Chapter 11.3 — Exercise 11.2 — Specific Angles

45°, From a Square Cut in Half

An isosceles right triangle with both legs equal to a has hypotenuse √(a²+a²)=√2·a — this is literally a square cut along its diagonal, with the 45° angle sitting at each of the two corners the diagonal doesn't pass through.

Sine and cosine come out identical here — both 1/√2 — precisely because the two legs are equal by construction. Tangent, opposite over adjacent, collapses to exactly 1 for the same reason: a÷a is always 1, regardless of what a itself happens to be.

30° and 60°, From an Equilateral Triangle's Altitude

In this construction, every side of the equilateral triangle is 2a, so BD=a — the altitude from a vertex of an equilateral triangle always bisects the opposite side exactly, and it also bisects the 60° angle it starts from, splitting it into two 30° halves. The altitude AD=√((2a)²−a²)=√3·a. Inside right triangle ABD (∠B=60°, ∠BAD=30°):

0° and 90°, the Limiting Cases

As an angle shrinks toward 0°, the "opposite" side shrinks toward nothing while the adjacent side grows to match the hypotenuse — giving sin0°=0, cos0°=1, tan0°=0. Push the same angle toward 90° instead, and the roles swap entirely: sin90°=1, cos90°=0, with tan90° left undefined, since it would require dividing by a side that's shrunk to zero.

These two extreme cases aren't genuine right triangles anymore in the usual sense — one angle has effectively vanished to zero, collapsing the triangle flat — which is exactly why tan0° and cot90° both come out as a clean 0 (a real side divided by a still-finite one) while cot0° and tan90° come out undefined (a real side divided by one that's shrunk away to nothing). Being comfortable with "undefined" as a genuine, legitimate answer, rather than assuming every trigonometric ratio must always produce some finite number, matters here.

The Complete Table

Sine and cosine mirror each other exactly across this table — sin's row read left-to-right is cos's row read right-to-left — which is worth using as a built-in memory check rather than memorising twelve separate values independently. Tangent and cotangent mirror each other the same way, and cosecant and secant do too, which means memorising just sin, tan, and cosec at all five angles is genuinely already enough on its own — the other three rows are just those same values read in reverse order.

Evaluating Expressions Directly

The last two rows aren't arbitrary substitution practice — they're the double-angle identities for tangent, verified here at one specific angle rather than proved in general: 2tan30°/(1+tan²45°) simplifying to tan30° previews tan(2A) in one disguised form, and 2tan30°/(1−tan²30°) landing exactly on tan60°=tan(2×30°) previews the genuine double-angle formula for tangent.

The first four rows are worth a second glance too, purely for how differently each one reaches its final answer. The first collapses two identical values into a doubled surd; the second and third both lean on a Pythagorean identity hiding inside a familiar-looking expression; the fourth is really just cot30° written out in an unfamiliar form. Spotting which shortcut applies before grinding through the raw substitution is a skill worth building deliberately, not just something that happens automatically with practice.

Testing the Sine and Cosine Addition Formulas

Two problems check whether sin(A+B) and cos(A+B) can be built from the separate individual ratios of A and B, using A=60° and B=30° as a concrete numeric test case.

Both formulas check out at this one pair of angles — sin(A+B)=sinA cosB+cosA sinB, and cos(A+B)=cosA cosB−sinA sinB. A third, tempting-looking formula gets tested immediately after and fails: sin(A+B) is not simply sinA+sinB. At A=60°, B=30°: sin90°=1, but sin60°+sin30°=(√3+1)/2≈1.366 — genuinely different numbers. Trigonometric ratios of a sum almost never distribute the way multiplication does over addition; the correct addition formulas above are the only reliable route.

It's worth being genuinely honest about what checking a formula at one single pair of angles actually proves and what it doesn't. Confirming sin(A+B)=sinA cosB+cosA sinB works at A=60°, B=30° shows the formula is consistent with that one case — it doesn't, on its own, prove the formula true for every possible pair of angles. A full proof needs a genuinely different argument, one that never substitutes specific numbers at all; a single numeric check is really only useful for building confidence, or for catching a formula that's already wrong, the way the sinA+sinB guess was caught here.

Two Triangles Solved at 30° and 60°

• △PQR, right angle at Q, PQ=6 cm, ∠RPQ=60°: tan60°=QR/6 gives QR=6√3 cm; cos60°=6/PR gives PR=12 cm — two entirely different ratios, each pulling one unknown length straight out of the table above.
• △XYZ, right angle at Y, YZ=x, XZ=2x: sinX=YZ/XZ=x/2x=1/2=sin30°, so ∠YXZ=30° and ∠YZX=60° (using the fact that a triangle's three angles always sum to 180°).

The second problem is worth noticing for what it doesn't need at all: no actual numeric value for x ever gets found, since the ratio x/2x=1/2 holds regardless of x's size — the angle is recovered from a ratio alone, with the specific length left completely unknown. The first problem, by contrast, needs both a specific angle and a specific length to reach actual measured lengths — a genuinely different kind of question dressed up in a similar-looking triangle.

From Fixed Angles to Any Angle's Complement

Every value in this exercise came from one of exactly five fixed angles. Exercise 11.3 generalises far beyond that short list, connecting the ratios of any angle to the ratios of whatever angle it's paired with to sum to 90° — a rule that applies to every angle, not just the five covered here. For the isosceles and equilateral triangle derivations behind this exercise's table, revisit the sections above, or return to the chapter introduction for the six ratio definitions this entire table, and every value inside it, is ultimately built from.