Radians
Angles can be measured in radians . One radian is the angle at the centre of a circle where the arc is as long as the radius, so a full turn (2 π r 2\pi r 2 π r 2 π 2\pi 2 π 180 ∘ = π 180^\circ = \pi 18 0 ∘ = π
θ r s = rθ A sector arc length s = rθ (θ in radians); 180° = π rad To change degrees to radians, multiply by π 180 \frac{\pi}{180} 180 π θ \theta θ s = r θ s = r\theta s = r θ 1 2 r 2 θ \frac12r^2\theta 2 1 r 2 θ
Compound angles
The sine of a sum is not the sum of the sines. The correct formulas are:
A B A + B The angle A + B sin(A ± B) = sin A cos B ± 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 tan ( A ± B ) = tan A ± tan B 1 ∓ tan A tan B \begin{aligned}
\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) &= \frac{\tan A \pm \tan B}{1 \mp \tan A\tan B}
\end{aligned} sin ( A ± B ) cos ( A ± B ) tan ( A ± B ) = sin A cos B ± cos A sin B = cos A cos B ∓ sin A sin B = 1 ∓ tan A tan B tan A ± tan B Notice the sign flips in the cosine formula: cos ( A + B ) \cos(A + B) cos ( A + B )
Compound angles Set A and B
A B 0.9659 sin(A + B) 0.2588 cos(A + B) 1.1887 sin A + sin B
A = 50° B = 25°
sin(A + B) = sin A cos B + cos A sin B = (0.766)(0.9063) + (0.6428)(0.4226) = 0.9659, the same as sin 75°. cos(A + B) = cos A cos B − sin A sin B = 0.2588, the same as cos 75°. But sin A + sin B = 1.1887: the sine of a sum is not the sum of the sines. When you are given one ratio, draw a right-angled triangle to find the others (see trigonometric ratios↺ ), then choose the sign from the quadrant.
Worked example · NECO 2023
NECO 2023 · Paper 1 · Q11
Given that cos A = 4 5 \cos A = \frac45 cos A = 5 4 cos B = 12 13 \cos B = \frac{12}{13} cos B = 13 12 sin ( A + B ) \sin(A + B) sin ( A + B ) A A A B B B
The other ratios
sin A = 3 5 {\sin A = \frac35} sin A = 5 3 sin B = 5 13 {\sin B = \frac{5}{13}} sin B = 13 5 Think first. cos A = 4/5 is a 3–4–5 triangle; cos B = 12/13 is 5–12–13.
The formula
sin ( A + B ) = sin A cos B + cos A sin B {\sin(A + B) = \sin A\cos B + \cos A\sin B} sin ( A + B ) = sin A cos B + cos A sin B = 3 5 × 12 13 + 4 5 × 5 13 = 36 65 + 20 65 = 56 65 {= \frac35 \times \frac{12}{13} + \frac45 \times \frac{5}{13} = \frac{36}{65} + \frac{20}{65} = \frac{56}{65}} = 5 3 × 13 12 + 5 4 × 13 5 = 65 36 + 65 20 = 65 56 E . ← Back Next step → Show all steps Start again
Worked example · WAEC 2019
WAEC 2019 · Paper 2 · Q10 (a)
If sin p = 1 2 \sin p = \frac12 sin p = 2 1 cos q = − 1 3 \cos q = -\frac13 cos q = − 3 1 sin ( p − q ) \sin(p - q) sin ( p − q ) 0 ∘ ≤ p ≤ 90 ∘ 0^\circ \le p \le 90^\circ 0 ∘ ≤ p ≤ 9 0 ∘ 90 ∘ ≤ q ≤ 180 ∘ 90^\circ \le q \le 180^\circ 9 0 ∘ ≤ q ≤ 18 0 ∘
The other ratios
cos p = 1 − 1 4 = 3 2 {\cos p = \sqrt{1 - \frac14} = \frac{\sqrt3}{2}} cos p = 1 − 4 1 = 2 3 sin q = 1 − 1 9 = 2 2 3 {\sin q = \sqrt{1 - \frac19} = \frac{2\sqrt2}{3}} sin q = 1 − 9 1 = 3 2 2 Think first. p is acute; q is obtuse, where sine is positive.
The formula
sin ( p − q ) = sin p cos q − cos p sin q {\sin(p - q) = \sin p\cos q - \cos p\sin q} sin ( p − q ) = sin p cos q − cos p sin q = 1 2 ( − 1 3 ) − 3 2 × 2 2 3 {= \frac12\left(-\frac13\right) - \frac{\sqrt3}{2} \times \frac{2\sqrt2}{3}} = 2 1 ( − 3 1 ) − 2 3 × 3 2 2 = − 1 6 − 2 6 6 = − 1 − 2 6 6 {= -\frac16 - \frac{2\sqrt6}{6} = \frac{-1 - 2\sqrt6}{6}} = − 6 1 − 6 2 6 = 6 − 1 − 2 6 ← Back Next step → Show all steps Start again
A formula can also be recognised in reverse:
Worked example · WAEC 2018
WAEC 2018 · Paper 2 · Q1
If tan θ + tan 30 ∘ 1 − tan θ tan 30 ∘ + 1 = 0 \dfrac{\tan\theta + \tan30^\circ}{1 - \tan\theta\tan30^\circ} + 1 = 0 1 − tan θ tan 3 0 ∘ tan θ + tan 3 0 ∘ + 1 = 0 tan θ \tan\theta tan θ
Spot the formula
tan ( θ + 30 ∘ ) + 1 = 0 {\tan(\theta + 30^\circ) + 1 = 0} tan ( θ + 3 0 ∘ ) + 1 = 0 tan ( θ + 30 ∘ ) = − 1 {\tan(\theta + 30^\circ) = -1} tan ( θ + 3 0 ∘ ) = − 1 Think first. The fraction is tan(θ + 30°).
Solve for tan θ
3 t + 1 3 − t = − 1 {\frac{\sqrt3\,t + 1}{\sqrt3 - t} = -1} 3 − t 3 t + 1 = − 1 3 t + 1 = t − 3 {\sqrt3\,t + 1 = t - \sqrt3} 3 t + 1 = t − 3 t ( 3 − 1 ) = − ( 3 + 1 ) {t(\sqrt3 - 1) = -(\sqrt3 + 1)} t ( 3 − 1 ) = − ( 3 + 1 ) t = − 3 + 1 3 − 1 {t = -\frac{\sqrt3 + 1}{\sqrt3 - 1}} t = − 3 − 1 3 + 1 Rationalise: t = − ( 3 + 1 ) 2 2 = − 4 + 2 3 2 = − 2 − 3 {t = -\frac{(\sqrt3 + 1)^2}{2} = -\frac{4 + 2\sqrt3}{2} = -2 - \sqrt3} t = − 2 ( 3 + 1 ) 2 = − 2 4 + 2 3 = − 2 − 3
Think first. Or write t = tan θ and tan 30° = 1/√3.
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Double angles
Put B = A B = A B = A
sin 2 A = 2 sin A cos A cos 2 A = cos 2 A − sin 2 A = 1 − 2 sin 2 A = 2 cos 2 A − 1 \begin{aligned}
\sin 2A &= 2\sin A\cos A \\
\cos 2A &= \cos^2 A - \sin^2 A \\
&= 1 - 2\sin^2 A \\
&= 2\cos^2 A - 1
\end{aligned} sin 2 A cos 2 A = 2 sin A cos A = cos 2 A − sin 2 A = 1 − 2 sin 2 A = 2 cos 2 A − 1 The three forms of cos 2 A \cos 2A cos 2 A
More: identities and the cosine rule
Your turn
(a) If sin X = p − q p + q \sin X = \dfrac{p - q}{p + q} sin X = p + q p − q 0 ∘ ≤ X ≤ 90 ∘ 0^\circ \le X \le 90^\circ 0 ∘ ≤ X ≤ 9 0 ∘ 1 − tan 2 X 1 - \tan^2 X 1 − tan 2 X
Worked solution (try it first) Draw a right-angled triangle with opposite side
p − q p - q p − q and hypotenuse
p + q p + q p + q .
Adjacent side:
( p + q ) 2 − ( p − q ) 2 = 4 p q \sqrt{(p + q)^2 - (p - q)^2} = \sqrt{4pq} ( p + q ) 2 − ( p − q ) 2 = 4 pq So
tan X = p − q 2 p q \tan X = \dfrac{p - q}{2\sqrt{pq}} tan X = 2 pq p − q and
tan 2 X = p 2 − 2 p q + q 2 4 p q \tan^2 X = \dfrac{p^2 - 2pq + q^2}{4pq} tan 2 X = 4 pq p 2 − 2 pq + q 2 .
1 − tan 2 X = 4 p q − p 2 + 2 p q − q 2 4 p q 1 - \tan^2 X = \dfrac{4pq - p^2 + 2pq - q^2}{4pq} 1 − tan 2 X = 4 pq 4 pq − p 2 + 2 pq − q 2 = 6 p q − p 2 − q 2 4 p q = \dfrac{6pq - p^2 - q^2}{4pq} = 4 pq 6 pq − p 2 − q 2 .
Watch out
( p + q ) 2 − ( p − q ) 2 = 4 p q (p + q)^2 - (p - q)^2 = 4pq ( p + q ) 2 − ( p − q ) 2 = 4 pq Change the sign of every term of ( p − q ) 2 (p - q)^2 ( p − q ) 2 Report a problem with this question