Projection and Half-Angle Formulas

Introduction

In addition to the Sine and Cosine rules, several other useful formulas connect the sides and angles of a triangle. These include the Projection formulas, which express a side in terms of others, and the Half-Angle formulas, which are powerful for finding angles when all three sides are known. 📐

1. Projection Formulas

The projection formula expresses the length of a side of a triangle in terms of the other two sides and the cosines of the angles they make with the first side.

• $$a = b\cos C + c\cos B$$

• $$b = a\cos C + c\cos A$$

• $$c = a\cos B + b\cos A$$

2. Half-Angle Formulas

These formulas are used to find the sine, cosine, or tangent of half an angle of a triangle when the lengths of the three sides are known. Let 's' be the semi-perimeter of the triangle, where $s = \frac{a+b+c}{2}$.

• Sine of Half-Angles:

  • $$\sin(A/2) = \sqrt{\frac{(s-b)(s-c)}{bc}}$$

  • $$\sin(B/2) = \sqrt{\frac{(s-a)(s-c)}{ac}}$$

  • $$\sin(C/2) = \sqrt{\frac{(s-a)(s-b)}{ab}}$$

• Cosine of Half-Angles:

  • $$\cos(A/2) = \sqrt{\frac{s(s-a)}{bc}}$$

  • $$\cos(B/2) = \sqrt{\frac{s(s-b)}{ac}}$$

  • $$\cos(C/2) = \sqrt{\frac{s(s-c)}{ab}}$$

• Tangent of Half-Angles:

  • $$\tan(A/2) = \sqrt{\frac{(s-b)(s-c)}{s(s-a)}}$$

  • $$\tan(B/2) = \sqrt{\frac{(s-a)(s-c)}{s(s-b)}}$$

  • $$\tan(C/2) = \sqrt{\frac{(s-a)(s-b)}{s(s-c)}}$$

3. Napier's Analogy (Tangent Rule)

This set of formulas relates the sum and difference of two sides to the tangent of half the sum and difference of their opposite angles.

$$\tan\left(\frac{B-C}{2}\right) = \frac{b-c}{b+c}\cot(A/2)$$

Similar formulas exist for $\tan((C-A)/2)$ and $\tan((A-B)/2)$.

Example 1: Proving the Projection Formula

Question: In any triangle ABC, prove that $a = b\cos C + c\cos B$.

Solution:

Using the Cosine Rule: $\cos C = \frac{a^2+b^2-c^2}{2ab}$ and $\cos B = \frac{a^2+c^2-b^2}{2ac}$.

RHS = $b\left(\frac{a^2+b^2-c^2}{2ab}\right) + c\left(\frac{a^2+c^2-b^2}{2ac}\right) = \frac{a^2+b^2-c^2}{2a} + \frac{a^2+c^2-b^2}{2a} = \frac{2a^2}{2a} = a$ = LHS. ✨

Example 2: Using Half-Angle Sine Formula

Question: In a triangle ABC, if $a=13, b=14, c=15$, find the value of $\sin(A/2)$.

Solution:

First, find the semi-perimeter: $s = \frac{13+14+15}{2} = \frac{42}{2}=21$.

$s-b = 21-14=7$. $s-c = 21-15=6$. $bc=14 \times 15=210$.

$$\sin(A/2) = \sqrt{\frac{(s-b)(s-c)}{bc}} = \sqrt{\frac{7 \times 6}{14 \times 15}} = \sqrt{\frac{42}{210}} = \sqrt{\frac{1}{5}} = \frac{1}{\sqrt{5}}$$

Example 3: Using Half-Angle Cosine Formula

Question: For the same triangle with sides 13, 14, 15, find $\cos(A/2)$.

Solution:

We have $s=21$. $s-a = 21-13=8$. $bc=210$.

$$\cos(A/2) = \sqrt{\frac{s(s-a)}{bc}} = \sqrt{\frac{21 \times 8}{14 \times 15}} = \sqrt{\frac{168}{210}} = \sqrt{\frac{4}{5}} = \frac{2}{\sqrt{5}}$$

Example 4: Using Half-Angle Tangent Formula

Question: For the same triangle, find $\tan(A/2)$.

Solution:

We can use the previous results: $\tan(A/2) = \frac{\sin(A/2)}{\cos(A/2)} = \frac{1/\sqrt{5}}{2/\sqrt{5}} = 1/2$.

Alternatively, using the formula: $\tan(A/2) = \sqrt{\frac{(s-b)(s-c)}{s(s-a)}} = \sqrt{\frac{7 \times 6}{21 \times 8}} = \sqrt{\frac{42}{168}} = \sqrt{\frac{1}{4}} = 1/2$.

Example 5: Proving an Identity

Question: In $\triangle ABC$, prove that $(b+c)\cos A + (c+a)\cos B + (a+b)\cos C = a+b+c$.

Solution:

Expand the LHS: $b\cos A + c\cos A + c\cos B + a\cos B + a\cos C + b\cos C$.

Group the terms according to the Projection Formula:

$(b\cos C + c\cos B) + (a\cos C + c\cos A) + (a\cos B + b\cos A)$.

This is equal to $a + b + c$ = RHS.

Example 6: Using Napier's Analogy

Question: In a $\triangle ABC$, if $a=13, b=7$ and $C=60^\circ$, find $\frac{A-B}{2}$.

Solution:

We use Napier's Analogy to find A-B:

$$\tan\left(\frac{A-B}{2}\right) = \frac{a-b}{a+b}\cot(C/2) = \frac{13-7}{13+7}\cot(30^\circ) = \frac{6}{20}\sqrt{3} = \frac{3\sqrt{3}}{10}$$

So, $\frac{A-B}{2} = \arctan(\frac{3\sqrt{3}}{10})$. Combined with $A+B=180^\circ-60^\circ=120^\circ$, we could solve for A and B.

Example 7: Relating Sides and Half-Angles

Question: In a $\triangle ABC$, prove that $b\cos^2(C/2)+c\cos^2(B/2)=s$.

Solution:

LHS = $b\left(\frac{s(s-c)}{ab}\right) + c\left(\frac{s(s-b)}{ac}\right) = \frac{s(s-c)}{a} + \frac{s(s-b)}{a}$.

$= \frac{s}{a}(s-c + s-b) = \frac{s}{a}(2s - b - c)$.

Since $2s=a+b+c$, then $2s-b-c = a$.

So, the expression becomes $\frac{s}{a}(a)=s$ = RHS.

Example 8: Application of Projection Formula

Question: If in a triangle ABC, $a=4, b=5, c=6$, find the length of the projection of side b onto side c.

Solution:

The projection of side b (AC) onto side c (AB) is given by the length $b\cos A$.

We first find $\cos A$ using the Cosine Rule:

$$\cos A = \frac{b^2+c^2-a^2}{2bc} = \frac{5^2+6^2-4^2}{2(5)(6)} = \frac{25+36-16}{60} = \frac{45}{60} = \frac{3}{4}$$

Therefore, the length of the projection is $b\cos A = 5 \times \frac{3}{4} = \frac{15}{4}$.

Example 9: Identity with Semi-perimeter

Question: In $\triangle ABC$, prove that $(b-c)\cot(A/2) + (c-a)\cot(B/2) + (a-b)\cot(C/2) = 0$.

Solution:

We know that $\cot(A/2) = \sqrt{\frac{s(s-a)}{(s-b)(s-c)}} = \frac{s(s-a)}{\Delta}$, where $\Delta$ is the area of the triangle.

Substituting this for all terms, the LHS becomes:

$$\frac{s}{\Delta} [ (b-c)(s-a) + (c-a)(s-b) + (a-b)(s-c) ]$$

Expanding the terms in the bracket: $[ (bs-ab-cs+ac) + (cs-bc-as+ab) + (as-ac-bs+bc) ]$. All terms cancel each other out, resulting in $0$. Thus, the expression is $0$.

Example 10: Solving a Triangle with Napier's Analogy

Question: In a triangle ABC, if $b=\sqrt{3}, c=1$ and $A=30^\circ$, find angle B.

Solution:

We have the sum of angles $B+C = 180^\circ-30^\circ=150^\circ$.

Using Napier's Analogy: $$\tan\left(\frac{B-C}{2}\right) = \frac{b-c}{b+c}\cot(A/2) = \frac{\sqrt{3}-1}{\sqrt{3}+1}\cot(15^\circ)$$

We know $\frac{\sqrt{3}-1}{\sqrt{3}+1} = 2-\sqrt{3}$ and $\cot(15^\circ)=2+\sqrt{3}$.

So, $$\tan\left(\frac{B-C}{2}\right) = (2-\sqrt{3})(2+\sqrt{3}) = 4-3=1$$

This means $\frac{B-C}{2}=45^\circ$, so $B-C=90^\circ$.

We solve the system of equations: $B+C=150^\circ$ and $B-C=90^\circ$. Adding them gives $2B=240^\circ \implies B=120^\circ$.