TRIGONOMETRY · HL
Identities
Prove LHS = RHS.
Section 1 of 7
What is an Identity?
An equation is true for some values of the variable — usually you solve to find them.
$2x + 1 = 7$ true only when $x = 3$
An identity is true for every value of the variable. You don't solve it — you prove it: show that the left-hand side really does equal the right-hand side.
Two methods — Must learn
(i) Tables. Rewrite everything in terms of $\sin A$ and/or $\cos A$ using the tables.
(ii) Algebra. Simplify what you've got — cancel, expand, factor, use Pythagoras.
(ii) Algebra. Simplify what you've got — cancel, expand, factor, use Pythagoras.
Almost every identity proof in HL trig is "tables, then algebra."
Here's the method on the simplest possible identity.
Prove $\cos A \tan A = \sin A$
$\cos A \tan A$
Tables: $\tan A = \dfrac{\sin A}{\cos A}$
$= \cos A \cdot \dfrac{\sin A}{\cos A}$
$= \sin A$ QED — the $\cos A$'s cancel
YOU TRY · 1
Prove: $\sin A \cot A = \cos A$
Tables: $\cot A = \dfrac{\cos A}{\sin A}$. Then multiply.
$\sin A \cot A$
$= \sin A \cdot \dfrac{\cos A}{\sin A}$
$= \cos A$ QED
$\sin A \cdot \dfrac{\cos A}{\sin A} = \cos A$
YOU TRY · 2
Prove: $\dfrac{\tan A}{\sin A} = \sec A$
Rewrite $\tan A$ and $\sec A$ in terms of $\sin A$ and $\cos A$.
$\dfrac{\tan A}{\sin A} = \dfrac{\sin A / \cos A}{\sin A}$
$= \dfrac{\sin A}{\cos A} \cdot \dfrac{1}{\sin A}$
$= \dfrac{1}{\cos A}$
$= \sec A$ QED
$\dfrac{\sin A / \cos A}{\sin A} = \dfrac{1}{\cos A} = \sec A$
Section 2 of 7
The Tables
On a right-angled triangle with angle $A$, hypotenuse $H$, adjacent side $a$ and opposite side $o$:
$\sin A = \dfrac{o}{H} \quad\quad \cos A = \dfrac{a}{H} \quad\quad \tan A = \dfrac{o}{a}$
The reciprocals get their own names:
The four rewrites — Must learn
$\tan A = \dfrac{\sin A}{\cos A}$ $\sec A = \dfrac{1}{\cos A}$
$\cot A = \dfrac{\cos A}{\sin A}$ $\csc A = \dfrac{1}{\sin A}$
$\cot A = \dfrac{\cos A}{\sin A}$ $\csc A = \dfrac{1}{\sin A}$
These four rewrites turn every identity into a sin/cos problem. Method (i): tables.
$\csc A$ is also written $\operatorname{cosec} A$ — same thing.
A quick check that the reciprocals are what we say. From the triangle, $\cos A = \dfrac{a}{H}$, so
$\dfrac{1}{\cos A} = \dfrac{1}{\,a/H\,} = \dfrac{H}{a} = \sec A$ (the sides flip)
Rewriting practice
Express each in $\sin A$ and $\cos A$ only.
$\sec A \tan A = \dfrac{1}{\cos A} \cdot \dfrac{\sin A}{\cos A} = \dfrac{\sin A}{\cos^{2} A}$
$\csc A \cot A = \dfrac{1}{\sin A} \cdot \dfrac{\cos A}{\sin A} = \dfrac{\cos A}{\sin^{2} A}$
YOU TRY · 3
Rewrite $\sec A - \cos A$ in $\sin A$ and $\cos A$ only, and simplify.
Replace $\sec A$ with $\dfrac{1}{\cos A}$, then common denominator.
$\sec A - \cos A = \dfrac{1}{\cos A} - \cos A$
$= \dfrac{1}{\cos A} - \dfrac{\cos^{2} A}{\cos A}$
$= \dfrac{1 - \cos^{2} A}{\cos A}$
$\dfrac{1 - \cos^{2} A}{\cos A}$
YOU TRY · 4
Rewrite $\dfrac{\sec A}{\tan A}$ in $\sin A$ and $\cos A$ only, and simplify.
$\sec A = \dfrac{1}{\cos A}$ and $\tan A = \dfrac{\sin A}{\cos A}$. Then divide.
$\dfrac{\sec A}{\tan A} = \dfrac{\,1/\cos A\,}{\,\sin A / \cos A\,}$
$= \dfrac{1}{\cos A} \cdot \dfrac{\cos A}{\sin A}$
$= \dfrac{1}{\sin A} = \csc A$
$\csc A$
YOU TRY · 5
Rewrite $\cot A \cdot \tan A$ in $\sin A$ and $\cos A$ only, and simplify.
$\cot A$ and $\tan A$ are reciprocals.
$\cot A \cdot \tan A = \dfrac{\cos A}{\sin A} \cdot \dfrac{\sin A}{\cos A}$
$= 1$
$1$
YOU TRY · 6
Rewrite $\csc A \cdot \tan A$ in $\sin A$ and $\cos A$ only, and simplify.
$\csc A = \dfrac{1}{\sin A}$ and $\tan A = \dfrac{\sin A}{\cos A}$.
$\csc A \cdot \tan A = \dfrac{1}{\sin A} \cdot \dfrac{\sin A}{\cos A}$
$= \dfrac{1}{\cos A}$
$= \sec A$
$\sec A$
Section 3 of 7
The Pythagorean Identity
Take a unit circle — radius $1$, centred at the origin. A point on the circle at angle $A$ from the positive $x$-axis has coordinates $(\cos A, \, \sin A)$.
Pythagoras on that right triangle (legs $\cos A$ and $\sin A$, hypotenuse $1$) gives:
$x^{2} + y^{2} = 1$
$\cos^{2} A + \sin^{2} A = 1$
Pythagorean identity — Must learn
$\sin^{2} A + \cos^{2} A = 1$
Rearrange: $\sin^{2} A = 1 - \cos^{2} A$ and $\cos^{2} A = 1 - \sin^{2} A$
Huge warning. $\cos^{2} A = 1 - \sin^{2} A$ is true. But you cannot square-root each piece and write $\cos A$ ≠ $1 - \sin A$.
Same mistake with numbers makes it obvious:
$5^{2} = 3^{2} + 4^{2}$ true: $25 = 9 + 16$ ✓
$5 = 3 + 4$ false: $5$ ≠ $7$
Squaring and adding is not the same as adding. Same here: $\cos^{2} A = 1 - \sin^{2} A$ ✓, but $\cos A$ ≠ $1 - \sin A$.
Worked: $\sin A \tan A + \cos A = \sec A$
$\sin A \tan A + \cos A$
Tables: $\tan A = \dfrac{\sin A}{\cos A}$, $\sec A = \dfrac{1}{\cos A}$
$= \sin A \cdot \dfrac{\sin A}{\cos A} + \cos A$
$= \dfrac{\sin^{2} A}{\cos A} + \dfrac{\cos A}{1}$ common denominator $\cos A$
$= \dfrac{\sin^{2} A + \cos^{2} A}{\cos A}$
$= \dfrac{1}{\cos A}$ (Pythagoras on the top)
$= \sec A$ QED
YOU TRY · 7
Prove: $\dfrac{\sin \theta}{\sqrt{1 - \sin^{2} \theta}} = \tan \theta$
$1 - \sin^{2} \theta = \cos^{2} \theta$. So $\sqrt{1 - \sin^{2}\theta} = \sqrt{\cos^{2}\theta} = \cos\theta$.
$\dfrac{\sin \theta}{\sqrt{1 - \sin^{2} \theta}}$
$= \dfrac{\sin \theta}{\sqrt{\cos^{2} \theta}}$
$= \dfrac{\sin \theta}{\cos \theta}$
$= \tan \theta$ QED
$\dfrac{\sin\theta}{\sqrt{\cos^{2}\theta}} = \dfrac{\sin\theta}{\cos\theta} = \tan\theta$
YOU TRY · 8
Prove: $(\sin A + \cos A)^{2} + (\sin A - \cos A)^{2} = 2$
Expand both squares. The cross terms cancel. Then Pythagoras.
$(\sin A + \cos A)^{2} + (\sin A - \cos A)^{2}$
$= \sin^{2}A + 2\sin A \cos A + \cos^{2}A$
$\;\;\; + \sin^{2}A - 2\sin A \cos A + \cos^{2}A$
$= 2\sin^{2}A + 2\cos^{2}A$
$= 2(\sin^{2}A + \cos^{2}A)$
$= 2 \cdot 1 = 2$ QED
Cross terms cancel; $2(\sin^{2}A + \cos^{2}A) = 2$
YOU TRY · 9
Show: $\sin^{2} A \, (1 + \cot^{2} A) = 1$
Rewrite $\cot A$ in $\sin A$ and $\cos A$, then multiply out.
$\sin^{2}A \left( 1 + \dfrac{\cos^{2}A}{\sin^{2}A} \right)$
$= \sin^{2}A + \sin^{2}A \cdot \dfrac{\cos^{2}A}{\sin^{2}A}$
$= \sin^{2}A + \cos^{2}A$
$= 1$ QED
$\sin^{2}A + \cos^{2}A = 1$
Section 4 of 7
The Difference-of-Squares Move
A move you'll use again and again in identity proofs. From plain algebra:
Difference of two squares — Must learn
$(a+b)(a-b) = a^{2} - b^{2}$
Works both ways — expand $(a+b)(a-b)$ to $a^{2}-b^{2}$, OR factor $a^{2}-b^{2}$ into $(a+b)(a-b)$.
With trig, the two patterns you'll meet most often:
$(1 + \cos\theta)(1 - \cos\theta) = 1 - \cos^{2}\theta = \sin^{2}\theta$
$(\cos A + \sin A)(\cos A - \sin A) = \cos^{2} A - \sin^{2} A$
Worked: $\dfrac{(1 + \cos\theta)(1 - \cos\theta)}{\cos^{2}\theta} = \tan^{2}\theta$
$\dfrac{(1 + \cos\theta)(1 - \cos\theta)}{\cos^{2}\theta}$
$(a+b)(a-b) = a^{2} - b^{2}$ on the top
$= \dfrac{1 - \cos^{2}\theta}{\cos^{2}\theta}$
Pythagoras: $1 - \cos^{2}\theta = \sin^{2}\theta$
$= \dfrac{\sin^{2}\theta}{\cos^{2}\theta}$
$= \tan^{2}\theta$ QED
Worked: $\dfrac{1}{1 - \sin A} + \dfrac{1}{1 + \sin A} = 2\sec^{2} A$
Two fractions on the left — common denominator first.
$\dfrac{1 + \sin A + 1 - \sin A}{(1 - \sin A)(1 + \sin A)}$
Top: $\sin A$'s cancel. Bottom: difference of squares.
$= \dfrac{2}{1 - \sin^{2} A}$
$= \dfrac{2}{\cos^{2} A}$ (Pythagoras)
$= 2 \sec^{2} A$ QED
YOU TRY · 10
Prove: $(1 + \sin A)(1 - \sin A) = \cos^{2} A$
$(a+b)(a-b) = a^{2} - b^{2}$, then Pythagoras.
$(1 + \sin A)(1 - \sin A) = 1 - \sin^{2} A$
$= \cos^{2} A$ QED
$1 - \sin^{2} A = \cos^{2} A$
YOU TRY · 11
Prove: $\dfrac{1 - \cos^{2} A}{1 + \cos A} = 1 - \cos A$
Factor the top as a difference of squares. Then cancel.
$\dfrac{1 - \cos^{2} A}{1 + \cos A}$
$= \dfrac{(1 + \cos A)(1 - \cos A)}{1 + \cos A}$
$= 1 - \cos A$ QED
$(1+\cos A)$ cancels: $1 - \cos A$
YOU TRY · 12
Prove: $(\sin A + \cos A)^{2} = 1 + 2 \sin A \cos A$
Expand the square. Then use Pythagoras on $\sin^{2} A + \cos^{2} A$.
$(\sin A + \cos A)^{2}$
$= \sin^{2} A + 2 \sin A \cos A + \cos^{2} A$
$= (\sin^{2} A + \cos^{2} A) + 2 \sin A \cos A$
$= 1 + 2 \sin A \cos A$ QED
$\sin^{2}A + \cos^{2}A = 1$, leaving $1 + 2\sin A \cos A$
YOU TRY · 13
Prove: $\dfrac{1}{1 - \cos A} + \dfrac{1}{1 + \cos A} = 2 \csc^{2} A$
Same method as the worked $2\sec^{2} A$ example — common denominator, difference of squares, Pythagoras.
$\dfrac{1 + \cos A + 1 - \cos A}{(1 - \cos A)(1 + \cos A)}$
$= \dfrac{2}{1 - \cos^{2} A}$
$= \dfrac{2}{\sin^{2} A}$
$= 2 \csc^{2} A$ QED
$\dfrac{2}{\sin^{2}A} = 2\csc^{2}A$
Section 5 of 7
Mixed Proofs — Pulling It Together
No new tools — just combining tables, common denominators, Pythagoras and difference-of-squares. The harder proofs are always built from these four moves.
Worked: $\dfrac{\tan\theta + \sin\theta}{\sec\theta + 1} = \sin\theta$
Both top and bottom have something + something. First job: tables — get everything in $\sin\theta$ and $\cos\theta$.
$\dfrac{\,\dfrac{\sin\theta}{\cos\theta} + \dfrac{\sin\theta}{1}\,}{\,\dfrac{1}{\cos\theta} + \dfrac{1}{1}\,}$
Top: common denominator $\cos\theta$. Bottom: same.
$= \dfrac{\,\dfrac{\sin\theta + \sin\theta \cos\theta}{\cos\theta}\,}{\,\dfrac{1 + \cos\theta}{\cos\theta}\,}$
The two $\cos\theta$ denominators cancel.
$= \dfrac{\sin\theta + \sin\theta \cos\theta}{1 + \cos\theta}$
Top: factor out $\sin\theta$.
$= \dfrac{\sin\theta\,(1 + \cos\theta)}{1 + \cos\theta}$
$= \sin\theta$ QED
YOU TRY · 14
Prove: $\dfrac{\sin A}{1 + \cos A} + \dfrac{1 + \cos A}{\sin A} = 2 \csc A$
Common denominator $\sin A (1 + \cos A)$. Top expands to $\sin^{2}A + (1+\cos A)^{2}$.
$\dfrac{\sin^{2}A + (1 + \cos A)^{2}}{\sin A\,(1 + \cos A)}$
$= \dfrac{\sin^{2}A + 1 + 2\cos A + \cos^{2}A}{\sin A\,(1 + \cos A)}$
$= \dfrac{1 + 1 + 2\cos A}{\sin A\,(1 + \cos A)}$ (Pythagoras)
$= \dfrac{2 + 2\cos A}{\sin A\,(1 + \cos A)}$
$= \dfrac{2(1 + \cos A)}{\sin A\,(1 + \cos A)}$
$= \dfrac{2}{\sin A}$
$= 2 \csc A$ QED
$\dfrac{2(1+\cos A)}{\sin A(1+\cos A)} = \dfrac{2}{\sin A} = 2\csc A$
YOU TRY · 15
Prove: $\dfrac{\cos A}{1 - \sin A} = \sec A + \tan A$
Multiply top and bottom by $1 + \sin A$. The bottom becomes a difference of squares.
$\dfrac{\cos A}{1 - \sin A} \cdot \dfrac{1 + \sin A}{1 + \sin A}$
$= \dfrac{\cos A\,(1 + \sin A)}{1 - \sin^{2} A}$
$= \dfrac{\cos A\,(1 + \sin A)}{\cos^{2} A}$ (Pythagoras)
$= \dfrac{1 + \sin A}{\cos A}$
$= \dfrac{1}{\cos A} + \dfrac{\sin A}{\cos A}$
$= \sec A + \tan A$ QED
Multiply by $\dfrac{1+\sin A}{1+\sin A}$, then split the fraction.
YOU TRY · 16
Prove: $\dfrac{\cot A - \tan A}{\cot A + \tan A} = \cos^{2} A - \sin^{2} A$
Replace $\cot$ and $\tan$ with sin/cos. Common denominator on top and bottom is $\sin A \cos A$.
$\dfrac{\,\dfrac{\cos A}{\sin A} - \dfrac{\sin A}{\cos A}\,}{\,\dfrac{\cos A}{\sin A} + \dfrac{\sin A}{\cos A}\,}$
$= \dfrac{\,\dfrac{\cos^{2}A - \sin^{2}A}{\sin A \cos A}\,}{\,\dfrac{\cos^{2}A + \sin^{2}A}{\sin A \cos A}\,}$
$= \dfrac{\cos^{2}A - \sin^{2}A}{\cos^{2}A + \sin^{2}A}$ ($\sin A \cos A$ cancels)
$= \dfrac{\cos^{2}A - \sin^{2}A}{1}$ (Pythagoras)
$= \cos^{2}A - \sin^{2}A$ QED
Bottom collapses to $1$ by Pythagoras.
Section 6 of 7
Double Angles
Three formulas you need at your fingertips. All on the formula sheet — but the proofs assume you can use them either way.
Double-angle formulas — Must learn
$\sin 2A = 2 \sin A \cos A$
$\cos 2A = \cos^{2} A - \sin^{2} A$
$\quad\quad\;\;\; = 1 - 2 \sin^{2} A$
$\quad\quad\;\;\; = 2 \cos^{2} A - 1$
$\tan 2A = \dfrac{2 \tan A}{1 - \tan^{2} A}$
$\cos 2A = \cos^{2} A - \sin^{2} A$
$\quad\quad\;\;\; = 1 - 2 \sin^{2} A$
$\quad\quad\;\;\; = 2 \cos^{2} A - 1$
$\tan 2A = \dfrac{2 \tan A}{1 - \tan^{2} A}$
$\cos 2A$ has three forms — pick whichever matches what's around it.
Worked: $\sin 2A = \dfrac{2 \tan A}{1 + \tan^{2} A}$
Start from the LHS and head for the RHS. RHS has $\tan A$, so tables.
$\sin 2A = 2 \sin A \cos A$
Aim: get $\tan A = \dfrac{\sin A}{\cos A}$ on top. Divide top and bottom by $\cos^{2} A$ (going the other way: from $2\tan A / (1 + \tan^{2}A)$ back to LHS).
$\dfrac{2 \tan A}{1 + \tan^{2} A} = \dfrac{\,2 \cdot \dfrac{\sin A}{\cos A}\,}{\,1 + \dfrac{\sin^{2} A}{\cos^{2} A}\,}$
$= \dfrac{\,\dfrac{2 \sin A}{\cos A}\,}{\,\dfrac{\cos^{2} A + \sin^{2} A}{\cos^{2} A}\,}$ (common denominator)
$= \dfrac{2 \sin A}{\cos A} \cdot \dfrac{\cos^{2} A}{\cos^{2} A + \sin^{2} A}$
$= \dfrac{2 \sin A \cos A}{1}$ (Pythagoras: bottom $=1$; one $\cos A$ cancels)
$= 2 \sin A \cos A = \sin 2A$ QED
Worked: $1 - (\cos x - \sin x)^{2} = \sin 2x$
$1 - (\cos x - \sin x)^{2}$
Expand the square first.
$= 1 - (\cos^{2} x - 2 \sin x \cos x + \sin^{2} x)$
$= 1 - (\sin^{2} x + \cos^{2} x) + 2 \sin x \cos x$
$= 1 - 1 + 2 \sin x \cos x$ (Pythagoras)
$= 2 \sin x \cos x = \sin 2x$ QED
Worked: $\dfrac{\cos 2A}{\cos A + \sin A} = \cos A - \sin A$
$\cos 2A$ has three forms. The bottom is $\cos A + \sin A$, so the matching form is $\cos^{2} A - \sin^{2} A$ — a difference of squares.
$\dfrac{\cos 2A}{\cos A + \sin A} = \dfrac{\cos^{2} A - \sin^{2} A}{\cos A + \sin A}$
Difference of squares on top: $p^{2} - q^{2} = (p+q)(p-q)$.
$= \dfrac{(\cos A + \sin A)(\cos A - \sin A)}{\cos A + \sin A}$
$= \cos A - \sin A$ QED
Worked (two-parter): $\dfrac{\sin 2A}{1 + \cos 2A} = \tan A$, hence show $\tan 22\tfrac{1}{2}^{\circ} = \sqrt{2} - 1$
Part 1 — Prove the identity. Top: $\sin 2A = 2\sin A \cos A$. Bottom: pick the form of $\cos 2A$ that makes $1 + \cos 2A$ collapse. Try $\cos 2A = 2\cos^{2}A - 1$.
$\dfrac{\sin 2A}{1 + \cos 2A} = \dfrac{2 \sin A \cos A}{1 + (2\cos^{2} A - 1)}$
$= \dfrac{2 \sin A \cos A}{2 \cos^{2} A}$ (the $1$'s cancel)
$= \dfrac{\sin A}{\cos A}$ (cancel $2 \cos A$)
$= \tan A$ QED
Part 2 — Hence $\tan 22\tfrac{1}{2}^{\circ}$. Let $A = 22\tfrac{1}{2}^{\circ}$, so $2A = 45^{\circ}$. Substitute into the identity:
$\tan 22\tfrac{1}{2}^{\circ} = \dfrac{\sin 45^{\circ}}{1 + \cos 45^{\circ}}$
From the tables: $\sin 45^{\circ} = \cos 45^{\circ} = \dfrac{1}{\sqrt{2}}$.
$= \dfrac{\,\dfrac{1}{\sqrt{2}}\,}{\,1 + \dfrac{1}{\sqrt{2}}\,}$
$= \dfrac{\,\dfrac{1}{\sqrt{2}}\,}{\,\dfrac{\sqrt{2} + 1}{\sqrt{2}}\,}$ (common denominator on bottom)
$= \dfrac{1}{\sqrt{2}} \cdot \dfrac{\sqrt{2}}{\sqrt{2} + 1} = \dfrac{1}{\sqrt{2} + 1}$
Rationalise: multiply top and bottom by $\sqrt{2} - 1$ (the conjugate).
$= \dfrac{1}{\sqrt{2} + 1} \cdot \dfrac{\sqrt{2} - 1}{\sqrt{2} - 1}$
$= \dfrac{\sqrt{2} - 1}{2 - 1} = \dfrac{\sqrt{2} - 1}{1}$
$= \sqrt{2} - 1$ QED
YOU TRY · 17
Prove: $\cos 2A = \dfrac{1 - \tan^{2} A}{1 + \tan^{2} A}$
Start from RHS. Tables: $\tan A = \dfrac{\sin A}{\cos A}$. Common denominator on top and bottom is $\cos^{2}A$.
$\dfrac{1 - \tan^{2} A}{1 + \tan^{2} A} = \dfrac{\,1 - \dfrac{\sin^{2}A}{\cos^{2}A}\,}{\,1 + \dfrac{\sin^{2}A}{\cos^{2}A}\,}$
$= \dfrac{\,\dfrac{\cos^{2}A - \sin^{2}A}{\cos^{2}A}\,}{\,\dfrac{\cos^{2}A + \sin^{2}A}{\cos^{2}A}\,}$
$= \dfrac{\cos^{2}A - \sin^{2}A}{\cos^{2}A + \sin^{2}A}$ ($\cos^{2}A$ cancels)
$= \dfrac{\cos^{2}A - \sin^{2}A}{1}$ (Pythagoras)
$= \cos 2A$ QED
$\cos^{2}A - \sin^{2}A = \cos 2A$
YOU TRY · 18
Prove: $\dfrac{1 - \cos 2A}{\sin 2A} = \tan A$
Bottom: $\sin 2A = 2 \sin A \cos A$. Top: choose the form of $\cos 2A$ that makes $1 - \cos 2A$ collapse — try $\cos 2A = 1 - 2\sin^{2}A$.
$\dfrac{1 - \cos 2A}{\sin 2A} = \dfrac{1 - (1 - 2\sin^{2}A)}{2\sin A \cos A}$
$= \dfrac{2 \sin^{2} A}{2 \sin A \cos A}$
$= \dfrac{\sin A}{\cos A}$
$= \tan A$ QED
$\dfrac{2\sin^{2}A}{2\sin A \cos A} = \tan A$
YOU TRY · 19
Show that $1 + \cos 2A = 2 \cos^{2} A$, and hence find $\cos 22\tfrac{1}{2}^{\circ}$ in surd form.
Use $\cos 2A = 2\cos^{2}A - 1$ on the LHS. Then for part 2, let $A = 22\tfrac{1}{2}^{\circ}$, so $2A = 45^{\circ}$, and solve for $\cos A$.
Part 1. $1 + \cos 2A = 1 + (2\cos^{2}A - 1)$
$\quad\quad\quad\quad\;\;\;\;\,= 2\cos^{2}A$ QED
Part 2. Let $A = 22\tfrac{1}{2}^{\circ}$, $2A = 45^{\circ}$:
$1 + \cos 45^{\circ} = 2 \cos^{2} 22\tfrac{1}{2}^{\circ}$
$1 + \dfrac{1}{\sqrt{2}} = 2 \cos^{2} 22\tfrac{1}{2}^{\circ}$
$\dfrac{\sqrt{2} + 1}{\sqrt{2}} = 2 \cos^{2} 22\tfrac{1}{2}^{\circ}$
$\cos^{2} 22\tfrac{1}{2}^{\circ} = \dfrac{\sqrt{2} + 1}{2\sqrt{2}} = \dfrac{2 + \sqrt{2}}{4}$
$\cos 22\tfrac{1}{2}^{\circ} = \dfrac{\sqrt{2 + \sqrt{2}}}{2}$
$\cos 22\tfrac{1}{2}^{\circ} = \dfrac{\sqrt{2 + \sqrt{2}}}{2}$
Section 7 of 7
Small Letters — Cosine Rule Identities
Up to now the identities used capital letters $A$, $B$, $C$ — just angle variables. Now lowercase $a$, $b$, $c$ enter the picture: these are the sides of a triangle opposite the angles $A$, $B$, $C$.
When you see a mix — $a$, $b$, $c$ and $\cos A$, $\cos B$, $\cos C$ — the tool is the cosine rule, on the formula sheet. Write out all three forms, then rearrange.
Cosine rule — Must learn (write all three)
$a^{2} = b^{2} + c^{2} - 2bc \cos A$
$b^{2} = a^{2} + c^{2} - 2ac \cos B$
$c^{2} = a^{2} + b^{2} - 2ab \cos C$
$b^{2} = a^{2} + c^{2} - 2ac \cos B$
$c^{2} = a^{2} + b^{2} - 2ab \cos C$
Write the three formulae first, every time. Then rearrange each to get $\cos$ on its own, and add or subtract.
Worked: prove $bc \cos A + ac \cos B = c^{2}$
Goal: get $bc\cos A$ on its own from the first cosine rule, and $ac\cos B$ from the second. Then add.
$a^{2} = b^{2} + c^{2} - 2bc \cos A$
$2bc \cos A = b^{2} + c^{2} - a^{2}$
$bc \cos A = \dfrac{b^{2} + c^{2} - a^{2}}{2}$
$b^{2} = a^{2} + c^{2} - 2ac \cos B$
$2ac \cos B = a^{2} + c^{2} - b^{2}$
$ac \cos B = \dfrac{a^{2} + c^{2} - b^{2}}{2}$
Now add the two expressions.
$bc \cos A + ac \cos B = \dfrac{b^{2} + c^{2} - a^{2}}{2} + \dfrac{a^{2} + c^{2} - b^{2}}{2}$
$= \dfrac{b^{2} + c^{2} - a^{2} + a^{2} + c^{2} - b^{2}}{2}$ ($a^{2}$'s cancel, $b^{2}$'s cancel)
$= \dfrac{2c^{2}}{2}$
$= c^{2}$ QED
Worked: prove $b \cos C + c \cos B = a$
Same method. This time we need $b\cos C$ and $c\cos B$ on their own.
$c^{2} = a^{2} + b^{2} - 2ab \cos C$
$2ab \cos C = a^{2} + b^{2} - c^{2}$
$b \cos C = \dfrac{a^{2} + b^{2} - c^{2}}{2a}$ (÷ by $2a$, not $2$)
$b^{2} = a^{2} + c^{2} - 2ac \cos B$
$2ac \cos B = a^{2} + c^{2} - b^{2}$
$c \cos B = \dfrac{a^{2} + c^{2} - b^{2}}{2a}$
$b \cos C + c \cos B = \dfrac{a^{2} + b^{2} - c^{2}}{2a} + \dfrac{a^{2} + c^{2} - b^{2}}{2a}$
$= \dfrac{a^{2} + b^{2} - c^{2} + a^{2} + c^{2} - b^{2}}{2a}$
$= \dfrac{2a^{2}}{2a}$
$= a$ QED
The pattern. Every cosine-rule identity is built the same way: write all three formulae, get the right $\cos$'s on their own, divide by whatever divisor makes the LHS appear, then add or subtract. Letters that aren't in the answer must cancel.
YOU TRY · 20
Prove: $ab \cos C + ac \cos B = a^{2}$
From the third cosine rule, get $ab\cos C$ on its own (divide by $2$). From the second, get $ac\cos B$. Then add.
$c^{2} = a^{2} + b^{2} - 2ab\cos C$
$ab\cos C = \dfrac{a^{2} + b^{2} - c^{2}}{2}$
$b^{2} = a^{2} + c^{2} - 2ac\cos B$
$ac\cos B = \dfrac{a^{2} + c^{2} - b^{2}}{2}$
Adding:
$= \dfrac{a^{2} + b^{2} - c^{2} + a^{2} + c^{2} - b^{2}}{2}$
$= \dfrac{2a^{2}}{2}$
$= a^{2}$ QED
$b^{2}$'s and $c^{2}$'s cancel; $\dfrac{2a^{2}}{2} = a^{2}$
YOU TRY · 21
Prove: $a \cos B + b \cos A = c$
From cosine rule 1, get $b\cos A$ on its own — divide by $2c$, not $2$. From rule 2, $a\cos B$, divide by $2c$.
$a^{2} = b^{2} + c^{2} - 2bc\cos A$
$b\cos A = \dfrac{b^{2} + c^{2} - a^{2}}{2c}$
$b^{2} = a^{2} + c^{2} - 2ac\cos B$
$a\cos B = \dfrac{a^{2} + c^{2} - b^{2}}{2c}$
Adding:
$= \dfrac{b^{2} + c^{2} - a^{2} + a^{2} + c^{2} - b^{2}}{2c}$
$= \dfrac{2c^{2}}{2c}$
$= c$ QED
$\dfrac{2c^{2}}{2c} = c$
YOU TRY · 22
Prove: $bc \cos A - ab \cos C = c^{2} - a^{2}$ (this one's a subtraction — careful with signs)
Get $bc\cos A$ from rule 1 (÷ $2$) and $ab\cos C$ from rule 3 (÷ $2$). Then subtract — remember to flip every sign in the second bracket.
$bc\cos A = \dfrac{b^{2} + c^{2} - a^{2}}{2}$
$ab\cos C = \dfrac{a^{2} + b^{2} - c^{2}}{2}$
$bc\cos A - ab\cos C = \dfrac{(b^{2} + c^{2} - a^{2}) - (a^{2} + b^{2} - c^{2})}{2}$
$= \dfrac{b^{2} + c^{2} - a^{2} - a^{2} - b^{2} + c^{2}}{2}$
$= \dfrac{2c^{2} - 2a^{2}}{2}$
$= c^{2} - a^{2}$ QED
$b^{2}$'s cancel; $\dfrac{2c^{2} - 2a^{2}}{2} = c^{2} - a^{2}$
SUM
Exam Strategy
Identity proofs scare students more than they should. They look hard, but the menu is small. Here's the checklist for any proof:
The proof checklist — Must learn
1.Tables — rewrite $\tan$, $\sec$, $\cot$, $\csc$ as $\sin$/$\cos$.
2.Common denominator — if you see fractions on top of each other.
3.Pythagoras — spot $\sin^{2} + \cos^{2}$ becoming $1$, or $1 - \sin^{2}$ becoming $\cos^{2}$.
4.Difference of squares — $(a+b)(a-b) = a^{2} - b^{2}$. Either direction.
5.Factor — pull out a common factor like $\sin\theta$.
6.For double angles — pick the form of $\cos 2A$ that matches what's nearby. $1 + \cos 2A \to 2\cos^{2}A$, $1 - \cos 2A \to 2\sin^{2}A$.
7.Small letters — write out all three cosine rule formulae, then rearrange.
The golden rule. Always work on the more complicated side, simplifying down to the simpler side. Don't try to make the simple side complicated.
Setting it out. Write LHS at the top of your page. Don't write LHS = RHS until you've actually finished — that's begging the question. Work down one column, with the next step under the previous one. Finish with QED.
The classic trap. $\cos^{2} A = 1 - \sin^{2} A$ does NOT mean $\cos A = 1 - \sin A$. Squaring then adding is not the same as adding. Same as $5^{2} = 3^{2} + 4^{2}$ being true while $5$ ≠ $3 + 4$. Never square-root piece-by-piece.
If you get stuck. Try the other side. Sometimes RHS is simpler to manipulate. Or meet in the middle — work both sides until they reach the same expression, then write the proof out top-down.
End of lesson
Identities — HL · Mathslive.ie