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Arcs, chords, rays and angles

Start with the drawing Mark distinct points A and B on a circle. They divide it into two arcs, both including A and B. The chord AB is the straight segment between the points. An arc is not a segment. If the extremes coincide, we distinguish the null arc, which is reduced to the point, and the arc of a turn, which covers the entire circumference. This distinction will be important when comparing the final position with the turn performed.

October 21, 2026 Read
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Triangles that grow without changing shape

A comic-drawn ramp On a mesh, draw a path 3 comics to the right and 4 comics to the top. Connect the beginning to the end by a segment. You have formed a triangle with a straight corner: a right triangle. Now make a copy with all lengths folded. The horizontal path will have 6 comics; the vertical, 8. The lengths changed, but the corner openings remained the same. $$3\cdot2=6,\qquad4\cdot2=8.$$Why does bending just one side change the shape? Compare two copies: in the first, fold the horizontal and vertical path; in the second, fold only the horizontal. In the second copy the ramp is more laid down. Growing without changing shape requires expanding in all directions in the same proportion.

October 21, 2026 Read
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Corners, spokes and paths on the wheel

Two paths between the same points Draw a wheel and mark two points on it, A and B. Connect them with a straight line: it is a rope. Now follow the edge of the wheel from A to B: you have covered an arc. There are two possible paths around the edge; indicate which one you are using. The center is the point in the middle of the wheel. A segment from the center to the edge is a radius. All spokes on the same wheel have the same length.

October 21, 2026 Read
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Whole turns, half turns and quarters

Take a turn with your body Look at a wall. Spin around until you’re looking at the back wall: you’ve made a u-turn. Another half turn takes you back to the start. Now try to do the same spin in four equal parts. Each part is a quarter turn. A lap can be split in several ways. Two halves and four quarters complete the same lap: $$\frac12+\frac12=1,\qquad \frac14+\frac14+\frac14+\frac14=1.$$Here the number 1 represents a full turn. We are not measuring the length of the path your feet travel, but the turn of your direction.

October 21, 2026 Read
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Solving triangles and discussing the ambiguous case

What data determines a triangle? Solving a triangle is getting its sides and angles from enough data. Always use the convention: $a,b,c$ opposite $A,B,C$. A non-degenerate triangle has positive sides and angles between 0 and $\pi$. One side and two angles Knowing $a,B,C$, calculate $A=\pi-B-C$. $B+C<\pi$ is required. The law of sines gives: $$b=\frac{a\sin B}{\sin A},\qquad c=\frac{a\sin C}{\sin A}.$$The data determines a triangle, minus position and reflection. Two sides and the angle between them Knowing $b,c,A$, with $0

October 21, 2026 Read
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Heights, Heron, Medians, angle bisectors and Radii

Get special measurements from the sides Consider sides $a,b,c$, area $K$ and semiperimeter $s=(a+b+c)/2$. We require a non-degenerate triangle: positive sides and each side smaller than the sum of the other two. Demonstrate Heron’s formula From the law of cosines, $\cos A=(b^2+c^2-a^2)/(2bc)$. In $K=bc\sin A/2$, use $\sin^2A=1-\cos^2A$: $$\begin{aligned} 16K^2&=4b^2c^2-(b^2+c^2-a^2)^2\\ &=((b+c)^2-a^2)(a^2-(b-c)^2)\\ &=(a+b+c)(b+c-a)\\ &\quad\cdot(a+b-c)(a-b+c)\\ &=16s(s-a)(s-b)(s-c). \end{aligned}$$The area is positive, so: $$K=\sqrt{s(s-a)(s-b)(s-c)}.$$Heights As $K=ah_a/2=bh_b/2=ch_c/2$: $$h_a=\frac{2K}{a},\qquad h_b=\frac{2K}{b},\qquad h_c=\frac{2K}{c}.$$Substituting Heron’s expression gives each height as a function of the sides only. $h_a=b\sin C=c\sin B$ and its permutations are also valid.

October 21, 2026 Read
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Laws of sines, cosines, projections and areas

Fix the notation In a non-degenerate triangle, the angles $A,B,C$ are opposite the sides $a,b,c$, respectively. All sides are positive and $A+B+C=\pi$. Demonstration of the law of cosines Place vertex A at the origin, B at $(c,0)$ and C at $(b\cos A,b\sin A)$. The distance between B and C is $a$. By Pythagoras’ theorem applied to coordinate differences: $$\begin{aligned} a^2&=(b\cos A-c)^2+(b\sin A)^2\\ &=b^2(\cos^2A+\sin^2A)+c^2-2bc\cos A\\ &=b^2+c^2-2bc\cos A. \end{aligned}$$The argument also applies to obtuse angles. If $A=\pi/2$, the cosine term is zero and we find Pythagoras again. Permuting the letters gives the other two cosine laws.

October 21, 2026 Read
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Sine, cosine and tangent inequalities

Choose arcs, not just points An equation like $\sin x=m$ marks meeting points with a horizontal line. The inequality $\sin x>m$ chooses the arcs above this line; $\sin x

October 21, 2026 Read
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Arcsine, arccosine and arctangent

Why is it necessary to restrict? An inverse needs to retrieve a single input from the output. Sine, cosine and tangent repeat values. To define inverses, we choose intervals in which each function is one-to-one and reaches its entire desired image. Restricted function Chosen domain Image Sine $[-\pi/2,\pi/2]$ $[-1,1]$ Cosine $[0,\pi]$ $[-1,1]$ Tangent $(-\pi/2,\pi/2)$ $\mathbb R$ In these intervals, sine and tangent are strictly increasing, and cosine is strictly decreasing. Each output appears only once; therefore, the inverses are well defined.

October 21, 2026 Read
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Intervals and classical trigonometric equations

Filter a family To solve $\sin x=1/2$ to $[0,4\pi)$, first obtain the families $x=\pi/6+2k\pi$ and $x=5\pi/6+2k\pi$. Then impose the limits of the range. The solutions are: $$\frac\pi6,\quad\frac{5\pi}6,\quad\frac{13\pi}6,\quad\frac{17\pi}6.$$The inclusion sign of the ends matters. After squaring or multiplying by a variable expression, substitute the answers into the initial equation. A combination of sine and cosine To solve $a\sin x+b\cos x=c$, with $(a,b)\ne(0,0)$, set $R=\sqrt{a^2+b^2}$ and choose $\varphi$ with: $$\cos\varphi=\frac aR,\qquad\sin\varphi=\frac bR.$$This is possible because the squares of these two numbers add up to 1. Using the addition formula:

October 21, 2026 Read
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Fundamental equations of sine, cosine and tangent

The equation looks for inputs A trigonometric equation asks for the values ​​of the variable that make an equality true. Before transforming the expression, determine its domain. Divisions by functions that can be zero can eliminate solutions. For $\sin x=m$ or $\cos x=m$, $-1\le m\le1$ is required. For the tangent, any real $m$ is possible, but entries where $\cos x=0$ remain prohibited. Demonstration of sine families A horizontal line cuts the cycle at a maximum of two points. If one of them corresponds to $\alpha$, the other corresponds to $\pi-\alpha$. As full turns repeat the stitch:

October 21, 2026 Read
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Transform sums into products and products into sums

Add two known formulas Write the formulas for $\sin(a+b)$ and $\sin(a-b)$. Adding them together, the terms with $\cos a\sin b$ cancel each other out: $$\sin(a+b)+\sin(a-b)=2\sin a\cos b.$$Now do $u=a+b$ and $v=a-b$. Then $a=(u+v)/2$ and $b=(u-v)/2$. This produces the first identity below; the other cases come from adding or subtracting the sine and cosine formulas: $$\sin u+\sin v=2\sin\frac{u+v}{2}\cos\frac{u-v}{2},$$ $$\sin u-\sin v=2\cos\frac{u+v}{2}\sin\frac{u-v}{2},$$ $$\cos u+\cos v=2\cos\frac{u+v}{2}\cos\frac{u-v}{2},$$ $$\cos u-\cos v=-2\sin\frac{u+v}{2}\sin\frac{u-v}{2}.$$These identities are valid for all reals $u,v$. In the latter case, the negative sign is essential.

October 21, 2026 Read