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.