Reciprocal and inverse circular functions: VCE Specialist Mathematics Units 3 & 4

1. Two operations that answer different questions

A reciprocal circular function reverses a numerical value by division: sec x = 1/cos x, cosec x = 1/sin x and cot x = cos x/sin x. An inverse circular function reverses a function relationship: sin⁻¹ u is the selected angle whose sine is u. Thus sec x and cos⁻¹ x have different inputs, outputs and graphs. The superscript −1 is conventional inverse-function notation here, while a reciprocal should be written as a fraction or named sec, cosec or cot.

All angles in this guide are in radians. A domain is the permitted set of inputs; a range is the set of outputs actually attained. Writing a reciprocal immediately imposes a denominator restriction. Writing an inverse demands a one-to-one branch of the original function. These restrictions are mathematical information, rather than an administrative detail to attach after solving an equation.

Think about what a calculator receives. In sec(π/3), an angle enters and the number 2 comes out. In cos⁻¹(1/2), a dimensionless ratio enters and the principal angle π/3 comes out. In cos⁻¹(2), there is no real answer because no real angle has cosine 2. Treating every symbol as an algebraic power hides this essential distinction.

2. Build reciprocal graphs from familiar graphs

Secant is defined when cos x ≠ 0, so exclude x = π/2 + kπ for every integer k. Its range is (−∞, −1] ∪ [1, ∞), because cosine lies between −1 and 1 and its nonzero values are being reciprocated. Secant has period 2π and is even. It passes through (0, 1), has another turning point at (π, −1), and develops vertical asymptotes at the excluded inputs.

Cosecant excludes x = kπ, has the same range as secant, has period 2π and is odd. Its positive arch has lowest point (π/2, 1). Cotangent excludes x = kπ, has range all real numbers, period π, and is odd. On each interval between adjacent asymptotes it decreases through a zero at x = π/2 + kπ. Do not draw a secant or cosecant arch crossing the horizontal axis: a reciprocal of a finite nonzero number cannot be zero.

Near an asymptote, determine the sign from the original denominator. Just left of π/2, cosine is small and positive, so secant becomes large and positive. Just right, cosine is small and negative, so secant becomes large and negative. This sign reasoning determines which way each branch rises or falls and prevents disconnected branches being joined across an undefined point.

For a transformed function a sec(b(x − h)) + k with nonzero a and b, the period is 2π/|b| and the centre of the transformation is controlled by h and k. Find excluded inputs by solving b(x − h) = π/2 + nπ. Obtain the range by transforming the original two intervals, taking care when a is negative. A vertical shift moves the entire pair of range intervals, not just their endpoint labels.

3. Why inverse functions need selected branches

Sine repeats values, so it has no inverse on all real numbers. Restrict sine to [−π/2, π/2], where it increases through every value from −1 to 1 exactly once. Reflect that restricted graph in y = x to obtain arcsin x, also written sin⁻¹ x. Its domain is [−1, 1] and its range is [−π/2, π/2]. The inverse graph swaps input and output restrictions.

For arccos x, restrict cosine to [0, π]. Its domain is [−1, 1] and range [0, π]; the graph decreases. For arctan x, restrict tangent to (−π/2, π/2). Its domain is all real numbers and range (−π/2, π/2), with horizontal asymptotes y = ±π/2. Those limiting angles are never attained at finite inputs. Arcsine and arctangent are odd; arccosine is not.

The statement sin(arcsin x) = x holds for −1 ≤ x ≤ 1. The reversed statement arcsin(sin θ) = θ holds only when θ lies on the selected sine branch. Outside that interval, return the angle on the principal branch with the same sine. This is why cancelling inverse symbols without checking domains can turn a correct-looking line into a false identity.

Transformations must act on the full function. For y = 3arcsin(2x − 1) + 4, require −1 ≤ 2x − 1 ≤ 1, giving 0 ≤ x ≤ 1. The range is [4 − 3π/2, 4 + 3π/2]. The inverse output is an angle before the vertical scale and translation are applied. When sketching, transform endpoint coordinates as well as the midpoint.

4. Identities are tools with domains

The basic identity sin²x + cos²x = 1 gives 1 + tan²x = sec²x after division by cos²x, valid where cos x ≠ 0. Division by sin²x gives 1 + cot²x = cosec²x where sin x ≠ 0. These restrictions matter when an identity is used to rewrite an equation. A convenient expression must still respect every restriction in the original question.

The compound-angle identities are sin(A ± B) = sin A cos B ± cos A sin B and cos(A ± B) = cos A cos B ∓ sin A sin B. The tangent version is tan(A ± B) = (tan A ± tan B)/(1 ∓ tan A tan B), when the displayed terms are defined. Setting A = B gives sin 2A = 2sin A cos A and cos 2A = cos²A − sin²A = 1 − 2sin²A = 2cos²A − 1. Also tan 2A = 2tan A/(1 − tan²A), wherever all displayed terms are defined.

Choose an identity according to the structure you need. An equation containing cos 2x and sin x becomes a polynomial in sin x using 1 − 2sin²x. A later integral containing sin²x can use sin²x = (1 − cos 2x)/2. An exact-value calculation involving 75° can use π/4 + π/6 in radians. The identity is a justified change of representation, not an instruction to introduce every possible circular function.

5. Change products, sums and phases into useful forms

Compound-angle identities also connect products and sums. Add or subtract the sine and cosine compound formulas to obtain the following product-to-sum identities. They hold for all real angles A and B; no denominator is introduced.

Reversing this relationship gives sum-to-product forms. Let M = (A + B)/2 and D = (A − B)/2, so A = M + D and B = M − D. The identities become:

A product form can turn an equation into separate factors; a sum form can make a later integral easier. For example, sin 5x + sin x = 2sin 3x cos 2x. On 0 ≤ x ≤ π/2, setting this expression to zero gives sin 3x = 0 or cos 2x = 0. The complete solution set is x = 0, π/4, π/3. Substitution checks all three; dividing by either factor at the start would lose some solutions.

An auxiliary angle rewrites a sin x + b cos x as R sin(x + α). Choose R = √(a² + b²), then match coefficients: cos α = a/R and sin α = b/R. These two conditions select the correct quadrant. Using α = arctan(b/a) without checking signs can select the wrong branch, and it fails when a = 0. If a = b = 0, the expression is identically zero and no phase choice is needed.

For an original check, 3sin x + 4cos x = 5sin(x + α), where α = arctan(4/3) lies in the first quadrant. Expanding gives 5sin x(3/5) + 5cos x(4/5), recovering both coefficients. The maximum value is 5, attained when x + α = π/2 + 2kπ. This gives both an exact bound and all maximum locations without differentiating. The equivalent cosine form R cos(x − β) uses cos β = b/R and sin β = a/R; again, determine the quadrant from both coefficients.

6. Precision, technology and a non-routine check

For an equation involving sec x, multiplying by cos x is allowed only after excluding its zeros. Squaring an equation can introduce extra candidates because it loses sign information. Applying an inverse function returns a principal value, which may need to be supplemented by periodic solutions. Keep these three issues separate and check candidates in the original equation.

For a CAS or graphing check, select radian mode, enter explicit parentheses and display the restricted inverse range. A plotter may draw near-vertical connections across reciprocal asymptotes; inspect the function's domain rather than trusting connecting pixels. Compare exact coordinates first and then use decimal values to assess the sketch. A numerical equality at one input does not prove an identity for an interval.

A useful investigation is to solve arcsin x = arccos x. Both functions share domain [−1, 1], and their sum is π/2 there. Equality therefore requires each to be π/4, giving x = √2/2. Their opposite monotonic behaviour proves uniqueness. This combines an identity, a domain restriction and a graphical argument rather than treating a decimal crossing as a complete solution. The same branch habits will be essential in advanced differentiation.