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1.5 Trigonometry

Syllabus
9709–2028–2029
Topic
1.5
Level
AS

Learning objectives

1.5.1Trig graphs• sketch and use graphs of the sine, cosine and tangent functions (for angles of any size, and using either degrees or radians) Including e.g. y = 3 sin x, y = 1 - cos 2x,.tany x 4 1 r= +`j1.5.2Exact trig values• use the exact values of the sine, cosine and tangent of 30°, 45°, 60°, and related angles e.g.,coss in1503 22 1 4 3 2 1c - r==.1.5.3Inverse trig notation• use the notations sin-1x, cos-1x, tan-1x to denote the principal values of the inverse trigonometric relations No specialised knowledge of these functions is required, but understanding of them as examples of inverse functions is expected.1.5.4Trig identities• use the identities cos sin tan/i i i and sinc os 122 /ii+ e.g. in proving identities, simplifying expressions and solving equations.1.5.5Trig equations• find all the solutions of simple trigonometrical equations lying in a specified interval (general forms of solution are not included). e.g. solve 3sin 21 0x += for x<<rr-, 3sin 5cos 102 =ii-- for 0° ⩽ i ⩽ 360°.

Build trig graphs from period, midline, amplitude or asymptotes

Graph Period Key vertical feature
y=asin⁡(bx+c)+dy=a\sin(bx+c)+d 2π/∣b∣2\pi/|b| amplitude ∣a∣|a|, midline y=dy=d
y=acos⁡(bx+c)+dy=a\cos(bx+c)+d 2π/∣b∣2\pi/|b| amplitude ∣a∣|a|, midline y=dy=d
y=atan⁡(bx+c)+dy=a\tan(bx+c)+d π/∣b∣\pi/|b| no amplitude; repeating vertical asymptotes, centre line y=dy=d

For each form, solve $bx+c=0$ to locate the horizontal reference shift $x=-c/b$. Horizontal scale is $1/|b|$.

Mark the requested domain, midline or tangent asymptotes, then place one cycle from known exact points before repeating by the period. A negative outside coefficient reflects in the midline.

y=2cos⁡(3x−π)+1y=2\cos(3x-\pi)+1 has amplitude 22, period 2π/32\pi/3, midline y=1y=1 and shift π/3\pi/3 right. Its greatest and least values are 33 and −1-1.

Tangent has no greatest/least value and no amplitude. Inside multiplication changes horizontal scale inversely: f(3x)f(3x) is compressed to one third of the original width.

Exact trigonometric values come from reference triangles and symmetry

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.

Inverse trig notation returns one principal value

sin⁡−1x\sin^{-1}x, cos⁡−1x\cos^{-1}x and tan⁡−1x\tan^{-1}x denote principal inverse-function values, not reciprocals. The original trig functions are restricted so each inverse returns exactly one angle.

Inverse Input domain Principal output range
sin⁡−1x\sin^{-1}x −1≤x≤1-1\le x\le1 −π/2≤θ≤π/2-\pi/2\le\theta\le\pi/2
cos⁡−1x\cos^{-1}x −1≤x≤1-1\le x\le1 0≤θ≤π0\le\theta\le\pi
tan⁡−1x\tan^{-1}x all real xx −π/2<θ<π/2-\pi/2<\theta<\pi/2

\cos^{-1}(-\sqrt3/2)=5\pi/6,because $5\pi/6$ lies in the principal cosine range and has cosine $-\sqrt3/2$.

When inverse notation appears inside an equation, evaluate the principal value with consistent degree/radian mode, then continue the algebra. Finding every solution of a trig equation is a separate next objective based on graphs and the stated interval.

sin⁡−1x≠1/sin⁡x\sin^{-1}x\ne1/\sin x; the reciprocal is csc⁡x\csc x. Do not append a general solution form here: Paper 1 requires principal notation and interval solutions, not general forms.

Use the two Paper 1 identities to change form without changing value

\tan\theta=\frac{\sin\theta}{\cos\theta}\quad(\cos\theta\ne0),\qquad \sin^2\theta+\cos^2\theta=1.

Use the ratio identity to replace tangent by sine/cosine, especially when forming a common denominator. Use the Pythagorean identity to replace 1−sin⁡2θ1-\sin^2\theta by cos⁡2θ\cos^2\theta or 1−cos⁡2θ1-\cos^2\theta by sin⁡2θ\sin^2\theta.

To prove an identity, work from one side and make valid algebraic substitutions until it matches the other. Do not begin by assuming both sides equal; show every cancellation or common denominator.

For $\sin\theta\ne0$:\frac{1-\cos^2\theta}{\sin\theta}=\frac{\sin^2\theta}{\sin\theta}=\sin\theta.

Cancellation can hide excluded denominator values, so retain conditions. Secant, cosecant, cotangent and their identities belong to later Pure Mathematics 2 and are not needed for this objective.

Solve trigonometric equations by finding all angles in the stated interval

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.