Reciprocal and inverse trigonometric functions flashcards

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sec θ
sec θ = 1/cos θ (secant). Undefined where cos θ = 0.
cosec θ
cosec θ = 1/sin θ (cosecant). Undefined where sin θ = 0.
cot θ
cot θ = 1/tan θ = cos θ/sin θ (cotangent). Undefined where sin θ = 0.
arcsin x
The angle in [−π/2, π/2] whose sine is x; defined for −1 ≤ x ≤ 1. Also written sin⁻¹ x.
arccos x
The angle in [0, π] whose cosine is x; defined for −1 ≤ x ≤ 1.
arctan x
The angle in (−π/2, π/2) whose tangent is x; defined for all real x.
Identity (≡)
An equation that is true for every value of the variable for which both sides are defined.
Pythagorean identity with sec and tan
sec²θ ≡ 1 + tan²θ (so tan²θ = sec²θ − 1). (No — learn it)
Pythagorean identity with cosec and cot
cosec²θ ≡ 1 + cot²θ (so cot²θ = cosec²θ − 1). (No — learn it)
Deriving sec²θ = 1 + tan²θ
Divide sin²θ + cos²θ ≡ 1 by cos²θ: tan²θ + 1 ≡ sec²θ. (No — derive it)
Deriving cosec²θ = 1 + cot²θ
Divide sin²θ + cos²θ ≡ 1 by sin²θ: 1 + cot²θ ≡ cosec²θ. (No — derive it)
cos(arcsin x) and sin(arccos x)
Both equal √(1 − x²) (positive root, from the ranges of arcsin and arccos). (No — derive it)
Key features of y = sec x on 0 ≤ x ≤ 2π
Asymptotes x = π/2, 3π/2; minima (0, 1), (2π, 1); maximum (π, −1); period 2π; range y ≤ −1 or y ≥ 1.
Key features of y = cosec x on 0 < x < 2π
Asymptotes x = 0, π, 2π; minimum (π/2, 1); maximum (3π/2, −1); period 2π; range y ≤ −1 or y ≥ 1.
Key features of y = cot x
Asymptotes x = nπ; zeros at x = π/2 + nπ; decreasing between asymptotes; period π; range all reals.
Graph of y = arcsin x
Increasing curve from (−1, −π/2) through (0, 0) to (1, π/2): sin on [−π/2, π/2] reflected in y = x.
Graph of y = arccos x
Decreasing curve from (−1, π) through (0, π/2) to (1, 0): cos on [0, π] reflected in y = x.
Graph of y = arctan x
Increasing through the origin, defined for all x, horizontal asymptotes y = −π/2 and y = π/2.
Why must sin, cos and tan be restricted before they have inverses?
They are many-one; an inverse exists only for a one-one function, so each is restricted to an interval where it takes every value once.
Where does a reciprocal trig graph have asymptotes?
Wherever the original function is zero: sec where cos x = 0, cosec and cot where sin x = 0.
Why does sec x = 0.5 have no solutions?
It needs cos x = 2, which is impossible. sec and cosec never take values strictly between −1 and 1.
Signs of sec, cosec and cot
Each has the same sign as cos, sin and tan respectively, so the quadrant (CAST) rules carry over.
Is arcsin(sin θ) = θ?
Only for −π/2 ≤ θ ≤ π/2. For example arcsin(sin(5π/6)) = π/6.
Exact value of sec(2π/3)?
cos(2π/3) = −½, so sec(2π/3) = −2.
Exact value of cosec 240°?
sin 240° = −√3/2, so cosec 240° = −2/√3 = −2√3/3.
Exact values of arccos(−½) and arctan(−1)?
2π/3 and −π/4.
θ obtuse, tan θ = −2√2. Find sec θ.
sec²θ = 1 + 8 = 9; cos θ < 0 for obtuse θ, so sec θ = −3.
Solve cot θ = −0.4 for 0° ≤ θ < 360°.
tan θ = −2.5 → θ = 111.8° or 291.8°.
Solve sec 2x = 3 for 0 ≤ x ≤ π.
cos 2x = 1/3, 0 ≤ 2x ≤ 2π → 2x = 1.231 or 5.052 → x = 0.615 or 2.53.
Solve 2 tan²θ + sec θ = 1 for 0° ≤ θ < 360°.
2 sec²θ + sec θ − 3 = 0 → sec θ = 1 or −3/2 → θ = 0°, 131.8°, 228.2°.
Exact value of sec(arctan(√5/2))?
sec² = 1 + 5/4 = 9/4; arctan gives an angle with cos > 0, so the answer is 3/2.
Pairing reciprocals by the first letter
Use the third letter: seC ↔ Cos, coSec ↔ Sin, coT ↔ Tan.
sin⁻¹ x versus 1/sin x
sin⁻¹ x = arcsin x (inverse function); 1/sin x = (sin x)⁻¹ = cosec x.
Choosing the identity in an equation
tan² with sec → sec² = 1 + tan²; cot² with cosec → cosec² = 1 + cot². Replace the squared term.
Dividing by a trig function
Never divide by (or cancel) a trig function such as sin θ, cos θ or tan θ that could be zero: factorise instead to keep every solution.
Proving an identity
Start from one side, work to the other, finish with “= RHS”. Do not work on both sides or assume the result.
Sign after using an identity
sec²θ = 9 gives sec θ = ±3: choose the sign from the quadrant or the range of the inverse function, and say why.

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