At CAPE, trigonometry stops being about triangles and becomes about identities. Almost every question is really asking: which identity turns this into something I can solve?
Radians and exact values
π radians = 180 ° \pi \text{ radians} = 180° π radians = 180°
so 1 1 1 ≈ 57.3 ° \approx 57.3° ≈ 57.3° π 180 \frac{\pi}{180} 180 π 180 π \frac{180}{\pi} π 180
60 ° = 60 × π 180 = π 3 3 π 4 = 3 π 4 × 180 π = 135 ° 60° = 60 \times \frac{\pi}{180} = \frac{\pi}{3} \qquad \frac{3\pi}{4} = \frac{3\pi}{4} \times \frac{180}{\pi} = 135° 60° = 60 × 180 π = 3 π 4 3 π = 4 3 π × π 180 = 135°
Radians make two formulae beautifully simple. For a sector of radius r r r θ \theta θ in radians :
arc length s = r θ sector area = 1 2 r 2 θ \text{arc length } s = r\theta \qquad \text{sector area} = \tfrac12 r^2\theta arc length s = r θ sector area = 2 1 r 2 θ
A sector of radius 5 and angle 0.8 rad has arc length 5 × 0.8 = 4 5 \times 0.8 = 4 5 × 0.8 = 4 1 2 ( 36 ) ( 0.5 ) = 9 \frac12(36)(0.5) = 9 2 1 ( 36 ) ( 0.5 ) = 9
Common mistake Both formulae are only valid in radians. With degrees you must convert first — using θ = 60 \theta = 60 θ = 60 π 3 \frac{\pi}{3} 3 π
The exact values worth knowing by heart:
θ \theta θ sin θ \sin\theta sin θ cos θ \cos\theta cos θ tan θ \tan\theta tan θ
0 0 0 0 0 0 1 1 1 0 0 0
π 6 = 30 ° \frac{\pi}{6} = 30° 6 π = 30° 1 2 \frac12 2 1 3 2 \frac{\sqrt3}{2} 2 3 1 3 \frac{1}{\sqrt3} 3 1
π 4 = 45 ° \frac{\pi}{4} = 45° 4 π = 45° 1 2 \frac{1}{\sqrt2} 2 1 1 2 \frac{1}{\sqrt2} 2 1 1 1 1
π 3 = 60 ° \frac{\pi}{3} = 60° 3 π = 60° 3 2 \frac{\sqrt3}{2} 2 3 1 2 \frac12 2 1 3 \sqrt3 3
π 2 = 90 ° \frac{\pi}{2} = 90° 2 π = 90° 1 1 1 0 0 0 undefined
Periods
y = sin x y = \sin x y = sin x y = cos x y = \cos x y = cos x 2 π 2\pi 2 π [ − 1 , 1 ] [-1, 1] [ − 1 , 1 ] y = tan x y = \tan x y = tan x π \pi π x = π 2 + n π x = \frac{\pi}{2} + n\pi x = 2 π + nπ
For y = a sin ( b x ) y = a\sin(bx) y = a sin ( b x ) ∣ a ∣ |a| ∣ a ∣ 2 π b \dfrac{2\pi}{b} b 2 π
The reciprocal functions
sec θ = 1 cos θ cosec θ = 1 sin θ cot θ = 1 tan θ = cos θ sin θ \sec\theta = \frac{1}{\cos\theta} \qquad \operatorname{cosec}\theta = \frac{1}{\sin\theta} \qquad \cot\theta = \frac{1}{\tan\theta} = \frac{\cos\theta}{\sin\theta} sec θ = cos θ 1 cosec θ = sin θ 1 cot θ = tan θ 1 = sin θ cos θ
Common mistake Note which pairs with which: sec goes with cos , and cosec goes with sin . The third letter is the reliable clue, since the first three do not match up.
Because ∣ sin θ ∣ ≤ 1 |\sin\theta| \le 1 ∣ sin θ ∣ ≤ 1 ∣ cosec θ ∣ ≥ 1 |\operatorname{cosec}\theta| \ge 1 ∣ cosec θ ∣ ≥ 1 − 1 -1 − 1 1 1 1
Pythagorean identities
The fundamental one:
sin 2 θ + cos 2 θ = 1 \sin^2\theta + \cos^2\theta = 1 sin 2 θ + cos 2 θ = 1
Dividing through by cos 2 θ \cos^2\theta cos 2 θ
tan 2 θ + 1 = sec 2 θ \tan^2\theta + 1 = \sec^2\theta tan 2 θ + 1 = sec 2 θ
Dividing instead by sin 2 θ \sin^2\theta sin 2 θ
1 + cot 2 θ = cosec 2 θ 1 + \cot^2\theta = \operatorname{cosec}^2\theta 1 + cot 2 θ = cosec 2 θ
Key result You only need to memorise the first. The other two are one division away, which is far safer than trying to recall which side the 1 sits on.
These are the tools for converting an equation into a single function. 2 sin 2 θ + 3 cos θ = 3 2\sin^2\theta + 3\cos\theta = 3 2 sin 2 θ + 3 cos θ = 3 sin 2 θ = 1 − cos 2 θ \sin^2\theta = 1 - \cos^2\theta sin 2 θ = 1 − cos 2 θ
2 − 2 cos 2 θ + 3 cos θ − 3 = 0 ⇒ 2 cos 2 θ − 3 cos θ + 1 = 0 2 - 2\cos^2\theta + 3\cos\theta - 3 = 0 \quad \Rightarrow \quad 2\cos^2\theta - 3\cos\theta + 1 = 0 2 − 2 cos 2 θ + 3 cos θ − 3 = 0 ⇒ 2 cos 2 θ − 3 cos θ + 1 = 0
— a quadratic in cos θ \cos\theta cos θ
sin ( A ± B ) = sin A cos B ± cos A sin B \sin(A \pm B) = \sin A\cos B \pm \cos A\sin B sin ( A ± B ) = sin A cos B ± cos A sin B cos ( A ± B ) = cos A cos B ∓ sin A sin B \cos(A \pm B) = \cos A\cos B \mp \sin A\sin B cos ( A ± B ) = cos A cos B ∓ sin A sin B tan ( A ± B ) = tan A ± tan B 1 ∓ tan A tan B \tan(A \pm B) = \frac{\tan A \pm \tan B}{1 \mp \tan A\tan B} tan ( A ± B ) = 1 ∓ tan A tan B tan A ± tan B
Common mistake The signs in the cosine formula are opposite to the ones in the question: cos ( A + B ) \cos(A+B) cos ( A + B ) ∓ \mp ∓
Find the exact value of sin 75 ° \sin 75° sin 75°
sin 75 ° = sin ( 45 ° + 30 ° ) = sin 45 ° cos 30 ° + cos 45 ° sin 30 ° \sin 75° = \sin(45° + 30°) = \sin45°\cos30° + \cos45°\sin30° sin 75° = sin ( 45° + 30° ) = sin 45° cos 30° + cos 45° sin 30° = 1 2 ⋅ 3 2 + 1 2 ⋅ 1 2 = 3 + 1 2 2 = 6 + 2 4 = \frac{1}{\sqrt2}\cdot\frac{\sqrt3}{2} + \frac{1}{\sqrt2}\cdot\frac12 = \frac{\sqrt3 + 1}{2\sqrt2} = \frac{\sqrt6 + \sqrt2}{4} = 2 1 ⋅ 2 3 + 2 1 ⋅ 2 1 = 2 2 3 + 1 = 4 6 + 2
Check numerically: 2.449 + 1.414 4 = 0.966 \frac{2.449 + 1.414}{4} = 0.966 4 2.449 + 1.414 = 0.966 sin 75 ° = 0.966 \sin 75° = 0.966 sin 75° = 0.966
Common mistake sin ( A + B ) \sin(A+B) sin ( A + B ) not sin A + sin B \sin A + \sin B sin A + sin B sin ( 30 ° + 60 ° ) = 1 \sin(30°+60°) = 1 sin ( 30° + 60° ) = 1 sin 30 ° + sin 60 ° = 0.5 + 0.866 = 1.366 \sin30° + \sin60° = 0.5 + 0.866 = 1.366 sin 30° + sin 60° = 0.5 + 0.866 = 1.366
Set B = A B = A B = A
sin 2 A = 2 sin A cos A \sin 2A = 2\sin A\cos A sin 2 A = 2 sin A cos A cos 2 A = cos 2 A − sin 2 A = 2 cos 2 A − 1 = 1 − 2 sin 2 A \cos 2A = \cos^2 A - \sin^2 A = 2\cos^2 A - 1 = 1 - 2\sin^2 A cos 2 A = cos 2 A − sin 2 A = 2 cos 2 A − 1 = 1 − 2 sin 2 A tan 2 A = 2 tan A 1 − tan 2 A \tan 2A = \frac{2\tan A}{1 - \tan^2 A} tan 2 A = 1 − tan 2 A 2 tan A
The three versions of cos 2 A \cos 2A cos 2 A
Key result Rearranged, cos 2 A = 1 − 2 sin 2 A \cos 2A = 1 - 2\sin^2 A cos 2 A = 1 − 2 sin 2 A sin 2 A = 1 − cos 2 A 2 \sin^2 A = \dfrac{1 - \cos 2A}{2} sin 2 A = 2 1 − cos 2 A cos 2 A = 1 + cos 2 A 2 \cos^2 A = \dfrac{1 + \cos 2A}{2} cos 2 A = 2 1 + cos 2 A sin 2 \sin^2 sin 2 cos 2 \cos^2 cos 2
Solve sin 2 θ = sin θ \sin 2\theta = \sin\theta sin 2 θ = sin θ 0 ≤ θ < 2 π 0 \le \theta < 2\pi 0 ≤ θ < 2 π
2 sin θ cos θ − sin θ = 0 2\sin\theta\cos\theta - \sin\theta = 0 2 sin θ cos θ − sin θ = 0 sin θ ( 2 cos θ − 1 ) = 0 \sin\theta(2\cos\theta - 1) = 0 sin θ ( 2 cos θ − 1 ) = 0
sin θ = 0 \sin\theta = 0 sin θ = 0 θ = 0 , π \theta = 0, \pi θ = 0 , π cos θ = 1 2 \cos\theta = \frac12 cos θ = 2 1 θ = π 3 , 5 π 3 \theta = \frac{\pi}{3}, \frac{5\pi}{3} θ = 3 π , 3 5 π
Four solutions in the interval.
Common mistake Never divide both sides by sin θ \sin\theta sin θ θ = 0 \theta = 0 θ = 0 θ = π \theta = \pi θ = π Factorise instead.
a cos θ + b sin θ = R cos ( θ − α ) a\cos\theta + b\sin\theta = R\cos(\theta - \alpha) a cos θ + b sin θ = R cos ( θ − α )
where
R = a 2 + b 2 tan α = b a R = \sqrt{a^2 + b^2} \qquad \tan\alpha = \frac{b}{a} R = a 2 + b 2 tan α = a b
Expanding the right side gives R cos θ cos α + R sin θ sin α R\cos\theta\cos\alpha + R\sin\theta\sin\alpha R cos θ cos α + R sin θ sin α R cos α = a R\cos\alpha = a R cos α = a R sin α = b R\sin\alpha = b R sin α = b
Express 3 cos θ + 4 sin θ 3\cos\theta + 4\sin\theta 3 cos θ + 4 sin θ R cos ( θ − α ) R\cos(\theta - \alpha) R cos ( θ − α )
R = 9 + 16 = 5 tan α = 4 3 ⇒ α = 53.13 ° R = \sqrt{9 + 16} = 5 \qquad \tan\alpha = \tfrac43 \Rightarrow \alpha = 53.13° R = 9 + 16 = 5 tan α = 3 4 ⇒ α = 53.13°
3 cos θ + 4 sin θ = 5 cos ( θ − 53.13 ° ) 3\cos\theta + 4\sin\theta = 5\cos(\theta - 53.13°) 3 cos θ + 4 sin θ = 5 cos ( θ − 53.13° )
This form makes maxima and minima immediate: since cos \cos cos [ − 1 , 1 ] [-1,1] [ − 1 , 1 ] maximum 5 (at θ = 53.13 ° \theta = 53.13° θ = 53.13° minimum − 5 -5 − 5 .
Solve 3 cos θ + 4 sin θ = 2 3\cos\theta + 4\sin\theta = 2 3 cos θ + 4 sin θ = 2 0 ° ≤ θ < 360 ° 0° \le \theta < 360° 0° ≤ θ < 360°
5 cos ( θ − 53.13 ° ) = 2 ⇒ cos ( θ − 53.13 ° ) = 0.4 5\cos(\theta - 53.13°) = 2 \quad \Rightarrow \quad \cos(\theta - 53.13°) = 0.4 5 cos ( θ − 53.13° ) = 2 ⇒ cos ( θ − 53.13° ) = 0.4 θ − 53.13 ° = ± 66.42 ° \theta - 53.13° = \pm 66.42° θ − 53.13° = ± 66.42° θ = 119.6 ° or θ = − 13.3 ° ≡ 346.7 ° \theta = 119.6° \quad \text{or} \quad \theta = -13.3° \equiv 346.7° θ = 119.6° or θ = − 13.3° ≡ 346.7°
Key result The R-formula turns a two-function equation into a single cosine, which is why it is the standard route for both "solve" and "find the maximum" questions.
Solving trigonometric equations
A method that does not lose solutions:
Reduce to one function using an identity.Solve as an ordinary equation — often a quadratic.Find every angle in the required interval, not just the calculator's principal value.Check the interval , especially after a substitution such as 2 θ 2\theta 2 θ
Solve 2 cos 2 θ + cos θ − 1 = 0 2\cos^2\theta + \cos\theta - 1 = 0 2 cos 2 θ + cos θ − 1 = 0 0 ° ≤ θ < 360 ° 0° \le \theta < 360° 0° ≤ θ < 360°
( 2 cos θ − 1 ) ( cos θ + 1 ) = 0 (2\cos\theta - 1)(\cos\theta + 1) = 0 ( 2 cos θ − 1 ) ( cos θ + 1 ) = 0
cos θ = 1 2 \cos\theta = \frac12 cos θ = 2 1 θ = 60 ° \theta = 60° θ = 60° θ = 300 ° \theta = 300° θ = 300° cos θ = − 1 \cos\theta = -1 cos θ = − 1 θ = 180 ° \theta = 180° θ = 180°
Three solutions.
Key result sin \sin sin cos \cos cos tan \tan tan All, Sine, Tangent, Cosine going anticlockwise from the first quadrant — the CAST diagram.
Common mistake If the equation involves 2 θ 2\theta 2 θ θ \theta θ 0 ° 0° 0° 360 ° 360° 360° 2 θ 2\theta 2 θ 0 ° 0° 0° 720 ° 720° 720° two full revolutions and halve them at the end. Forgetting this loses half the answers.