Trigonometry · Paper 2
Double & Compound Angles
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Section 1 of 11
Doubling doesn't work
Be careful — doubling the angle does not double the sine, cosine, or tan. Check it on the calculator:
$\cos 60° = 0.5$
$\cos 30° = 0.866$
Since $2 \times 30° = 60°$, does $\;2\cos 30° = \cos 60°\;$?
$2 \times 0.866 = 1.732$
$2\cos 30°$ ≠ $\cos 60°$
So we need proper formulas. They're in your tables.
You try
Does $\;2\sin 30° = \sin 60°$? Use $\sin 30° = 0.5$ and $\sin 60° \approx 0.866$ to check.
Work out $2\sin 30°$ and compare with $\sin 60°$.
$2\sin 30° = 2 \times 0.5 = 1$
$\sin 60° \approx 0.866$
$2\sin 30°$ ≠ $\sin 60°$ — doubling fails again.
No — $1$ ≠ $0.866$.
Section 2 of 11
The $\sin 2A$ formulas
Two formulas for $\sin 2A$ from the tables — you'll use both:
Must learn — from tables
$\sin 2A = 2\sin A \cos A$
$\sin 2A = \dfrac{2\tan A}{1 + \tan^{2}\!A}$
Note: $\tan^{2}\!A$ means $(\tan A)^{2}$ — square the value of $\tan A$.
Section 3 of 11
Type 1: $\sin A$ given, find $\sin 2A$
Find $\sin 2A$ given $\sin A = \dfrac{3}{5}$.
$\sin 2A = 2\sin A \cos A$
We have $\sin A$, but we need $\cos A$ too. Draw the triangle:
$\sin A = \dfrac{3}{5}$ (opp $= 3$, hyp $= 5$, so adj $= 4$ by Pythagoras)
$\cos A = \dfrac{4}{5}$
$\sin 2A = 2\!\left(\dfrac{3}{5}\right)\!\!\left(\dfrac{4}{5}\right)$
$\sin 2A = \dfrac{24}{25}$
You try
Find $\sin 2A$ given $\cos A = \dfrac{5}{13}$.
Use $\sin 2A = 2\sin A \cos A$. You need $\sin A$ — get it from the triangle.
$\cos A = \dfrac{5}{13}$ (adj $= 5$, hyp $= 13$, so opp $= 12$)
$\sin A = \dfrac{12}{13}$
$\sin 2A = 2\!\left(\dfrac{12}{13}\right)\!\!\left(\dfrac{5}{13}\right)$
$\sin 2A = \dfrac{120}{169}$
$\dfrac{120}{169}$
You try
Find $\sin 2A$ given $\sin A = \dfrac{8}{17}$.
$\sin A = \dfrac{8}{17}$ → opp $= 8$, hyp $= 17$. Find adj.
$\sin A = \dfrac{8}{17}$ (opp $= 8$, hyp $= 17$, so adj $= 15$)
$\cos A = \dfrac{15}{17}$
$\sin 2A = 2\!\left(\dfrac{8}{17}\right)\!\!\left(\dfrac{15}{17}\right)$
$\sin 2A = \dfrac{240}{289}$
$\dfrac{240}{289}$
You try
Find $\sin 2A$ given $\tan A = \dfrac{3}{4}$. (Use the $\tan A$ formula.)
$\sin 2A = \dfrac{2\tan A}{1+\tan^{2}\!A}$ — plug in $\tan A = \dfrac{3}{4}$.
$\tan^{2}\!A = \left(\dfrac{3}{4}\right)^{2} = \dfrac{9}{16}$
$\sin 2A = \dfrac{2 \cdot \dfrac{3}{4}}{1 + \dfrac{9}{16}} = \dfrac{\dfrac{3}{2}}{\dfrac{25}{16}}$
$= \dfrac{3}{2} \times \dfrac{16}{25}$
$\sin 2A = \dfrac{24}{25}$ (same answer as the worked example — makes sense, $\tan A = \dfrac{3}{4}$ matches the 3-4-5 triangle)
$\dfrac{24}{25}$
Section 4 of 11
Type 2: $\sin 2A$ given, find $\sin A$
Given $\sin 2A = \dfrac{3}{5}$, find $\sin A$.
First, two attempts that do not work:
$\dfrac{\sin 2A}{2} = \dfrac{3/5}{2} \;\;\Rightarrow\;\; \sin A = \dfrac{3}{10}$ — rubbish, you can't just divide both sides by $2$.
$2\sin A \cos A = \dfrac{3}{5}$ — no good, two unknowns.
Use the other $\sin 2A$ formula — the one with $\tan A$ — because that has only one unknown:
$\sin 2A = \dfrac{2\tan A}{1 + \tan^{2}\!A} = \dfrac{3}{5}$
Let $t = \tan A$ to clean it up:
$\dfrac{2t}{1 + t^{2}} = \dfrac{3}{5}$
$10t = 3(1 + t^{2})$
$10t = 3 + 3t^{2}$
$3t^{2} - 10t + 3 = 0$
Factor (split the middle: $-10t = -9t - t$):
$3t^{2} - 9t - t + 3 = 0$
$3t(t-3) - 1(t-3) = 0$
$(t-3)(3t-1) = 0$
$t = 3 \quad\;\;\text{or}\quad\;\; t = \dfrac{1}{3}$
So there are two answers. Convert each $\tan A$ back to $\sin A$ via the triangle:
Case 1: $\tan A = 3 = \dfrac{3}{1}$
opp $= 3$, adj $= 1$, hyp $= \sqrt{3^{2}+1^{2}} = \sqrt{10}$
$\sin A = \dfrac{3}{\sqrt{10}}$
Case 2: $\tan A = \dfrac{1}{3}$
opp $= 1$, adj $= 3$, hyp $= \sqrt{1^{2}+3^{2}} = \sqrt{10}$
$\sin A = \dfrac{1}{\sqrt{10}}$
You try
Given $\sin 2A = \dfrac{24}{25}$, find both possible values of $\sin A$.
Set $\dfrac{2t}{1+t^{2}} = \dfrac{24}{25}$, let $t = \tan A$, solve the quadratic.
$\dfrac{2t}{1+t^{2}} = \dfrac{24}{25}$
$50t = 24(1+t^{2})$
$24t^{2} - 50t + 24 = 0$
$12t^{2} - 25t + 12 = 0$
$(3t-4)(4t-3) = 0$
$t = \dfrac{4}{3} \;\;\text{or}\;\; t = \dfrac{3}{4}$
Case 1: $\tan A = \dfrac{4}{3}$ (3-4-5 triangle, opp $=4$, hyp $=5$)
$\sin A = \dfrac{4}{5}$
Case 2: $\tan A = \dfrac{3}{4}$ (3-4-5 triangle, opp $=3$, hyp $=5$)
$\sin A = \dfrac{3}{5}$
$\sin A = \dfrac{4}{5}\;$ or $\;\dfrac{3}{5}$
Section 5 of 11
Compound angles — same trap
Same warning for $\cos(A+B)$: it does not split into $\cos A + \cos B$.
$\cos(A+B)$ ≠ $\cos A + \cos B$
Check with $A = 30°$, $B = 60°$:
$\cos(30°+60°) = \cos 90° = 0$
$\cos 30° + \cos 60° = 0.866 + 0.5 = 1.366$
$0$ ≠ $1.366$
So again — use the tables. These four are the compound angle formulas:
Compound angle formulas (from tables)
$\sin(A \pm B) = \sin A \cos B \pm \cos A \sin B$
$\cos(A \pm B) = \cos A \cos B \mp \sin A \sin B$
$\tan(A \pm B) = \dfrac{\tan A \pm \tan B}{1 \mp \tan A \tan B}$
Watch the signs: in $\cos$ and $\tan$, the second sign flips.
You try
Show that $\sin(A+B)$ ≠ $\sin A + \sin B$ by checking with $A = B = 30°$.
Compute $\sin 60°$ and $\sin 30° + \sin 30°$ separately.
$\sin(30°+30°) = \sin 60° \approx 0.866$
$\sin 30° + \sin 30° = 0.5 + 0.5 = 1$
$0.866$ ≠ $1$ — the formula doesn't distribute.
$0.866$ ≠ $1$, so $\sin(A+B)$ ≠ $\sin A + \sin B$.
Section 6 of 11
Reference triangles for $30°, 45°, 60°$
Without a calculator you need exact values for $\sin$, $\cos$, $\tan$ of $30°, 45°, 60°$. Two triangles give them all.
The $45°$ triangle — isoceles
Half a unit square. Legs $= 1$, hypotenuse $= \sqrt{1^{2}+1^{2}} = \sqrt{2}$:
$\sin 45° = \dfrac{1}{\sqrt{2}}, \quad \cos 45° = \dfrac{1}{\sqrt{2}}, \quad \tan 45° = 1$
The $30°/60°$ triangle — equilateral split
Take an equilateral triangle with side $2$. Drop a perpendicular — it splits into two right-angled triangles with hyp $= 2$ and base $= 1$. By Pythagoras the height $x$ is:
$x^{2} + 1^{2} = 2^{2}$
$x^{2} = 3$
$x = \sqrt{3}$
Reading off the $60°$ angle (opp $= \sqrt{3}$, adj $= 1$, hyp $= 2$):
$\sin 60° = \dfrac{\sqrt{3}}{2}, \quad \cos 60° = \dfrac{1}{2}, \quad \tan 60° = \sqrt{3}$
Reading off the $30°$ angle (opp $= 1$, adj $= \sqrt{3}$, hyp $= 2$):
$\sin 30° = \dfrac{1}{2}, \quad \cos 30° = \dfrac{\sqrt{3}}{2}, \quad \tan 30° = \dfrac{1}{\sqrt{3}}$
Must learn — the six exact values
$\sin 30° = \dfrac{1}{2} \quad \cos 30° = \dfrac{\sqrt{3}}{2} \quad \tan 30° = \dfrac{1}{\sqrt{3}}$
$\sin 45° = \dfrac{1}{\sqrt{2}} \quad \cos 45° = \dfrac{1}{\sqrt{2}} \quad \tan 45° = 1$
$\sin 60° = \dfrac{\sqrt{3}}{2} \quad \cos 60° = \dfrac{1}{2} \quad \tan 60° = \sqrt{3}$
You try
Without a calculator, evaluate: $\sin^{2} 30° + \cos^{2} 30°$.
Square each value, then add.
$\sin 30° = \dfrac{1}{2}, \;\; \cos 30° = \dfrac{\sqrt{3}}{2}$
$\sin^{2} 30° = \dfrac{1}{4}, \;\; \cos^{2} 30° = \dfrac{3}{4}$
$\dfrac{1}{4} + \dfrac{3}{4} = 1$
$= 1$ (matches the identity $\sin^{2}\!\theta + \cos^{2}\!\theta = 1$)
$1$
You try
Without a calculator, evaluate: $\tan 60° \cdot \tan 30°$.
$\tan 60° = \sqrt{3}, \;\; \tan 30° = \dfrac{1}{\sqrt{3}}$.
$\sqrt{3} \times \dfrac{1}{\sqrt{3}}$
$= 1$
$1$
Section 7 of 11
Type 3: $\sin 15°$ with no calculator
Find $\sin 15°$ without a calculator.
$15°$ isn't a $30°/45°/60°$ value — but it's the difference of two of them. Use the angles you know: $60°, 45°, 30°$.
$15° = 45° - 30°$
$\sin 15° = \sin(45° - 30°)$
Apply the compound formula $\sin(A-B) = \sin A \cos B - \cos A \sin B$:
$\sin(45° - 30°) = \sin 45° \cos 30° - \cos 45° \sin 30°$
$= \dfrac{1}{\sqrt{2}} \cdot \dfrac{\sqrt{3}}{2} \;-\; \dfrac{1}{\sqrt{2}} \cdot \dfrac{1}{2}$
$= \dfrac{\sqrt{3}}{2\sqrt{2}} \;-\; \dfrac{1}{2\sqrt{2}}$
$= \dfrac{\sqrt{3} - 1}{2\sqrt{2}}$
Rationalise the denominator (multiply top and bottom by $\sqrt{2}$):
$= \dfrac{\sqrt{3} - 1}{2\sqrt{2}} \cdot \dfrac{\sqrt{2}}{\sqrt{2}}$
$= \dfrac{\sqrt{2}\!\left(\sqrt{3} - 1\right)}{2 \cdot 2}$
$\sin 15° = \dfrac{\sqrt{6} - \sqrt{2}}{4}$
You try
Find $\cos 15°$ without a calculator. Use $15° = 45° - 30°$.
$\cos(A-B) = \cos A \cos B + \sin A \sin B$. Same triangles, different formula.
$\cos 15° = \cos(45° - 30°) = \cos 45° \cos 30° + \sin 45° \sin 30°$
$= \dfrac{1}{\sqrt{2}} \cdot \dfrac{\sqrt{3}}{2} + \dfrac{1}{\sqrt{2}} \cdot \dfrac{1}{2}$
$= \dfrac{\sqrt{3} + 1}{2\sqrt{2}}$
$= \dfrac{\sqrt{3} + 1}{2\sqrt{2}} \cdot \dfrac{\sqrt{2}}{\sqrt{2}}$
$\cos 15° = \dfrac{\sqrt{6} + \sqrt{2}}{4}$
$\dfrac{\sqrt{6} + \sqrt{2}}{4}$
You try
Find $\sin 75°$ without a calculator. Use $75° = 45° + 30°$.
$\sin(A+B) = \sin A \cos B + \cos A \sin B$.
$\sin 75° = \sin(45° + 30°) = \sin 45° \cos 30° + \cos 45° \sin 30°$
$= \dfrac{1}{\sqrt{2}} \cdot \dfrac{\sqrt{3}}{2} + \dfrac{1}{\sqrt{2}} \cdot \dfrac{1}{2}$
$= \dfrac{\sqrt{3} + 1}{2\sqrt{2}} \cdot \dfrac{\sqrt{2}}{\sqrt{2}}$
$\sin 75° = \dfrac{\sqrt{6} + \sqrt{2}}{4}$ (same as $\cos 15°$ — makes sense, they're complementary)
$\dfrac{\sqrt{6} + \sqrt{2}}{4}$
Section 8 of 11
$\tan 105°$ with no calculator
Find $\tan 105°$ without a calculator.
$105°$ isn't standard — but $105° = 45° + 60°$, both of which we know:
$\tan 105° = \tan(45° + 60°)$
Use $\tan(A+B) = \dfrac{\tan A + \tan B}{1 - \tan A \tan B}$:
$= \dfrac{\tan 45° + \tan 60°}{1 - \tan 45° \tan 60°}$
$= \dfrac{1 + \sqrt{3}}{1 - \sqrt{3}}$
Rationalise — multiply top and bottom by $1 + \sqrt{3}$ (the conjugate of the bottom):
$= \dfrac{1 + \sqrt{3}}{1 - \sqrt{3}} \cdot \dfrac{1 + \sqrt{3}}{1 + \sqrt{3}}$
Top:
$(1+\sqrt{3})(1+\sqrt{3}) = 1 + \sqrt{3} + \sqrt{3} + 3 = 4 + 2\sqrt{3}$
Bottom:
$(1-\sqrt{3})(1+\sqrt{3}) = 1 - 3 = -2$
$\tan 105° = \dfrac{4 + 2\sqrt{3}}{-2}$
$\tan 105° = -2 - \sqrt{3}$
It's negative — which makes sense, since $105°$ is in the second quadrant where $\tan$ is negative.
You try
Find $\tan 75°$ without a calculator. Use $75° = 45° + 30°$.
$\tan(A+B) = \dfrac{\tan A + \tan B}{1 - \tan A \tan B}$. Use $\tan 45° = 1, \;\tan 30° = \dfrac{1}{\sqrt{3}}$.
$\tan 75° = \dfrac{1 + \dfrac{1}{\sqrt{3}}}{1 - \dfrac{1}{\sqrt{3}}}$
Multiply top and bottom by $\sqrt{3}$:
$= \dfrac{\sqrt{3} + 1}{\sqrt{3} - 1}$
Rationalise by $\sqrt{3} + 1$:
$= \dfrac{(\sqrt{3}+1)^{2}}{(\sqrt{3})^{2} - 1^{2}} = \dfrac{3 + 2\sqrt{3} + 1}{3 - 1} = \dfrac{4 + 2\sqrt{3}}{2}$
$\tan 75° = 2 + \sqrt{3}$
$2 + \sqrt{3}$
Section 9 of 11
Type 4: Given $\sin A$ and $\cos B$, find $\cos(A+B)$
Given $\sin A = \dfrac{3}{5}$ and $\cos B = \dfrac{12}{13}$, find $\cos(A+B)$.
$\cos(A+B) = \cos A \cos B - \sin A \sin B$
The formula needs all four — $\sin A, \cos A, \sin B, \cos B$. We have two; get the other two from triangles.
Triangle for $A$ (given $\sin A = \dfrac{3}{5}$)
opp $= 3$, hyp $= 5$, adj $= 4$ (Pythagoras)
$\cos A = \dfrac{4}{5}$
Triangle for $B$ (given $\cos B = \dfrac{12}{13}$)
adj $= 12$, hyp $= 13$, opp $= 5$ (Pythagoras)
$\sin B = \dfrac{5}{13}$
Now sub in
$\cos(A+B) = \dfrac{4}{5} \cdot \dfrac{12}{13} \;-\; \dfrac{3}{5} \cdot \dfrac{5}{13}$
$= \dfrac{48}{65} - \dfrac{15}{65}$
$\cos(A+B) = \dfrac{33}{65}$
You try
Given $\sin A = \dfrac{5}{13}$ and $\cos B = \dfrac{3}{5}$, find $\sin(A+B)$.
$\sin(A+B) = \sin A \cos B + \cos A \sin B$. Get $\cos A$ and $\sin B$ from triangles.
Triangle for $A$: opp $= 5$, hyp $= 13$ → adj $= 12$, so $\cos A = \dfrac{12}{13}$
Triangle for $B$: adj $= 3$, hyp $= 5$ → opp $= 4$, so $\sin B = \dfrac{4}{5}$
$\sin(A+B) = \dfrac{5}{13} \cdot \dfrac{3}{5} + \dfrac{12}{13} \cdot \dfrac{4}{5}$
$= \dfrac{15}{65} + \dfrac{48}{65}$
$\sin(A+B) = \dfrac{63}{65}$
$\dfrac{63}{65}$
You try
Given $\cos A = \dfrac{4}{5}$ and $\sin B = \dfrac{5}{13}$, find $\cos(A-B)$.
$\cos(A-B) = \cos A \cos B + \sin A \sin B$. The sign flips compared to $\cos(A+B)$.
Triangle for $A$: adj $=4$, hyp $=5$ → opp $=3$, so $\sin A = \dfrac{3}{5}$
Triangle for $B$: opp $=5$, hyp $=13$ → adj $=12$, so $\cos B = \dfrac{12}{13}$
$\cos(A-B) = \dfrac{4}{5} \cdot \dfrac{12}{13} + \dfrac{3}{5} \cdot \dfrac{5}{13}$
$= \dfrac{48}{65} + \dfrac{15}{65}$
$\cos(A-B) = \dfrac{63}{65}$
$\dfrac{63}{65}$
Section 10 of 11
Type 5: Given $\tan A$ and $\tan(A+B)$, find $\tan B$
Given $\tan A = 5$ and $\tan(A+B) = 8$, find $\tan B$.
$\tan(A+B) = \dfrac{\tan A + \tan B}{1 - \tan A \tan B}$
Let $t = \tan B$ and sub in:
$8 = \dfrac{5 + t}{1 - 5t}$
Cross-multiply:
$8(1 - 5t) = 5 + t$
$8 - 40t = 5 + t$
$3 = 41t$
$\tan B = \dfrac{3}{41}$
You try
Given $\tan A = 2$ and $\tan(A+B) = 3$, find $\tan B$.
Same setup — let $t = \tan B$, plug into $\tan(A+B) = \dfrac{\tan A + \tan B}{1 - \tan A \tan B}$, cross-multiply.
$3 = \dfrac{2 + t}{1 - 2t}$
$3(1 - 2t) = 2 + t$
$3 - 6t = 2 + t$
$1 = 7t$
$\tan B = \dfrac{1}{7}$
$\dfrac{1}{7}$
You try
Given $\tan B = 4$ and $\tan(A-B) = 1$, find $\tan A$.
$\tan(A-B) = \dfrac{\tan A - \tan B}{1 + \tan A \tan B}$. Let $t = \tan A$ this time.
$1 = \dfrac{t - 4}{1 + 4t}$
$1 + 4t = t - 4$
$3t = -5$
$\tan A = -\dfrac{5}{3}$
$-\dfrac{5}{3}$
Section 11 of 11
Type 6: Prove $\cos 3A = 4\cos^{3}\!A - 3\cos A$
There's no $\cos 3A$ formula in the tables — so we build it by writing $3A$ as a compound angle:
$3A = A + 2A$
$\cos 3A = \cos(A + 2A)$
Use $\cos(A+B) = \cos A \cos B - \sin A \sin B$ with $B = 2A$:
$= \cos A \cos 2A - \sin A \sin 2A$
Now expand $\cos 2A$ and $\sin 2A$. From tables:
$\cos 2A = \cos^{2}\!A - \sin^{2}\!A$ and $\sin 2A = 2\sin A \cos A$
$= \cos A\!\left(\cos^{2}\!A - \sin^{2}\!A\right) \;-\; \sin A \cdot 2\sin A \cos A$
Multiply out:
$= \cos^{3}\!A - \cos A \sin^{2}\!A - 2\cos A \sin^{2}\!A$
$= \cos^{3}\!A - 3\cos A \sin^{2}\!A$
Now we need everything in terms of $\cos A$. Use $\sin^{2}\!A = 1 - \cos^{2}\!A$:
$= \cos^{3}\!A - 3\cos A\!\left(1 - \cos^{2}\!A\right)$
$= \cos^{3}\!A - 3\cos A + 3\cos^{3}\!A$
$\cos 3A = 4\cos^{3}\!A - 3\cos A \;\;\checkmark$
You try
Prove that $\sin 3A = 3\sin A - 4\sin^{3}\!A$.
Same strategy: $\sin 3A = \sin(A + 2A)$. Expand with the compound formula, then sub in $\sin 2A$ and $\cos 2A$.
$\sin 3A = \sin(A + 2A)$
$= \sin A \cos 2A + \cos A \sin 2A$
Sub $\cos 2A = 1 - 2\sin^{2}\!A$ and $\sin 2A = 2\sin A \cos A$:
$= \sin A\!\left(1 - 2\sin^{2}\!A\right) + \cos A \cdot 2\sin A \cos A$
$= \sin A - 2\sin^{3}\!A + 2\sin A \cos^{2}\!A$
Use $\cos^{2}\!A = 1 - \sin^{2}\!A$:
$= \sin A - 2\sin^{3}\!A + 2\sin A\!\left(1 - \sin^{2}\!A\right)$
$= \sin A - 2\sin^{3}\!A + 2\sin A - 2\sin^{3}\!A$
$\sin 3A = 3\sin A - 4\sin^{3}\!A \;\;\checkmark$
Proved — $\sin 3A = 3\sin A - 4\sin^{3}\!A$.
That's Double & Compound Angles.
Six types covered — given a single ratio find the double, given the double work back to the single, exact values without a calculator, $\cos(A+B)$ from two triangles, finding a missing $\tan$, and proving triple-angle identities.