1.5 Trigonometry

For y=a sin(bx+c)+d, |a| is amplitude, period is 2π/|b|, −c/b is the horizontal shift, and d is the midline. Equivalent rules apply to cosine.

Mark the midline and one cycle before sketching. Check the sign of a and the requested domain; transformations change coordinates, not just the appearance.

y=2cos(3x−π)+1 has amplitude 2, period 2π/3, midline y=1 and shift π/3 to the right.

The coefficient inside the bracket changes period inversely; multiplying x by 3 does not stretch the graph by 3.

The standard exact values sin, cos and tan at 0, π/6, π/4, π/3 and π/2 follow from 30–60–90 and 45–45–90 triangles, with signs set by the quadrant.

Reduce angles using periodicity and reference angles before applying the table. Keep radicals exact until a decimal is explicitly requested.

sin(5π/6)=sin(π−π/6)=1/2, while cos(5π/6)=−√3/2 because cosine is negative in quadrant II.

The reference angle gives a magnitude, not automatically the sign; tan is undefined where cos is zero.

sin⁻¹, cos⁻¹ and tan⁻¹ are functions with restricted principal ranges: typically [−π/2,π/2], [0,π] and (−π/2,π/2). They undo the trig function only on those ranges.

For an equation, use the principal value first, then generate other angles with symmetry and periodicity in the stated interval.

tan⁻¹(1)=π/4, but tan x=1 also has x=π/4+nπ; the inverse notation alone gives only the principal answer.

sin⁻¹x is not 1/sin x, and swapping sin with sin⁻¹ without checking the domain changes the problem.

Identities such as sin²x+cos²x=1, tanx=sinx/cosx and 1+tan²x=sec²x follow from definitions and hold wherever both sides are defined.

To prove an identity, simplify one side to the other using a common denominator or a fundamental identity; do not assume the result you are trying to prove.

(1−cos²x)/sinx simplifies to sinx when sinx≠0 because 1−cos²x=sin²x.

Algebraic cancellation can remove values where a denominator is zero; record the excluded angles even if the simplified expression looks defined.

Solve a trig equation by reducing it to a principal angle, applying quadrant symmetry, then adding periods. The interval determines which solutions survive.

Factorise or use a substitution when expressions such as 2sin²x−sinx−1 appear. Check every candidate in the original equation, especially after squaring.

2sinx−1=0 gives sinx=1/2, so on [0,2π] the solutions are π/6 and 5π/6.

One inverse-trig answer is not the complete solution, and the period of tan is π rather than 2π.