for an exact value of a non-standard angle. Express it as a sum or difference of π/6, π/4, or π/3.
⚠️ Power-reducing identity proofs test whether you can chain multiple identities together. Start with the square, apply once, then again.
⚠️ Rectangular-to-polar conversion is mechanical but easy to lose marks on through algebra slips. Write every step.
⚠️ Ellipse standard form questions test whether you can divide correctly and identify a, b, and the centre.
True or false: sin(π/4) = √2/2. ___
Fill in the blank: sin²u + cos²u = ___.
True or false: the solutions to sin x = 1/2 in [0, 2π) are x = π/6 and x = π/3. ___
Fill in the blank: in polar coordinates, x = r ___ θ.
True or false: in the standard form of an ellipse, the right-hand side must equal 1. ___
Answers: 1. True. 2. 1 (the Pythagorean identity). 3. False (x = π/6 and x = 5π/6). 4. cos. 5. True.
Q: Find sec θ if sin θ = −3/5 and tan θ > 0.
A: sin θ < 0 places θ in QIII or QIV. tan θ > 0 places θ in QI or QIII. Overlap: QIII. In QIII, y = −3, r = 5, x = −√(25 − 9) = −4. sec θ = r/x = 5/(−4) = −5/4.
Q: Find the exact value of cos(π/12).
A: π/12 = π/3 − π/4. cos(π/3 − π/4) = cos(π/3) cos(π/4) + sin(π/3) sin(π/4) = (1/2)(1/√2) + (√3/2)(1/√2) = (1 + √3)/(2√2).
Q: Solve 2sin²x − sin x − 1 = 0 in [0, 2π).
A: Let y = sin x. 2y² − y − 1 = 0 factors as (2y + 1)(y − 1) = 0. So y = −1/2 or y = 1. sin x = 1 gives x = π/2. sin x = −1/2 gives x = 7π/6 and x = 11π/6. Solutions: π/2, 7π/6, 11π/6.
Q: Convert x² + y² − 4x = 0 to a polar equation.
A: Substitute x = r cos θ, y = r sin θ: r² − 4r cos θ = 0. Factor: r(r − 4 cos θ) = 0. Since r ≠ 0: r = 4 cos θ.
Q: Find the centre and semi-axes of the ellipse (x − 1)²/9 + (y + 2)²/4 = 1.
A: Centre (1, −2), a = 3 (horizontal), b = 2 (vertical).
Trigonometric identities are used constantly in integral calculus, particularly for trig substitution and integration of powers of sine and cosine. The power-reducing identities, for instance, are exactly what you use to integrate sin²x or cos⁴x.
Polar coordinates reappear in multivariable calculus for double integrals over circular regions, and in physics for describing orbital motion.
Conic sections connect to both analytic geometry and physics (planetary orbits are ellipses by Kepler's first law).
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