E5, E8 - Trigonometric identities and proofs

E5, E8 - Trigonometric identities and proofs

Use the quotient and Pythagorean identities in E5 to find exact values and simplify expressions. Then build the general arguments required in E8, checking domains and explaining why each algebraic step is valid.

Identities and their domains

If sin⁡θ=5/13\sin\theta=5/13, must tan⁡θ\tan\theta be positive? The sine value tells you the height on a unit circle, but there are two points at that height. Their horizontal coordinates have opposite signs. A missing sign can change an exact answer completely.

You already know

A unit-circle point at angle θ\theta has coordinates (cos⁡θ,sin⁡θ)(\cos\theta,\sin\theta). Pythagoras relates its squared coordinates, and the quadrant determines their signs.

For example, (12/13,5/13)(12/13,5/13) and (−12/13,5/13)(-12/13,5/13) both lie on the unit circle because 144/169+25/169=1144/169+25/169=1. This picture explains both the identity and the sign choice.

Identity and common domain

An identity is an equality true for every permitted value of its variable. Its common domain consists of values for which both expressions exist. The symbol ≡\equiv emphasises an identity; == is also accepted when that meaning is clear. An equation such as sin⁡θ=1/2\sin\theta=1/2 holds only at particular angles, so it is not an identity.

The unit circle has radius 11, so Pythagoras gives the first result below. The second expresses tangent as vertical projection divided by horizontal projection.

Quotient and Pythagorean identities

sin⁡2θ+cos⁡2θ≡1\sin^2\theta+\cos^2\theta\equiv1

for every real angle, and

tan⁡θ≡sin⁡θcos⁡θ(cos⁡θ≠0)\tan\theta\equiv\frac{\sin\theta}{\cos\theta}\qquad(\cos\theta\ne0)

Here sin⁡2θ=(sin⁡θ)2\sin^2\theta=(\sin\theta)^2, not sin⁡(θ2)\sin(\theta^2). Use degrees or radians consistently.

Not in the formula booklet: learn it.

Finding a square gives a magnitude. Only the quadrant chooses the signed value.

Worked example 1: an obtuse angle

Question. Given sin⁡θ=5/13\sin\theta=5/13 and π/2<θ<π\pi/2<\theta<\pi, find cos⁡θ\cos\theta and tan⁡θ\tan\theta exactly.

Plan. Find cosine's square, use quadrant II for its sign, then take the sine-to-cosine ratio.

cos⁡2θ=1−25169=144169Pythagorascos⁡θ=−1213quadrant IItan⁡θ=5/13−12/13=−512quotient identity\begin{aligned} \cos^2\theta&=1-\frac{25}{169}=\frac{144}{169} &&\text{Pythagoras}\\ \cos\theta&=-\frac{12}{13} &&\text{quadrant II}\\ \tan\theta&=\frac{5/13}{-12/13}=-\frac5{12} &&\text{quotient identity} \end{aligned}

Why this step? The square alone allows either sign; the stated interval excludes the positive cosine.

Answer. cos⁡θ=−12/13\cos\theta=-12/13, tan⁡θ=−5/12\tan\theta=-5/12.

Check. 25/169+144/169=125/169+144/169=1; positive sine divided by negative cosine gives negative tangent.

The second example changes only the quadrant. The magnitude calculation stays the same, but the signs change.

Worked example 2: an acute angle

Question. Given sin⁡θ=5/13\sin\theta=5/13 and 0<θ<π/20<\theta<\pi/2, find cos⁡θ\cos\theta and tan⁡θ\tan\theta exactly.

Plan. Repeat the square calculation and choose the quadrant I sign.

cos⁡2θ=1−25169=144169Pythagorascos⁡θ=1213quadrant Itan⁡θ=5/1312/13=512take the ratio\begin{aligned} \cos^2\theta&=1-\frac{25}{169}=\frac{144}{169} &&\text{Pythagoras}\\ \cos\theta&=\frac{12}{13} &&\text{quadrant I}\\ \tan\theta&=\frac{5/13}{12/13}=\frac5{12} &&\text{take the ratio} \end{aligned}

Why this step? The positive root follows from the interval, not from a rule that square roots always give the required trig value.

Answer. cos⁡θ=12/13\cos\theta=12/13, tan⁡θ=5/12\tan\theta=5/12.

Check. Both coordinates and their ratio are positive in quadrant I.

The opening question now has its answer: the same sine value can give positive or negative tangent. A quadrant is useful information, not decoration.

Watch out

Wrong: cos⁡2θ=144/169⇒cos⁡θ=12/13\cos^2\theta=144/169\Rightarrow\cos\theta=12/13 without checking the interval. Squaring erased the sign. Right: cos⁡θ=±12/13\cos\theta=\pm12/13 first, then select the sign using the quadrant.

Now change the sine value while keeping the acute interval.

Deriving the reciprocal identities

Dividing the unit-circle identity creates useful identities for the reciprocal functions. The divisor must be nonzero.

For cos⁡θ≠0\cos\theta\ne0, divide every term by cos⁡2θ\cos^2\theta. For sin⁡θ≠0\sin\theta\ne0, divide instead by sin⁡2θ\sin^2\theta:

sin⁡2θcos⁡2θ+1=1cos⁡2θdivide by cosine squared1+cos⁡2θsin⁡2θ=1sin⁡2θdivide by sine squared\begin{aligned} \frac{\sin^2\theta}{\cos^2\theta}+1&=\frac1{\cos^2\theta} &&\text{divide by cosine squared}\\ 1+\frac{\cos^2\theta}{\sin^2\theta}&=\frac1{\sin^2\theta} &&\text{divide by sine squared} \end{aligned}

Recognise sec⁡θ=1/cos⁡θ\sec\theta=1/\cos\theta, cosec⁡θ=1/sin⁡θ\operatorname{cosec}\theta=1/\sin\theta and cot⁡θ=cos⁡θ/sin⁡θ\cot\theta=\cos\theta/\sin\theta. These reciprocal definitions are not printed in the formula booklet: learn them.

Reciprocal Pythagorean identities

sec⁡2θ≡1+tan⁡2θ(cos⁡θ≠0)\sec^2\theta\equiv1+\tan^2\theta\qquad(\cos\theta\ne0) cosec⁡2θ≡1+cot⁡2θ(sin⁡θ≠0)\operatorname{cosec}^2\theta\equiv1+\cot^2\theta\qquad(\sin\theta\ne0)

Not in the formula booklet: learn it.

For instance, cot⁡γ=3\cot\gamma=3 with acute γ\gamma gives cosec⁡2γ=10\operatorname{cosec}^2\gamma=10, hence cosec⁡γ=10\operatorname{cosec}\gamma=\sqrt{10}. The reciprocal has the same sign as its denominator.

Worked example 3: finding secant

Question. Given tan⁡β=−2\tan\beta=-2 and π/2<β<π\pi/2<\beta<\pi, find sec⁡β\sec\beta exactly.

Plan. Find secant's square, then use the cosine sign.

sec⁡2β=1+(−2)2=5reciprocal identitysec⁡β=−5cos⁡β<0\begin{aligned} \sec^2\beta&=1+(-2)^2=5 &&\text{reciprocal identity}\\ \sec\beta&=-\sqrt5 &&\cos\beta<0 \end{aligned}

Why this step? Secant is the reciprocal of a negative cosine, so it must be negative too.

Answer. sec⁡β=−5\sec\beta=-\sqrt5.

Check. (−5)2−(−2)2=5−4=1(-\sqrt5)^2-(-2)^2=5-4=1.

Change only the tangent value from −2-2 to −3-3. The quadrant rule stays in place.

Worked example 4: a different tangent value

Question. Given tan⁡β=−3\tan\beta=-3 and π/2<β<π\pi/2<\beta<\pi, find sec⁡β\sec\beta exactly.

Plan. Use the same identity and sign decision.

sec⁡2β=1+(−3)2=10reciprocal identitysec⁡β=−10quadrant II\begin{aligned} \sec^2\beta&=1+(-3)^2=10 &&\text{reciprocal identity}\\ \sec\beta&=-\sqrt{10} &&\text{quadrant II} \end{aligned}

Why this step? A negative tangent does not itself determine secant's sign; the given quadrant does.

Answer. sec⁡β=−10\sec\beta=-\sqrt{10}.

Check. 10−9=110-9=1, as required by sec⁡2β−tan⁡2β=1\sec^2\beta-\tan^2\beta=1.

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