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P3.2 - Trigonometry

Syllabus
2019
Topic
P3.2
Level
A2

Learning objectives

P3.2.1Secant, cosecant and cotangentKnowledge of secant, cosecant and Angles measured in both degrees and radians. cotangent and of arcsin, arccos and arctan.; Their relationships to sine, cosine and tangent.; Understanding of their graphs and appropriate restricted domains.P3.2.2Further trigonometric identitiesKnowledge and use of sec2 θ = 1 + tan2 θ and cosec2 θ = 1 + cot2 θ.P3.2.3KnowledgeKnowledge and use of double angle To include application to half angles.; Knowledge of the formulae; use of formulae for t (tan 1 θ) formulae will not be required.; Students should be sin (A ± B), cos (A ± B) and able to solve equations such as a cos θ + b sin θ = c in a tan (A ± B) and of expressions for given interval, and to prove identities such as a cos θ + b sin θ in the equivalent cos x cos 2x + sin x sin 2x ≡ cos x. forms of r cos (θ ± a) or r sin (θ ± a).

Connect reciprocal and inverse trigonometric functions

Secant, cosecant and cotangent are reciprocal or quotient functions, so their graphs inherit zeros, signs and forbidden inputs from cosine and sine. Inverse trigonometric functions instead return a principal angle from a restricted one-one branch.

Function Definition Undefined when Range Period
sec⁡x\sec x 1/cos⁡x1/\cos x cos⁡x=0\cos x=0 (−∞,−1]∪[1,∞)(-\infty,-1]\cup[1,\infty) 2π2\pi
cosec⁡x\operatorname{cosec}x 1/sin⁡x1/\sin x sin⁡x=0\sin x=0 (−∞,−1]∪[1,∞)(-\infty,-1]\cup[1,\infty) 2π2\pi
cot⁡x\cot x cos⁡x/sin⁡x\cos x/\sin x sin⁡x=0\sin x=0 R\mathbb R π\pi

A forbidden input is a vertical asymptote. Secant and cosecant form branches outside the horizontal band between -1 and 1; cotangent decreases between consecutive asymptotes. The same structure can be read in degrees by replacing a full turn with 360 degrees.

Inverse Input domain Principal output range
arcsin⁡x\arcsin x [−1,1][-1,1] [−π/2,π/2][-\pi/2,\pi/2]
arccos⁡x\arccos x [−1,1][-1,1] [0,π][0,\pi]
arctan⁡x\arctan x R\mathbb R (−π/2,π/2)(-\pi/2,\pi/2)

For 0≤x<2π0\le x<2\pi, sec⁡x=−2\sec x=-2 becomes cos⁡x=−12\cos x=-\frac12, giving x=2π3,4π3x=\frac{2\pi}{3},\frac{4\pi}{3}. The calculator value arccos⁡(−12)=2π3\arccos(-\frac12)=\frac{2\pi}{3} is the principal angle; symmetry supplies the second interval solution.

sec⁡−1x\sec^{-1}x may denote inverse secant in some contexts, but 1/sec⁡x1/\sec x is its reciprocal. Keep inverse-function notation distinct from reciprocal identities, and match calculator mode to radians or degrees.

Derive and use the secant-cosecant identities

The two further identities are versions of the Pythagorean identity written entirely in tangent/secant or cotangent/cosecant. Deriving them reveals both the algebra and their domain restrictions.

sec⁡2θ=1+tan⁡2θ(cos⁡θ≠0),cosec⁡2θ=1+cot⁡2θ(sin⁡θ≠0)\sec^2\theta=1+\tan^2\theta\quad(\cos\theta\ne0),\qquad \operatorname{cosec}^2\theta=1+\cot^2\theta\quad(\sin\theta\ne0)

Divide sin⁡2θ+cos⁡2θ=1\sin^2\theta+\cos^2\theta=1 by cos⁡2θ\cos^2\theta to obtain tan⁡2θ+1=sec⁡2θ\tan^2\theta+1=\sec^2\theta. Dividing instead by sin⁡2θ\sin^2\theta gives 1+cot⁡2θ=cosec⁡2θ1+\cot^2\theta=\operatorname{cosec}^2\theta.

For example, tan⁡2θ=3sec⁡θ−3\tan^2\theta=3\sec\theta-3 becomes sec⁡2θ−1=3sec⁡θ−3,\sec^2\theta-1=3\sec\theta-3, so (sec⁡θ−1)(sec⁡θ−2)=0(\sec\theta-1)(\sec\theta-2)=0. Thus sec⁡θ=1\sec\theta=1 or 22 before any stated interval is applied.

An identity changes form but not domain. Do not use either formula at an angle where its original denominator is zero, and remember that sec⁡2θ=k\sec^2\theta=k gives both signs of sec⁡θ\sec\theta when k>0k>0.

Choose a compound-angle form that exposes the solution

Compound-angle formulae convert sums and differences of angles into products, or combine a sine-cosine pair into one shifted function. Choose the direction that reduces the number of trigonometric terms.

Function Sum/difference formula
sin⁡(A±B)\sin(A\pm B) sin⁡Acos⁡B±cos⁡Asin⁡B\sin A\cos B\pm\cos A\sin B
cos⁡(A±B)\cos(A\pm B) cos⁡Acos⁡B∓sin⁡Asin⁡B\cos A\cos B\mp\sin A\sin B
tan⁡(A±B)\tan(A\pm B) tan⁡A±tan⁡B1∓tan⁡Atan⁡B\dfrac{\tan A\pm\tan B}{1\mp\tan A\tan B}

sin⁡2θ=2sin⁡θcos⁡θ,cos⁡2θ=2cos⁡2θ−1=1−2sin⁡2θ,tan⁡2θ=2tan⁡θ1−tan⁡2θ;cos⁡2θ=1+cos⁡2θ2, sin⁡2θ=1−cos⁡2θ2\sin2\theta=2\sin\theta\cos\theta,\quad \cos2\theta=2\cos^2\theta-1=1-2\sin^2\theta,\quad \tan2\theta=\frac{2\tan\theta}{1-\tan^2\theta};\qquad \cos^2\theta=\frac{1+\cos2\theta}{2},\ \sin^2\theta=\frac{1-\cos2\theta}{2}

To combine acos⁡θ+bsin⁡θ=Rcos⁡(θ−α),a\cos\theta+b\sin\theta=R\cos(\theta-\alpha), expand the right side and match coefficients: Rcos⁡α=aR\cos\alpha=a, Rsin⁡α=bR\sin\alpha=b, so R=a2+b2R=\sqrt{a^2+b^2}. Choose the sign and quadrant of α\alpha from the coefficients; an equivalent shifted sine form is also valid.

For 0≤θ<2π0\le\theta<2\pi, 3cos⁡θ+4sin⁡θ=23\cos\theta+4\sin\theta=2 becomes 5cos⁡(θ−α)=25\cos(\theta-\alpha)=2, where α=arctan⁡(4/3)\alpha=\arctan(4/3). Hence θ−α=±arccos⁡(2/5)+2πn\theta-\alpha=\pm\arccos(2/5)+2\pi n, giving θ≈2.09\theta\approx2.09 or 6.056.05 radians in the interval.

For an identity, work from one side and look for a compound angle. For example, cos⁡xcos⁡2x+sin⁡xsin⁡2x=cos⁡(2x−x)=cos⁡x.\cos x\cos2x+\sin x\sin2x=\cos(2x-x)=\cos x. This uses the cosine-difference formula rather than checking selected values.

Do not divide by a trigonometric factor before preserving its zero branch, and filter every periodic solution against the stated interval and angle unit. The t=tan⁡(θ/2)t=\tan(\theta/2) formula is not required in this specification.