Difficulty: Intermediate | Prerequisites: Arithmetic and Algebraic Operations, Inequalities.
AP Calculus assumes you can evaluate trig functions at standard angles without a calculator, know the reciprocal and inverse trig functions, and can verify or disprove identities. The unit circle and the Pythagorean identity are the backbone of everything here.
Reciprocal trig functions
csc θ = 1/sin θ, sec θ = 1/cos θ, cot θ = cos θ/sin θ. Think of them as the "flipped" versions of sine, cosine, and tangent.
Inverse trig functions (sin⁻¹x, cos⁻¹x, tan⁻¹x)
Functions that return the angle whose sine, cosine, or tangent is x. sin⁻¹x is defined only for -1 ≤ x ≤ 1 and returns values in [-π/2, π/2]. cos⁻¹x is defined for -1 ≤ x ≤ 1 and returns values in [0, π]. tan⁻¹x is defined for all real x and returns values in (-π/2, π/2). In simple terms, they answer: "what angle gives me this ratio?"
Even and odd trig functions
Cosine is even: cos(-x) = cos(x). Sine and tangent are odd: sin(-x) = -sin(x), tan(-x) = -tan(x). Think of it as: the cosine graph is symmetric about the y-axis; sine and tangent are symmetric about the origin.
Quadrant analysis
Determining the sign of each trig function based on the quadrant of the angle. In quadrant I all are positive; in II sine is positive; in III tangent is positive; in IV cosine is positive. The mnemonic "All Students Take Calculus" (A-S-T-C) is widely used.
If sin θ = 1/5, then csc θ = 5 (the reciprocal).
If sec θ = 5/4 and tan θ > 0, θ is in quadrant I. From sec θ = 5/4, cos θ = 4/5, so sin θ = 3/5 (using sin²θ + cos²θ = 1, and sin θ positive in QI).
If cot θ = 3/4 and cos θ < 0, θ is in quadrant III (both sin and cos negative). Build a 3-4-5 right triangle: sin θ = -4/5.
If tan θ = √3 and sin θ > 0, θ is in quadrant I. The reference angle is π/3, so sec θ = 1/cos(π/3) = 1/(1/2) = 2.
If cos θ = -1/2 with θ in [0, π], then θ = 2π/3 (so α = 2/3 when θ = απ).
tan(sin⁻¹(1/√2)): sin⁻¹(1/√2) = π/4, so tan(π/4) = 1.
sin x is defined for all real x. True.
sin²x = cos²x - 1 for all x. False. The correct identity is sin²x = 1 - cos²x.
tan x · cos x = sin x on (-π/2, π/2). True.
tan x · cot x = 1 on (0, π/2). True.
cos x · sec x = 1 on (-π/2, π/2). True.
tan²x - sec²x = 1 on (-π/2, π/2). False. The correct identity is tan²x + 1 = sec²x, so tan²x - sec²x = -1.
csc x · sin x = 1 on (0, π). True.
0 ≤ sin x ≤ 1 for all x. False. sin x ranges from -1 to 1.
sin⁻¹x is defined for all real x. False. Only for -1 ≤ x ≤ 1.
1 + tan x > 1 on (-π/2, π/2). False. tan x can be negative.
sin(sin⁻¹x) = x for all x. False. Only when -1 ≤ x ≤ 1 (the domain of sin⁻¹).
tan⁻¹x is defined for all real x. True.
1 + tan²x = sec²x on (-π/2, π/2). True.
sin(x + π/2) = cos x for all x. True.
sin²A + cos²A = 1
1 + tan²A = sec²A
1 + cot²A = csc²A
sin(A + B) = sin A cos B + cos A sin B
cos(A + B) = cos A cos B - sin A sin B
sin(2A) = 2 sin A cos A
cos(2A) = cos²A - sin²A = 2cos²A - 1 = 1 - 2sin²A
cos(-x) = cos(x) (cosine is even)
sin(-x) = -sin(x) (sine is odd)
sin(x + π/2) = cos(x)
sin(6θ) = 2 sin(3θ) cos(3θ) (double-angle with A = 3θ)
Students often think sin⁻¹x means 1/sin x. It does not. sin⁻¹x is the inverse sine function (arcsin), which returns an angle. The reciprocal 1/sin x is csc x.
Students often think sin²x = cos²x - 1. The correct rearrangement of the Pythagorean identity is sin²x = 1 - cos²x.
Students often think tan²x - sec²x = 1. The sign is wrong: tan²x - sec²x = -1.
Students often think 0 ≤ sin x ≤ 1 for all x. Sine ranges from -1 to 1. The statement would need |sin x| ≤ 1.
⚠️ Trig substitution in integrals requires you to move fluently between trig identities.
⚠️ Derivatives of trig functions (d/dx sin x = cos x, d/dx tan x = sec²x, etc.) rely on these identities in their proofs.
⚠️ Inverse trig functions appear in derivative and integral formulas: d/dx(sin⁻¹x) = 1/√(1 - x²), d/dx(tan⁻¹x) = 1/(1 + x²).
⚠️ The domain restrictions of inverse trig functions are tested directly.
If sin θ = 1/5, what is csc θ? (Answer: 5.)
True or False: cos⁻¹x is defined for all real x. (Answer: False. Only for -1 ≤ x ≤ 1.)
What is cos(-x) if cos(x) = -2/3? (Answer: -2/3. Cosine is even.)
True or False: sin(6θ) = 2 sin(3θ) cos(3θ). (Answer: True. Double-angle formula with A = 3θ.)
True or False: tan x · cot x = 1. (Answer: True, where both are defined.)
Q: If sec θ = 5/4 and tan θ > 0, find sin θ.
A: 3/5. Since sec θ = 5/4, cos θ = 4/5. Both sec and tan positive means QI. sin²θ = 1 - 16/25 = 9/25, so sin θ = 3/5.
Q: If cot θ = 3/4 and cos θ < 0, find sin θ.
A: -4/5. Cot positive and cos negative means QIII. Build a 3-4-5 triangle; both sin and cos are negative in QIII.
Q: For θ in [0, π], cos θ = -1/2. Express θ as απ and find α.
A: α = 2/3. The angle is 2π/3.
Q: Evaluate tan(sin⁻¹(1/√2)).
A: 1. sin⁻¹(1/√2) = π/4, and tan(π/4) = 1.
Q: True or False: for all real x, sin(x + π/2) = cos x.
A: True. This follows from the sum formula: sin(x + π/2) = sin x cos(π/2) + cos x sin(π/2) = cos x.
Trig functions are differentiated and integrated constantly in calculus. The Pythagorean identity drives trig substitution (a key integration technique). Inverse trig functions produce important antiderivatives: ∫ dx/√(1 - x²) = sin⁻¹x + C, and ∫ dx/(1 + x²) = tan⁻¹x + C. Quadrant analysis is needed whenever you evaluate definite integrals involving trig.
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