3 Geometry and trigonometry

Use distance and midpoint in three dimensions.

Treat a 3D point as coordinates (x,y,z). Distance extends Pythagoras and midpoint averages corresponding coordinates.

Worked example

Between A(1,−2,3) and B(5,4,−1), distance is √68 and midpoint is (3,1,1).

The z-coordinate contributes just like x and y.

A midpoint is a point; divide each coordinate by 2 after adding.

Solid and spatial checks: $V_{pyramid}=Bh/3$, $V_{cone}=\pi r^2h/3$, $V_{sphere}=4\pi r^3/3$; total sphere area is $4\pi r^2$, while a solid hemisphere including its base has area $3\pi r^2$. In a $3\times4\times12$ cuboid, the base diagonal is $5$ and the space diagonal is $\sqrt{5^2+12^2}=13$. Its angle $\alpha$ with the base satisfies $\tan\alpha=12/5$, so $\alpha\approx67.38^\circ$.

Choose the right triangle rule for a triangle problem.

Use Pythagoras for a right triangle; use sine/cosine rules or 1/2ab sin C when the triangle is not right-angled.

Worked example

With a=7,b=9 and C=60°, area is 1/2(7)(9)sin60°.

Name the known sides and included angle before choosing a formula.

The sine rule is not a universal replacement for the cosine rule.

Formula map: in a right triangle, $\sin A=opposite/hypotenuse$, $\cos A=adjacent/hypotenuse$, and $\tan A=opposite/adjacent$. For any triangle, $a/\sin A=b/\sin B=c/\sin C$, $c^2=a^2+b^2-2ab\cos C$, and $Area=\tfrac12ab\sin C$. With $a=7$, $b=9$, $C=60^\circ$, the cosine rule gives $c^2=49+81-126(1/2)=67$, hence $c=\sqrt{67}$, while the area is $63\sqrt3/4$ square units.

Translate a written context into a trigonometric diagram.

Draw a labelled triangle, mark the reference direction and identify elevation, depression or bearing before calculating.

Worked example

A bearing of 060° is clockwise from north; a 20 m line at 30° elevation gives height 10 m.

The diagram fixes which angle and side the calculator should use.

Bearings are not measured from the positive x-axis unless converted.

Use radians for arc length and sector area.

Radians link angle to arc length: s=rθ and sector area=1/2r²θ when θ is in radians.

Worked example

For r=6 and θ=π/3, arc length is 2π and sector area is 6π.

Convert degrees before using the radian formulas.

Using degrees directly in s=rθ gives the wrong scale.

Use the unit circle for exact trigonometric values.

On the unit circle, cosθ is x and sinθ is y; quadrant signs determine exact values and tanθ=sinθ/cosθ.

Worked example

At 5π/6, sinθ=1/2 and cosθ=−√3/2, so tanθ=−1/√3.

Reference angles provide magnitude; the quadrant supplies the sign.

Do not make sine and cosine positive in every quadrant.

Ambiguous sine-rule case: if $a=8$, $b=10$ and $A=30^\circ$, then $\sin B=10\sin30^\circ/8=0.625$. Thus $B\approx38.68^\circ$ or $180^\circ-38.68^\circ=141.32^\circ$; both are valid because each gives $A+B<180^\circ$. Always test the supplementary angle against the triangle angle sum instead of accepting only the calculator's principal value.

Transform identities instead of guessing angles.

The identity sin²θ+cos²θ=1 and double-angle formulas rewrite expressions without solving for θ.

Worked example

If sinθ=3/5 and θ is acute, cos2θ=1−2sin²θ=7/25.

Choose an identity that matches the information already given.

An identity is true for every allowed angle, not a numerical approximation.

Complete double-angle set: $\sin2\theta=2\sin\theta\cos\theta$ and $\cos2\theta=\cos^2\theta-\sin^2\theta=1-2\sin^2\theta=2\cos^2\theta-1$. If $\sin\theta=3/5$ and $\theta$ is acute, then $\cos\theta=4/5$, so $\sin2\theta=24/25$ and $\cos2\theta=7/25$. Without a quadrant condition, the sign of the missing ratio may not be unique.

Read amplitude, period and shifts from a trigonometric model.

In y=a sin(b(x+c))+d, |a| is amplitude, 2π/|b| period, c horizontal shift and d the midline.

Worked example

For y=3sin(2x−π)+4, amplitude=3, period=π, shift right π/2 and midline y=4.

Rewrite the inside as b(x+c) before reading the shift.

The inside coefficient changes period and horizontal shift, not vertical amplitude.

Graph family boundaries: sine and cosine have amplitude $|a|$ and period $2\pi/|b|$ (or $360^\circ/|b|$). Tangent has no amplitude, period $\pi/|b|$ (or $180^\circ/|b|$), zeros at integer multiples of its period and vertical asymptotes halfway between. A periodic model must state units, midline and the time represented by one complete period.

Solve trigonometric equations on a stated interval.

Find a reference angle, use quadrant symmetry, and list only solutions inside the requested interval.

Worked example

sin x=1/2 on [0,2π] gives x=π/6 and 5π/6.

The interval determines which repeated solutions are included.

A calculator principal value is not the complete interval solution.

Quadratic reduction example on $0\le x\le2\pi$: $2\sin^2x-3\sin x+1=0$ factors as $(2\sin x-1)(\sin x-1)=0$. Hence $\sin x=1/2$ or $\sin x=1$, giving $x=\pi/6,\pi/2,5\pi/6$. Check every candidate in the original equation and report only values in the stated finite interval; no general solution is required.

Reciprocal and inverse trigonometric functions undo different operations.

Reciprocal functions invert a value, while inverse trigonometric functions return an angle; the notation and domain restrictions are different.

Example

For sin θ=0.6 with θ acute, θ=sin⁻¹(0.6)≈36.9°. By contrast, csc θ=1/sin θ, so csc θ≈1.67.

Choose inverse trig when the unknown is an angle; choose a reciprocal when the operation is division by the trig value.

sin⁻¹x is not 1/sin x, and inverse-trig answers must be checked against the stated interval.

Reciprocal definitions: $\sec\theta=1/\cos\theta$, $\cosec\theta=1/\sin\theta$, and $\cot\theta=1/\tan\theta$, wherever the denominator is non-zero. Hence $1+\tan^2\theta=\sec^2\theta$ and $1+\cot^2\theta=\cosec^2\theta$. Principal inverse ranges are $\arcsin x\in[-\pi/2,\pi/2]$ for $x\in[-1,1]$, $\arccos x\in[0,\pi]$ for $x\in[-1,1]$, and $\arctan x\in(-\pi/2,\pi/2)$ for all real $x$.

Compound-angle identities rewrite a difficult angle.

The addition and subtraction identities express sin(A±B) and cos(A±B) using known values of A and B, so an unfamiliar angle can be decomposed into familiar ones.

Example

sin 75°=sin(45°+30°)=sin45°cos30°+cos45°sin30°=(√6+√2)/4.

Choose a decomposition that gives exact known angles, expand once, then simplify and check the sign from the quadrant.

An identity is true for every allowed angle; a numerical equation may have only selected solutions.

Identity set: $\sin(A\pm B)=\sin A\cos B\pm\cos A\sin B$ and $\cos(A\pm B)=\cos A\cos B\mp\sin A\sin B$. Dividing the sine expansion by the cosine expansion gives $\tan(A\pm B)=(\tan A\pm\tan B)/(1\mp\tan A\tan B)$ when defined; setting $A=B=\theta$ gives $\tan2\theta=2\tan\theta/(1-\tan^2\theta)$. Denominators and quadrant signs remain part of the answer.

Supplementary angles reveal trigonometric graph symmetry.

The AHL 3.11 identities are $\sin(\pi-\theta)=\sin\theta$, $\cos(\pi-\theta)=-\cos\theta$, and $\tan(\pi-\theta)=-\tan\theta$. They follow from the unit circle: supplementary angles have the same vertical coordinate and opposite horizontal coordinates.

Example

If $\theta=\pi/6$, then $\sin(5\pi/6)=1/2$, $\cos(5\pi/6)=-\sqrt3/2$, and $\tan(5\pi/6)=-1/\sqrt3$.

Locate $\pi-\theta$ in quadrant II, then keep the sine sign and reverse the cosine and tangent signs.

The identity describes supplementary-angle symmetry; it does not say that every trigonometric function keeps its value under $\theta\mapsto\pi-\theta$.

Vectors encode direction and magnitude together.

A vector separates how far something moves from where it points, allowing displacement, velocity and geometric relationships to be combined component by component.

Example

The displacement from A(1,2,0) to B(4,0,6) is (3,−2,6), with length √49=7.

Subtract initial coordinates from final coordinates, then use components for addition and the norm for magnitude.

A position vector is not the same object as a displacement; keep the origin and direction convention explicit.

Component toolkit: for $\mathbf v=(v_1,v_2,v_3)=v_1\mathbf i+v_2\mathbf j+v_3\mathbf k$, $|\mathbf v|=\sqrt{v_1^2+v_2^2+v_3^2}$ and the unit vector in its direction is $\mathbf v/|\mathbf v|$ when $\mathbf v\ne\mathbf0$. If $\overrightarrow{OA}=\mathbf a$ and $\overrightarrow{OB}=\mathbf b$, then $\overrightarrow{AB}=\mathbf b-\mathbf a$; vector equalities can prove midpoint or parallelogram properties component by component.

The scalar product tests angles and perpendicularity.

a·b=|a||b|cosθ converts two vectors into a number, so its sign and value reveal the angle between them.

Example

(2,1,0)·(1,−2,0)=2−2=0, so the vectors are perpendicular.

Use components for calculation, then interpret zero, positive or negative as right, acute or obtuse angle evidence.

The scalar product is commutative and produces a scalar; it is not the vector product and cannot give a direction.

For non-zero vectors $\mathbf a=(a_1,a_2,a_3)$ and $\mathbf b=(b_1,b_2,b_3)$, $\mathbf a\cdot\mathbf b=a_1b_1+a_2b_2+a_3b_3$ and $\cos\theta=(\mathbf a\cdot\mathbf b)/(|\mathbf a||\mathbf b|)$. Zero dot product is equivalent to perpendicularity; parallel vectors satisfy $|\mathbf a\cdot\mathbf b|=|\mathbf a||\mathbf b|$, with the sign distinguishing the same and opposite directions.

A vector line is a point plus a direction.

The equation r=a+λd generates every point reached from a known point a by scaling the direction vector d.

Example

The line through (1,2,0) parallel to (2,−1,3) is r=(1,2,0)+λ(2,−1,3); λ=2 gives (5,0,6).

Match coordinates to solve for a parameter, then compare direction vectors to test parallelism or intersection.

Different equations can describe the same line; a shared point alone does not prove two lines intersect unless the parameters agree.

Equivalent forms for a line through $(x_0,y_0,z_0)$ with direction $(l,m,n)$ are $\mathbf r=\mathbf a+\lambda\mathbf d$, $x=x_0+\lambda l$, $y=y_0+\lambda m$, $z=z_0+\lambda n$, and $(x-x_0)/l=(y-y_0)/m=(z-z_0)/n$ where the displayed denominators are non-zero. For two lines, use their direction vectors in the dot-product formula for the acute angle. In a motion model, $\lambda$ may represent time, $\mathbf d$ velocity and $|\mathbf d|$ speed.

Classify 3D lines by solving their parameter conditions.

Two lines may intersect, be parallel, or be skew; comparing direction vectors and solving all three coordinate equations distinguishes the cases.

Example

If a+λd=b+μe has one consistent pair (λ,μ), the lines intersect; if d is a scalar multiple of e they are parallel or coincident.

Solve coordinates together rather than checking one projection, then verify the resulting point in all three components.

Crossing in an x–y sketch does not prove a 3D intersection; skew lines can have matching projections.

The vector product creates a perpendicular direction.

a×b is perpendicular to both vectors and has magnitude |a||b|sinθ, making it useful for normals and areas.

Example

(1,0,0)×(0,2,0)=(0,0,2), whose magnitude 2 is the area of the rectangle spanned by the vectors.

Use the determinant pattern, then apply the right-hand rule to interpret orientation and check the magnitude.

Reversing the order changes the sign: a×b=−(b×a); parallel vectors give the zero vector.

For $\mathbf a=(a_1,a_2,a_3)$ and $\mathbf b=(b_1,b_2,b_3)$, compute $\mathbf a\times\mathbf b=(a_2b_3-a_3b_2,\ a_3b_1-a_1b_3,\ a_1b_2-a_2b_1)$. Its magnitude is the area of the spanned parallelogram, so the corresponding triangle area is $\tfrac12|\mathbf a\times\mathbf b|$. The zero vector occurs for non-zero parallel inputs, while order controls the normal's direction.

A plane is fixed by a point and a normal vector.

A normal n is perpendicular to every direction in the plane, so n·(r−a)=0 gives a compact equation for the plane through a.

Example

Through (1,0,2) with normal (2,1,−1), the plane is 2(x−1)+y−(z−2)=0.

Substitute the known point to determine the constant, then test any proposed point with the dot product equation.

A direction vector lying in the plane is perpendicular to the normal; it is not itself the plane’s normal.

A plane through point vector $\mathbf a$ can be written $\mathbf r=\mathbf a+\lambda\mathbf b+\mu\mathbf c$ using two non-parallel directions in the plane, or $\mathbf r\cdot\mathbf n=\mathbf a\cdot\mathbf n$ using a normal $\mathbf n$. If $\mathbf n=(A,B,C)$, this becomes $Ax+By+Cz=d$. To convert from two in-plane directions to normal form, use $\mathbf n=\mathbf b\times\mathbf c$.

3D intersections and angles come from shared equations.

An intersection is a point satisfying both objects; angles are then found from direction vectors or normals rather than from a misleading 2D sketch.

Example

Substitute a line into a plane equation to solve λ, then use the resulting point to verify the intersection and compute any requested angle.

Solve the shared condition first, check it in the original equations, and use the relevant dot-product angle formula only afterwards.

A line parallel to a plane has no intersection unless it lies in the plane; supplementary angle conventions must be stated.

Intersection workflow: substitute a line into a plane, or solve the simultaneous Cartesian equations for two or three planes, then interpret no solution, one solution or a family geometrically. If $\alpha$ is the acute angle between line direction $\mathbf d$ and plane normal $\mathbf n$, the line-plane angle is $90^\circ-\alpha$, equivalently $\sin\phi=|\mathbf d\cdot\mathbf n|/(|\mathbf d||\mathbf n|)$. For planes, use their normals in $\cos\theta=|\mathbf n_1\cdot\mathbf n_2|/(|\mathbf n_1||\mathbf n_2|)$.