FP1 1.1 - The t-formulae

FP1 1.1 - The t-formulae

Derive and use the half-angle formulae that express sine, cosine and tangent as rational functions of t=tan⁡(θ/2)t=\tan(\theta/2) (FP1 1.1). You will track their domains and recover the correct angles, including the points the substitution cannot represent.

Sine and cosine from the half-angle

How can knowing tan⁡(θ/2)=1/2\tan(\theta/2)=1/2 tell you that sin⁡θ=4/5\sin\theta=4/5 and cos⁡θ=3/5\cos\theta=3/5, without first finding θ\theta? The half-angle parameter turns both functions into fractions involving only one real number.

Start with u=θ/2u=\theta/2. The double-angle identities give sin⁡θ=2sin⁡ucos⁡u\sin\theta=2\sin u\cos u and cos⁡θ=cos⁡2u−sin⁡2u\cos\theta=\cos^2u-\sin^2u. Divide each by the identity 1=sin⁡2u+cos⁡2u1=\sin^2u+\cos^2u, then divide numerator and denominator by cos⁡2u\cos^2u. This last step requires cos⁡u≠0\cos u\ne0, exactly the condition needed to define tan⁡u\tan u.

You already know

From A-level Mathematics, sin⁡2u=2sin⁡ucos⁡u\sin2u=2\sin u\cos u, cos⁡2u=cos⁡2u−sin⁡2u\cos2u=\cos^2u-\sin^2u, and sin⁡2u+cos⁡2u=1\sin^2u+\cos^2u=1. The compound-angle identities are in the formula booklet. The variable t=tan⁡(θ/2)t=\tan(\theta/2) is the tangent of the half-angle, not half of tan⁡θ\tan\theta.

Writing t=tan⁡ut=\tan u gives the derivation:

sin⁡θ=2sin⁡ucos⁡ucos⁡2u+sin⁡2udouble angle and identity=2tan⁡u1+tan⁡2u=2t1+t2divide by cos⁡2ucos⁡θ=cos⁡2u−sin⁡2ucos⁡2u+sin⁡2udouble angle and identity=1−tan⁡2u1+tan⁡2u=1−t21+t2same division\begin{aligned} \sin\theta &=\frac{2\sin u\cos u}{\cos^2u+\sin^2u} && \text{double angle and identity}\\ &=\frac{2\tan u}{1+\tan^2u} =\frac{2t}{1+t^2} && \text{divide by }\cos^2u\\ \cos\theta &=\frac{\cos^2u-\sin^2u}{\cos^2u+\sin^2u} && \text{double angle and identity}\\ &=\frac{1-\tan^2u}{1+\tan^2u} =\frac{1-t^2}{1+t^2} && \text{same division} \end{aligned}

The common denominator is positive for every finite real tt, so neither of these rational expressions has a real denominator zero.

The sine and cosine t-formulae

If t=tan⁡(θ/2)t=\tan(\theta/2) is defined, then

sin⁡θ=2t1+t2,cos⁡θ=1−t21+t2\sin\theta=\frac{2t}{1+t^2}, \qquad \cos\theta=\frac{1-t^2}{1+t^2}

The substitution requires θ≠(2k+1)π\theta\ne(2k+1)\pi, where k∈Zk\in\mathbb Z. Sine and cosine themselves still exist at those omitted angles.

Not in the formula booklet: learn it.

The formulae also explain a number pattern. Taking t=2/3t=2/3 gives (cos⁡θ,sin⁡θ)=(5/13,12/13)(\cos\theta,\sin\theta)=(5/13,12/13): a point built from the 5,12,135,12,13 right triangle. More generally, t=m/nt=m/n gives

cos⁡θ=n2−m2n2+m2,sin⁡θ=2mnn2+m2\cos\theta=\frac{n^2-m^2}{n^2+m^2}, \qquad \sin\theta=\frac{2mn}{n^2+m^2}

for integers m,nm,n with n≠0n\ne0. The identity (n2−m2)2+(2mn)2=(n2+m2)2(n^2-m^2)^2+(2mn)^2=(n^2+m^2)^2 is the same unit-circle identity after clearing denominators. A rational half-angle parameter naturally builds integer right triangles.

[DIAGRAM: asset_name: A rational half-angle point on the unit circle; asset_slug: 9fm0-fp1-t-unit-circle; caption: The parameter t equals one half gives the point (three fifths, four fifths) on the unit circle; description: Equal-scale coordinate axes show the unit circle, O at the origin and P at (3/5,4/5). A blue radius OP and a sage line from the open point A at (minus 1,0) to P are shown. The sage line has slope one half and equation y equals (x plus 1) over 2. A small arc at O marks theta and a separate small arc at A marks theta over 2. Labels are offset from axes and curves, with a side note giving t equals one half, cos theta equals three fifths and sin theta equals four fifths.]
The parameter t equals one half gives the point (three fifths, four fifths) on the unit circle

Worked example 1: exact values from a rational parameter

Question. Given t=tan⁡(θ/2)=1/2t=\tan(\theta/2)=1/2, find sin⁡θ\sin\theta and cos⁡θ\cos\theta exactly.

Plan. Substitute the half-angle tangent directly into the two rational formulae.

1+t2=1+14=54common denominatorsin⁡θ=2(1/2)5/4=45sine formulacos⁡θ=1−1/45/4=35cosine formula\begin{aligned} 1+t^2&=1+\frac14=\frac54 && \text{common denominator}\\ \sin\theta&=\frac{2(1/2)}{5/4}=\frac45 && \text{sine formula}\\ \cos\theta&=\frac{1-1/4}{5/4}=\frac35 && \text{cosine formula} \end{aligned}

Why this step? The known value is already tt; finding θ\theta first would introduce an unnecessary inverse tangent and rounding.

Answer. sin⁡θ=4/5\sin\theta=4/5 and cos⁡θ=3/5\cos\theta=3/5.

Check. (4/5)2+(3/5)2=1(4/5)^2+(3/5)^2=1, and both values are positive, consistent with the first-quadrant representative of this direction.

Change only the sign of tt. Its square stays the same, while the numerator of sine reverses sign.

Worked example 2: reflecting the direction

Question. Given t=tan⁡(θ/2)=−1/2t=\tan(\theta/2)=-1/2, find sin⁡θ\sin\theta and cos⁡θ\cos\theta exactly.

Plan. Keep the signed value in 2t2t, and square it in both other places.

1+t2=54unchanged squaresin⁡θ=2(−1/2)5/4=−45signed numeratorcos⁡θ=1−1/45/4=35unchanged cosine\begin{aligned} 1+t^2&=\frac54 && \text{unchanged square}\\ \sin\theta&=\frac{2(-1/2)}{5/4}=-\frac45 && \text{signed numerator}\\ \cos\theta&=\frac{1-1/4}{5/4}=\frac35 && \text{unchanged cosine} \end{aligned}

Why this step? Squaring tt removes its sign only in the squared terms. The sine numerator is linear in tt, so its sign remains.

Answer. sin⁡θ=−4/5\sin\theta=-4/5 and cos⁡θ=3/5\cos\theta=3/5.

Check. The direction reflects the previous point in the horizontal axis: cosine is unchanged and sine reverses.

A common denominator does not mean that every numerator has the same structure.

Watch out

Wrong: cos⁡θ=(1−t2)/(1−t2)\cos\theta=(1-t^2)/(1-t^2). The cosine denominator is 1+t21+t^2, which never vanishes for real tt. Keep the minus sign in its numerator alone.

Use a new rational parameter.

The tangent formula and its denominator

Dividing the sine t-formula by the cosine t-formula gives tangent, but its extra denominator zeros must be kept distinct from the half-angle substitution's own exclusions.

For cos⁡θ≠0\cos\theta\ne0,

tan⁡θ=sin⁡θcos⁡θ=2t/(1+t2)(1−t2)/(1+t2)=2t1−t2\tan\theta =\frac{\sin\theta}{\cos\theta} =\frac{2t/(1+t^2)}{(1-t^2)/(1+t^2)} =\frac{2t}{1-t^2}

The common factors 1+t21+t^2 cancel. The remaining denominator vanishes at t=±1t=\pm1, corresponding to directions where cos⁡θ=0\cos\theta=0.

The tangent t-formula

If t=tan⁡(θ/2)t=\tan(\theta/2) is defined and t≠±1t\ne\pm1, then

tan⁡θ=2t1−t2\tan\theta=\frac{2t}{1-t^2}

The conditions for tt and for tan⁡θ\tan\theta are separate. At t=±1t=\pm1, the sine and cosine t-formulae are still valid, but tan⁡θ\tan\theta is undefined.

Not in the formula booklet: learn it.

Do not reject t=±1t=\pm1 from a question involving only sine or cosine. For example, t=1t=1 gives sin⁡θ=1\sin\theta=1, cos⁡θ=0\cos\theta=0, and t=−1t=-1 gives sin⁡θ=−1\sin\theta=-1, cos⁡θ=0\cos\theta=0. These are valid circle points even though their tangent does not exist.

Worked example 3: a positive tangent

Question. Given t=tan⁡(θ/2)=1/2t=\tan(\theta/2)=1/2, find tan⁡θ\tan\theta.

Plan. Check the denominator is non-zero, then use the tangent formula.

1−t2=1−14=34≠0domain checktan⁡θ=2(1/2)3/4=43tangent formula\begin{aligned} 1-t^2&=1-\frac14=\frac34\ne0 && \text{domain check}\\ \tan\theta&=\frac{2(1/2)}{3/4}=\frac43 && \text{tangent formula} \end{aligned}

Why this step? A tangent value needs a non-zero cosine, reflected here in 1−t2≠01-t^2\ne0.

Answer. tan⁡θ=4/3\tan\theta=4/3.

Check. The earlier exact sine and cosine values give (4/5)/(3/5)=4/3(4/5)/(3/5)=4/3.

Continue this lesson.

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