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P2.6 - Trigonometry

Syllabus
2019
Topic
P2.6
Level
AS

Use identities to connect sine, cosine and tangent

A trigonometric identity is true for every angle where both sides are defined. These two identities let you replace one function by another without changing the relationship.

tan⁡θ=sin⁡θcos⁡θ(cos⁡θ≠0),sin⁡2θ+cos⁡2θ=1\tan\theta=\frac{\sin\theta}{\cos\theta}\quad(\cos\theta\ne0),\qquad \sin^2\theta+\cos^2\theta=1

Useful rearrangements are sin⁡2θ=1−cos⁡2θ\sin^2\theta=1-\cos^2\theta and cos⁡2θ=1−sin⁡2θ\cos^2\theta=1-\sin^2\theta. Choose the version that leaves only one trigonometric function.

If cos⁡θ=−35\cos\theta=-\frac35 and π2<θ<π\frac\pi2<\theta<\pi, then sin⁡2θ=1−925=1625.\sin^2\theta=1-\frac9{25}=\frac{16}{25}. Sine is positive in this interval, so sin⁡θ=45\sin\theta=\frac45 and tan⁡θ=−43\tan\theta=-\frac43. The identity gives the magnitude; the interval determines the sign.

For example, 8tan⁡θ=3cos⁡θ8\tan\theta=3\cos\theta becomes 8sin⁡θ=3cos⁡2θ=3(1−sin⁡2θ)8\sin\theta=3\cos^2\theta=3(1-\sin^2\theta), hence 3sin⁡2θ+8sin⁡θ−3=0.3\sin^2\theta+8\sin\theta-3=0.

Do not take a square root without choosing its sign from the angle interval. Also retain cos⁡θ≠0\cos\theta\ne0 whenever the tangent identity is used.

Solve every trigonometric branch inside an interval

A trigonometric equation is complete only when every valid branch in the stated interval has been found. Keep the angle unit and interval visible throughout.

Use this sequence: rewrite in one trigonometric function; factor or solve the resulting algebraic equation; reject values outside the function's range; generate every periodic angle; undo any shift or multiple; then apply the original endpoints.

Equation for uu Degree solutions Radian solutions
sin⁡u=k\sin u=k u=α+360∘nu=\alpha+360^\circ n or 180∘−α+360∘n180^\circ-\alpha+360^\circ n u=α+2πnu=\alpha+2\pi n or π−α+2πn\pi-\alpha+2\pi n
cos⁡u=k\cos u=k u=±α+360∘nu=\pm\alpha+360^\circ n u=±α+2πnu=\pm\alpha+2\pi n
tan⁡u=k\tan u=k u=α+180∘nu=\alpha+180^\circ n u=α+πnu=\alpha+\pi n

Here n∈Zn\in\mathbb Z and α\alpha is the appropriate inverse-trigonometric value. If u=2xu=2x or u=x+π2u=x+\frac\pi2, solve over the corresponding uu-interval before converting back to xx.

Solve 6cos⁡2x+sin⁡x−5=06\cos^2x+\sin x-5=0 for 0≤x<360∘0\le x<360^\circ. Let s=sin⁡xs=\sin x: 6(1−s2)+s−5=0⟹(3s+1)(2s−1)=0.6(1-s^2)+s-5=0\quad\Longrightarrow\quad(3s+1)(2s-1)=0. Thus sin⁡x=12\sin x=\frac12 or sin⁡x=−13\sin x=-\frac13, giving x=30∘, 150∘, 199.5∘, 340.5∘x=30^\circ,\ 150^\circ,\ 199.5^\circ,\ 340.5^\circ to one decimal place where needed.

Do not divide by a trigonometric factor before preserving its zero branch. Match calculator mode to degrees or radians, and include an endpoint only when the stated inequality includes it.