1.05j - Fundamental trigonometric identities

1.05j - Fundamental trigonometric identities

Derive the two fundamental identities in 1.05j from unit-circle coordinates. You will recover exact values with the correct signs, prove equalities on their allowed domains and solve equations without dividing away valid solutions.

Two identities from coordinates

If sin⁡θ=3/5\sin\theta=3/5, must cos⁡θ\cos\theta be 4/54/5? The unit circle gives the magnitude, but not the sign: (4/5,3/5)(4/5,3/5) and (−4/5,3/5)(-4/5,3/5) both lie on it. We need an identity and a quadrant to distinguish them.

You already know

The unit-circle point at angle θ\theta is (cos⁡θ,sin⁡θ)(\cos\theta,\sin\theta). Its distance from the origin is 11, and its coordinate signs identify the quadrant.

For example, (3/5)2+(4/5)2=9/25+16/25=1(3/5)^2+(4/5)^2=9/25+16/25=1. The same Pythagoras argument works at every angle, including those on the axes.

Two unit-circle points share sine 3/5 but have opposite cosine signs

Trigonometric identity

An identity is an equality valid for every input in its stated domain. An equation usually holds only for particular inputs. For example, sin⁡2θ+cos⁡2θ=1\sin^2\theta+\cos^2\theta=1 is an identity for all real angles; sin⁡θ=1/2\sin\theta=1/2 is an equation with selected solutions.

The tangent of the direction is the ratio of its signed vertical and horizontal coordinates. That ratio needs a nonzero horizontal coordinate.

The fundamental identities

sin⁡2θ+cos⁡2θ=1(θ∈R)\sin^2\theta+\cos^2\theta=1\qquad(\theta\in\mathbb R) tan⁡θ=sin⁡θcos⁡θ(cos⁡θ≠0)\tan\theta=\frac{\sin\theta}{\cos\theta}\qquad(\cos\theta\ne0)

Not in the formula sheet: learn it.

For integer kk, the tangent identity excludes θ=π/2+kπ\theta=\pi/2+k\pi. At π/2\pi/2, Pythagoras still gives 1+0=11+0=1, while tangent is undefined. Also sin⁡2θ\sin^2\theta means (sin⁡θ)2(\sin\theta)^2, not sin⁡(θ2)\sin(\theta^2).

A useful surprise is that the sum without squares behaves differently: at π/4\pi/4, sin⁡θ+cos⁡θ=2\sin\theta+\cos\theta=\sqrt2, rather than 11. Squaring the coordinates is essential, not a typographical detail.

Key idea

The unit circle gives a sum of squares for every angle; a ratio identity additionally needs its denominator to be nonzero.

One exact counterexample is enough to reject a claimed identity, but one successful substitution cannot prove it.

Recovering values without losing signs

An identity determines a squared magnitude; the given interval determines the sign of the function itself.

Rearranging gives cos⁡2θ=1−sin⁡2θ\cos^2\theta=1-\sin^2\theta or sin⁡2θ=1−cos⁡2θ\sin^2\theta=1-\cos^2\theta. These are algebraic forms of the unsupplied fundamental identity. Take both possible square-root signs seriously until the quadrant decides between them.

Worked example 1: positive sine, negative cosine

Question. Given sin⁡θ=3/5\sin\theta=3/5 and π/2<θ<π\pi/2<\theta<\pi, find cos⁡θ\cos\theta and tan⁡θ\tan\theta exactly.

Plan. Find cosine squared, choose its quadrant-II sign, then use the tangent ratio.

cos⁡2θ=1−(3/5)2=16/25Pythagorean identitycos⁡θ=−4/5negative in quadrant IItan⁡θ=3/5−4/5=−3/4sine over cosine\begin{aligned} \cos^2\theta&=1-(3/5)^2=16/25 && \text{Pythagorean identity}\\ \cos\theta&=-4/5 && \text{negative in quadrant II}\\ \tan\theta&=\frac{3/5}{-4/5}=-3/4 && \text{sine over cosine} \end{aligned}

Why this step? The equation cos⁡2θ=16/25\cos^2\theta=16/25 alone permits both signs; the interval removes the positive option.

Answer. cos⁡θ=−4/5\cos\theta=-4/5, tan⁡θ=−3/4\tan\theta=-3/4.

Check. The squares sum to 11, and tangent is negative in quadrant II.

Change only the quadrant to quadrant I. The given sine stays 3/53/5, but the sign decision changes.

Worked example 2: the same sine in quadrant I

Question. Given sin⁡θ=3/5\sin\theta=3/5 and 0<θ<π/20<\theta<\pi/2, find cos⁡θ\cos\theta and tan⁡θ\tan\theta exactly.

Plan. Reuse the squared magnitude with the positive quadrant-I sign.

cos⁡2θ=16/25same sine magnitudecos⁡θ=4/5positive in quadrant Itan⁡θ=3/54/5=3/4ratio identity\begin{aligned} \cos^2\theta&=16/25 && \text{same sine magnitude}\\ \cos\theta&=4/5 && \text{positive in quadrant I}\\ \tan\theta&=\frac{3/5}{4/5}=3/4 && \text{ratio identity} \end{aligned}

Why this step? Equal sine values can belong to different quadrants, so they do not force equal cosine values.

Answer. cos⁡θ=4/5\cos\theta=4/5, tan⁡θ=3/4\tan\theta=3/4.

Check. Both coordinates and their ratio are positive, as the interval requires.

You can start from another function too. If cos⁡θ=−5/13\cos\theta=-5/13 in quadrant III, then sin⁡2θ=144/169\sin^2\theta=144/169, so sin⁡θ=−12/13\sin\theta=-12/13 and tan⁡θ=12/5\tan\theta=12/5. If instead tan⁡θ=2\tan\theta=2 in quadrant III, write sin⁡θ=2cos⁡θ\sin\theta=2\cos\theta; substitution gives 5cos⁡2θ=15\cos^2\theta=1, hence cos⁡θ=−1/5\cos\theta=-1/\sqrt5 and sin⁡θ=−2/5\sin\theta=-2/\sqrt5.

Watch out

Wrong: conclude cos⁡θ=1−sin⁡2θ\cos\theta=\sqrt{1-\sin^2\theta} without checking signs. Right: the square root is a non-negative magnitude; cosine may be its negative.

Use a new sine value with the quadrant-II decision.

Simplifying and proving with domains

Start with one side of a proposed identity and transform it by valid steps, preserving every denominator restriction.

For example,

1−sin⁡2θcos⁡θ=cos⁡2θcos⁡θ=cos⁡θ(cos⁡θ≠0)\frac{1-\sin^2\theta}{\cos\theta} =\frac{\cos^2\theta}{\cos\theta} =\cos\theta\qquad(\cos\theta\ne0)

The simplified expression has more allowed inputs than the quotient. Cancellation does not define the starting quotient at its missing points. Likewise (1−cos⁡2θ)/sin⁡θ=sin⁡θ(1-\cos^2\theta)/\sin\theta=\sin\theta only on sin⁡θ≠0\sin\theta\ne0.

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