5 Trigonometry — printable coverage pack

5.1 · Understand and use the definitions of sine, cosine and tangent for all arguments; the sine and cosine rules; the area of a triangle in the form ½ab sin C; work with radian measure, including use for arc length and area of sector.

For a unit-circle angle $\theta$ , $\cos\theta$ and $\sin\theta$ are the point's horizontal and vertical coordinates, while $\tan\theta=\sin\theta/\cos\theta$ where $\cos\theta\neq0$ ; this extends the ratios beyond acute angles.
For a triangle, use $a/\sin A=b/\sin B=c/\sin C$ when an opposite side-angle pair is known, $a^2=b^2+c^2-2bc\cos A$ for three sides or an included angle, and $\tfrac12ab\sin C$ for area.
Radian measure makes arc and sector formulae direct: an angle $\theta$ radians in a circle of radius $r$ gives arc length $s=r\theta$ and sector area $A=\tfrac12r^2\theta$ .
Keep the calculator in the required angle mode and label sides opposite their matching angles; using degrees in $s=r\theta$ or pairing the wrong side and angle is a common error.

Tier 1 · Easy

1. A circular arc has radius $7.5\,\text{cm}$ and subtends $1.2$ radians at the centre. Find its length.[2 marks]
Answer
- $9.0\,\text{cm}$
Method: Use $s=r\theta$ with the angle already in radians: $s=7.5(1.2)=9.0\,\text{cm}$ .

Tier 2 · Standard

1. Two sides of a triangular sail are $8\,\text{m}$ and $11\,\text{m}$ , with included angle $0.9$ radians. Calculate the third side and the area of the sail, giving each answer to $3$ significant figures.[4 marks]
Answer
- Third side $=8.69\,\text{m}$
- Area $=34.5\,\text{m}^2$
Method: The cosine rule gives $c^2=8^2+11^2-2(8)(11)\cos(0.9)=75.597\ldots$ , so $c=8.694\ldots\,\text{m}=8.69\,\text{m}$ . The area is $\tfrac12(8)(11)\sin(0.9)=34.466\ldots\,\text{m}^2$ , hence $34.5\,\text{m}^2$ .

Tier 3 · Hard

1. A minor segment is cut from a circle of radius $6\,\text{cm}$ by a chord whose endpoints subtend $1.4$ radians at the centre. Determine the perimeter and area of the segment, giving both to $3$ significant figures.[5 marks]
Answer
- Perimeter $=16.1\,\text{cm}$
- Area $=7.46\,\text{cm}^2$
Method: The arc length is $6(1.4)=8.4\,\text{cm}$ . Splitting the isosceles triangle in half gives chord length $2(6)\sin(0.7)=7.730\ldots\,\text{cm}$ , so the perimeter is $16.130\ldots\,\text{cm}$ . The sector area is $\tfrac12(6^2)(1.4)=25.2\,\text{cm}^2$ and the triangle area is $\tfrac12(6^2)\sin(1.4)=17.738\ldots\,\text{cm}^2$ . Their difference is $7.461\ldots\,\text{cm}^2$ , giving the stated answers.

5.2 · Understand and use the standard small angle approximations of sine, cosine and tangent: sin θ ≈ θ, cos θ ≈ 1 − θ²/2, tan θ ≈ θ.

For $|\theta|$ close to zero and measured in radians, $\sin\theta\approx\theta$ , $\tan\theta\approx\theta$ and $\cos\theta\approx1-\theta^2/2$ .
Replace each trigonometric function by its stated approximation, simplify algebraically, and retain only a solution whose magnitude is small enough for the approximation to be credible.
For example, $1-\cos\theta\approx\theta^2/2$ , so $(1-\cos\theta)/\theta^2\approx1/2$ for a small non-zero $\theta$ .
The approximations are radian results, not degree results; another common error is to accept a large root created by the approximate polynomial.

Tier 1 · Easy

1. Use a standard small-angle approximation to estimate $\sin(0.064)$ .[1 mark]
Answer
- $0.064$
Method: Since $0.064$ is a small angle in radians, use $\sin\theta\approx\theta$ to obtain $\sin(0.064)\approx0.064$ .

Tier 2 · Standard

1. Without using a calculator's trigonometric keys, estimate $\dfrac{1-\cos(0.08)}{(0.08)^2}$ by a small-angle approximation.[2 marks]
Answer
- $0.5$
Method: Use $\cos\theta\approx1-\theta^2/2$ . Then $1-\cos(0.08)\approx(0.08)^2/2$ , so division by $(0.08)^2$ gives $1/2=0.5$ .

Tier 3 · Hard

1. A small positive angle $x$ satisfies $\sin x+\cos x=1.08$ . Use the standard small-angle approximations to estimate $x$ , giving $4$ decimal places, and explain which algebraic root is admissible.[4 marks]
Answer
- $x\approx0.0835$ radians
- The other root is not a small angle.
Method: Substitution gives $x+1-x^2/2\approx1.08$ , hence $x^2-2x+0.16\approx0$ . Therefore $x\approx1\pm\sqrt{0.84}$ , giving $0.08348\ldots$ or $1.9165\ldots$ . Only $0.08348\ldots$ is small, so $x\approx0.0835$ radians; the larger root lies outside the approximation's intended range.

5.3 · Understand and use the sine, cosine and tangent functions; their graphs, symmetries and periodicity; know and use exact values of sin, cos and tan for standard angles and their multiples.

$\sin x$ and $\cos x$ have range $[-1,1]$ and period $2\pi$ , while $\tan x$ has range $\mathbb{R}$ , period $\pi$ and vertical asymptotes at $x=\pi/2+k\pi$ .
Use $\sin(-x)=-\sin x$ , $\cos(-x)=\cos x$ and $\tan(-x)=-\tan x$ , then reduce an angle by a whole period before using its reference angle and quadrant.
The exact first-quadrant values come from the $45^\circ$ and $30^\circ$ - $60^\circ$ triangles; quadrant signs then determine standard-angle values around the unit circle.
Do not read the period from the amplitude: in $a\cos(bx)+c$ , the amplitude is $|a|$ , the period is $2\pi/|b|$ , and $c$ moves the midline.

Tier 1 · Easy

1. Find the exact value of $\tan(5\pi/6)$ .[2 marks]
Answer
- $-\dfrac{1}{\sqrt3}$
- $-\dfrac{\sqrt3}{3}$
Method: The reference angle is $\pi/6$ . Tangent is negative in the second quadrant, so $\tan(5\pi/6)=-\tan(\pi/6)=-1/\sqrt3=-\sqrt3/3$ .

Tier 2 · Standard

1. For $y=3\cos(2x)-1$ , state the amplitude, period, maximum value and minimum value.[4 marks]
Answer
- Amplitude $3$
- Period $\pi$
- Maximum $2$
- Minimum $-4$
Method: The coefficient outside cosine gives amplitude $|3|=3$ . The factor $2$ inside gives period $2\pi/2=\pi$ . Since $-1\leq\cos(2x)\leq1$ , multiplying by $3$ and subtracting $1$ gives $-4\leq y\leq2$ .

Tier 3 · Hard

1. Evaluate exactly $\sin(-11\pi/6)$ , $\cos(13\pi/3)$ and $\tan(7\pi/4)$ .[4 marks]
Answer
- $\sin(-11\pi/6)=\dfrac12$
- $\cos(13\pi/3)=\dfrac12$
- $\tan(7\pi/4)=-1$
Method: Add $2\pi$ to $-11\pi/6$ to get $\pi/6$ , so its sine is $1/2$ . Subtract $4\pi$ from $13\pi/3$ to get $\pi/3$ , so its cosine is $1/2$ . The angle $7\pi/4$ has reference angle $\pi/4$ in quadrant IV, where tangent is negative, giving $-1$ .

5.4 · Understand and use the definitions of secant, cosecant and cotangent and of arcsin, arccos and arctan; their relationships to sine, cosine and tangent; understanding of their graphs; their ranges and domains.

The reciprocal functions are $\sec x=1/\cos x$ , $\cosec x=1/\sin x$ and $\cot x=\cos x/\sin x$ ; secant and cosecant have range $(-\infty,-1]\cup[1,\infty)$ and period $2\pi$ , while cotangent has range $\mathbb{R}$ and period $\pi$ .
The principal ranges are $-\pi/2\leq\arcsin x\leq\pi/2$ , $0\leq\arccos x\leq\pi$ and $-\pi/2<\arctan x<\pi/2$ ; $\arcsin x$ and $\arccos x$ require $-1\leq x\leq1$ .
To determine a composite inverse-trigonometric domain, first restrict its input to the inverse function's domain; for example, $\arccos(2x-1)$ requires $-1\leq2x-1\leq1$ .
The notation $\sin^{-1}x$ means the inverse function $\arcsin x$ , not the reciprocal $1/\sin x$ ; the reciprocal is $\cosec x$ .

Tier 1 · Easy

1. Given $\cos\theta=-4/5$ , write down $\sec\theta$ .[1 mark]
Answer
- $\sec\theta=-\dfrac54$
Method: Secant is the reciprocal of cosine, so $\sec\theta=1/(-4/5)=-5/4$ .

Tier 2 · Standard

1. Give the principal values, in radians, of $\arcsin(-1/2)$ and $\arctan(-1)$ .[2 marks]
Answer
- $\arcsin(-1/2)=-\pi/6$
- $\arctan(-1)=-\pi/4$
Method: Within the principal sine-inverse range, $\sin(-\pi/6)=-1/2$ , so $\arcsin(-1/2)=-\pi/6$ . Within the principal tangent-inverse range, $\tan(-\pi/4)=-1$ , so $\arctan(-1)=-\pi/4$ .

Tier 3 · Hard

1. For $y=2\sec x-1$ , state the period and range, then give all vertical asymptotes in $-\pi\leq x\leq\pi$ .[5 marks]
Answer
- Period $2\pi$
- Range $y\leq-3$ or $y\geq1$
- Vertical asymptotes $x=-\pi/2$ and $x=\pi/2$
Method: Secant has period $2\pi$ , so the transformation does not change the period. Since $\sec x\leq-1$ or $\sec x\geq1$ , multiplying by $2$ and subtracting $1$ gives $y\leq-3$ or $y\geq1$ . Vertical asymptotes occur where $\cos x=0$ , namely $x=\pi/2+k\pi$ ; in the stated interval these are $x=-\pi/2$ and $x=\pi/2$ .

5.5 · Understand and use tan θ = sin θ / cos θ; understand and use sin²θ + cos²θ = 1, sec²θ = 1 + tan²θ and cosec²θ = 1 + cot²θ.

The quotient identity is $\tan\theta=\sin\theta/\cos\theta$ , and the three Pythagorean identities are $\sin^2\theta+\cos^2\theta=1$ , $\sec^2\theta=1+\tan^2\theta$ and $\cosec^2\theta=1+\cot^2\theta$ .
Choose an identity containing the known and required functions, rearrange it, and use the stated quadrant to select the correct sign after taking a square root.
If $\tan\theta=-7/24$ in quadrant II, a reference triangle has side magnitudes $7$ , $24$ and $25$ , so $\sin\theta=7/25$ and $\cos\theta=-24/25$ .
From $\sin^2\theta+\cos^2\theta=1$ one obtains two possible signs; ignoring the quadrant or silently choosing the positive root can change every reciprocal ratio that follows.

Tier 1 · Easy

1. An acute angle $\theta$ satisfies $\sin\theta=5/13$ . Find $\tan\theta$ and $\sec\theta$ exactly.[3 marks]
Answer
- $\tan\theta=5/12$
- $\sec\theta=13/12$
Method: Because $\theta$ is acute, $\cos\theta$ is positive. From $\cos^2\theta=1-25/169=144/169$ , $\cos\theta=12/13$ . Hence $\tan\theta=(5/13)/(12/13)=5/12$ and $\sec\theta=1/(12/13)=13/12$ .

Tier 2 · Standard

1. Given that $\tan\theta=-7/24$ and $\pi/2<\theta<\pi$ , determine $\sin\theta$ , $\cos\theta$ and $\cosec\theta$ .[4 marks]
Answer
- $\sin\theta=7/25$
- $\cos\theta=-24/25$
- $\cosec\theta=25/7$
Method: Use $\sec^2\theta=1+\tan^2\theta=1+49/576=625/576$ . In quadrant II cosine and secant are negative, so $\sec\theta=-25/24$ and $\cos\theta=-24/25$ . Then $\sin\theta=\tan\theta\cos\theta=(-7/24)(-24/25)=7/25$ , giving $\cosec\theta=25/7$ .

Tier 3 · Hard

1. An angle $\theta$ lies in the first quadrant and satisfies $\sec\theta+\tan\theta=5$ . Find $\sin\theta$ exactly.[5 marks]
Answer
- $\sin\theta=12/13$
Method: Since $(\sec\theta+\tan\theta)(\sec\theta-\tan\theta)=\sec^2\theta-\tan^2\theta=1$ , it follows that $\sec\theta-\tan\theta=1/5$ . Adding the two equations gives $2\sec\theta=26/5$ , so $\sec\theta=13/5$ and $\tan\theta=12/5$ . Thus $\cos\theta=5/13$ and $\sin\theta=\tan\theta\cos\theta=(12/5)(5/13)=12/13$ .

5.6 · Understand and use double angle formulae; formulae for sin(A ± B), cos(A ± B), tan(A ± B) with geometrical proofs; express a cos θ + b sin θ in the form r cos(θ ± α) or r sin(θ ± α).

The compound-angle formulae are $\sin(A\pm B)=\sin A\cos B\pm\cos A\sin B$ , $\cos(A\pm B)=\cos A\cos B\mp\sin A\sin B$ and $\tan(A\pm B)=(\tan A\pm\tan B)/(1\mp\tan A\tan B)$ ; a geometrical proof can calculate the same unit-circle chord by coordinates and by the cosine rule.
Setting $A=B=\theta$ gives $\sin2\theta=2\sin\theta\cos\theta$ , $\cos2\theta=\cos^2\theta-\sin^2\theta=1-2\sin^2\theta=2\cos^2\theta-1$ and $\tan2\theta=2\tan\theta/(1-\tan^2\theta)$ .
To write $a\cos\theta+b\sin\theta=R\cos(\theta-\alpha)$ , compare coefficients to obtain $R\cos\alpha=a$ , $R\sin\alpha=b$ , hence $R=\sqrt{a^2+b^2}$ with the quadrant of $\alpha$ set by the signs.
For a compound-angle expression, the sign in the cosine formula reverses; for an $R$ -form, expanding the proposed form before choosing $\alpha$ prevents a wrong sign.

Tier 1 · Easy

1. Use a compound-angle formula to find the exact value of $\sin(75^\circ)$ .[3 marks]
Answer
- $\dfrac{\sqrt6+\sqrt2}{4}$
Method: Write $75^\circ=45^\circ+30^\circ$ . Then $\sin75^\circ=\sin45^\circ\cos30^\circ+\cos45^\circ\sin30^\circ=(1/\sqrt2)(\sqrt3/2)+(1/\sqrt2)(1/2)=(\sqrt6+\sqrt2)/4$ .

Tier 2 · Standard

1. Express $5\cos\theta-12\sin\theta$ as $R\cos(\theta+\alpha)$ , where $R>0$ and $0<\alpha<\pi/2$ . Hence state its maximum and minimum values.[5 marks]
Answer
- $13\cos(\theta+\alpha)$ , where $\alpha=\arctan(12/5)$
- Maximum $13$
- Minimum $-13$
Method: Expand $R\cos(\theta+\alpha)=R\cos\theta\cos\alpha-R\sin\theta\sin\alpha$ . Comparing coefficients gives $R\cos\alpha=5$ and $R\sin\alpha=12$ , so $R=\sqrt{25+144}=13$ and $\tan\alpha=12/5$ . Since cosine ranges from $-1$ to $1$ , the expression ranges from $-13$ to $13$ .

Tier 3 · Hard

1. Use two points on the unit circle and the cosine rule to prove geometrically that $\cos(A-B)=\cos A\cos B+\sin A\sin B$ .[5 marks]
Answer
- Equate the coordinate and cosine-rule expressions for the squared chord joining $(\cos A,\sin A)$ and $(\cos B,\sin B)$ .
Method: Let $P=(\cos A,\sin A)$ and $Q=(\cos B,\sin B)$ on the unit circle. By coordinates, $PQ^2=(\cos A-\cos B)^2+(\sin A-\sin B)^2=2-2(\cos A\cos B+\sin A\sin B)$ . In triangle $OPQ$ , $OP=OQ=1$ . If $\phi\in[0,\pi]$ is the smaller central angle, then $\cos\phi=\cos(A-B)$ , so the cosine rule gives $PQ^2=2-2\cos(A-B)$ . Equating the two expressions for $PQ^2$ and dividing by $-2$ proves $\cos(A-B)=\cos A\cos B+\sin A\sin B$ .

5.7 · Solve simple trigonometric equations in a given interval, including quadratic equations in sin, cos and tan and equations involving multiples of the unknown angle.

Solve first for the trigonometric ratio, use a reference angle and the signs in each quadrant, then list only solutions in the stated interval.
For a quadratic in one trigonometric function, substitute a temporary variable, factorise or use the quadratic formula, and reject any ratio outside its possible range before solving each remaining branch.
When the equation involves $kx$ , transform the given interval for $x$ into the corresponding interval for $kx$ , find every solution there, and divide only at the end.
Inverse-trigonometric buttons return a principal value rather than the full solution set; endpoints, excluded endpoints and degree-versus-radian mode must all be checked explicitly.

Tier 1 · Easy

1. Solve $\sin\theta=0.4$ for $0\leq\theta\leq2\pi$ , giving solutions to $3$ decimal places.[3 marks]
Answer
- $\theta=0.412$ or $2.730$
Method: The principal value is $\arcsin(0.4)=0.4115\ldots$ . Sine is positive in quadrants I and II, so the second solution is $\pi-0.4115\ldots=2.7301\ldots$ . Thus $\theta=0.412$ or $2.730$ .

Tier 2 · Standard

1. Determine all $x$ satisfying $2\cos^2x-3\cos x+1=0$ for $0\leq x<2\pi$ .[4 marks]
Answer
- $x=0,\ \pi/3,\ 5\pi/3$
Method: Factorise to $(2\cos x-1)(\cos x-1)=0$ . Hence $\cos x=1/2$ or $\cos x=1$ . In the interval, $\cos x=1/2$ at $x=\pi/3$ and $5\pi/3$ , while $\cos x=1$ at $x=0$ ; $2\pi$ is excluded.

Tier 3 · Hard

1. Find every solution of $\tan(2x)=-\sqrt3$ in the interval $-\pi/2\leq x\leq\pi$ .[5 marks]
Answer
- $x=-\pi/6,\ \pi/3,\ 5\pi/6$
Method: The transformed interval is $-\pi\leq2x\leq2\pi$ . Since tangent has period $\pi$ and reference angle $\pi/3$ , the solutions for $2x$ in this interval are $-\pi/3$ , $2\pi/3$ and $5\pi/3$ . Dividing each by $2$ gives $x=-\pi/6$ , $\pi/3$ and $5\pi/6$ .

5.8 · Construct proofs involving trigonometric functions and identities.

A trigonometric identity is true for every value in its domain, so a proof transforms one side into the other using exact algebra and established identities rather than testing selected angles.
Usually begin with the more complicated side, replace secant, cosecant, cotangent or tangent by sine and cosine when helpful, and factor or take a common denominator before cancelling.
Multiplying numerator and denominator by a conjugate can expose $1-\sin^2x=\cos^2x$ or $1-\cos^2x=\sin^2x$ and complete the proof cleanly.
Never cancel terms across addition, and record domain restrictions: algebra such as division by $\sin x$ is valid only where that denominator is non-zero.

Tier 1 · Easy

1. Prove that $\sin x\cot x=\cos x$ wherever the left-hand side is defined.[2 marks]
Answer
- Replace $\cot x$ by $\cos x/\sin x$ and cancel $\sin x$ .
Method: $\sin x\cot x=\sin x(\cos x/\sin x)=\cos x$ . The cancellation is valid wherever $\cot x$ is defined, namely where $\sin x\neq0$ .

Tier 2 · Standard

1. Prove that $\dfrac{1-\cos(2x)}{\sin(2x)}=\tan x$ for values at which both sides are defined.[3 marks]
Answer
- Use $1-\cos(2x)=2\sin^2x$ and $\sin(2x)=2\sin x\cos x$ .
Method: Apply the double-angle forms to the left-hand side: $(1-\cos2x)/\sin2x=(2\sin^2x)/(2\sin x\cos x)=\sin x/\cos x=\tan x$ , with cancellation only where the original expressions are defined.

Tier 3 · Hard

1. Prove that $\dfrac{1}{1-\sin x}-\dfrac{1}{1+\sin x}=2\tan x\sec x$ wherever the expressions exist.[5 marks]
Answer
- The left-hand side simplifies to $2\sin x/\cos^2x=2\tan x\sec x$ .
Method: Combine the fractions: the numerator is $(1+\sin x)-(1-\sin x)=2\sin x$ and the denominator is $(1-\sin x)(1+\sin x)=1-\sin^2x=\cos^2x$ . Thus the left-hand side is $2\sin x/\cos^2x=2(\sin x/\cos x)(1/\cos x)=2\tan x\sec x$ .

5.9 · Use trigonometric functions to solve problems in context, including problems involving vectors, kinematics and forces.

Resolve a vector of magnitude $V$ at angle $\theta$ to the positive horizontal into components $V\cos\theta$ and $V\sin\theta$ , changing signs to match its actual direction.
For resultant or equilibrium problems, form separate equations in two perpendicular directions; a zero resultant requires both component sums to equal zero.
In a kinematics model, trigonometric functions can describe direction or periodic motion, and the mathematical solution must be interpreted using the stated time interval, units and physical constraints.
A calculator angle without a quadrant check can point in the opposite direction; draw and label a diagram, then state bearings or directions in the form the context requests.

Tier 1 · Easy

1. A drone travels at $12\,\text{m s}^{-1}$ on a path $35^\circ$ above the horizontal. Calculate its horizontal and vertical velocity components to $3$ significant figures.[3 marks]
Answer
- Horizontal $=9.83\,\text{m s}^{-1}$
- Vertical $=6.88\,\text{m s}^{-1}$
Method: Resolve the velocity vector: the horizontal component is $12\cos35^\circ=9.829\ldots\,\text{m s}^{-1}$ and the vertical component is $12\sin35^\circ=6.882\ldots\,\text{m s}^{-1}$ .

Tier 2 · Standard

1. Two horizontal forces act on a crate: $8\,\text{N}$ due east and $11\,\text{N}$ at $60^\circ$ north of east. Find the magnitude and direction of their resultant, to $3$ significant figures and the nearest degree respectively.[5 marks]
Answer
- Magnitude $=16.5\,\text{N}$
- Direction $35^\circ$ north of east
Method: The east component is $8+11\cos60^\circ=13.5\,\text{N}$ and the north component is $11\sin60^\circ=9.526\ldots\,\text{N}$ . Hence $R=\sqrt{13.5^2+9.526^2}=16.523\ldots\,\text{N}$ . Its direction is $\arctan(9.526/13.5)=35.21\ldots^\circ$ north of east.

Tier 3 · Hard

1. A ring is in equilibrium under its weight of $18\,\text{N}$ , a tension $P$ directed $25^\circ$ above the horizontal, and a horizontal tension $Q$ acting oppositely. Calculate $P$ and $Q$ to $3$ significant figures.[5 marks]
Answer
- $P=42.6\,\text{N}$
- $Q=38.6\,\text{N}$
Method: Vertical equilibrium gives $P\sin25^\circ=18$ , so $P=18/\sin25^\circ=42.592\ldots\,\text{N}$ . Horizontal equilibrium then gives $Q=P\cos25^\circ=38.601\ldots\,\text{N}$ . Therefore $P=42.6\,\text{N}$ and $Q=38.6\,\text{N}$ .