Triangle Trigonometry

Sine, Cosine and Tangent #

Each ratio pairs the angle with two sides.

Each ratio pairs the angle with two particular sides of a right triangle

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Name the sides from the angle you care about: opposite, adjacent, hypotenuse.

Opposite over hypotenuse never changes for that angle. That is sine: sin = opp ÷ hyp.

Adjacent over hypotenuse is cosine: cos = adj ÷ hyp. Same idea, different pair.

Tangent skips the hypotenuse: tan = opp ÷ adj.

On a circle of radius 1 the height is the sine and the distance across is the cosine.

Exact Sine, Cosine and Tangent at 30, 45 and 60° #

Nine values, cut from two familiar shapes.

Cutting a square and an equilateral triangle in half gives ratios that stay exact

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Cut a square of side 1 corner to corner. Each half has two 45° corners.

Both legs are 1, so by Pythagoras the hypotenuse is $\sqrt{2}$√2.

So $\sin 45^{\circ }=\cos 45^{\circ }=\frac{1}{\sqrt{2}}$sin 45° = cos 45° = 1/√2, and tan 45° = 1.

Fold an equilateral triangle of side 2 in half. The base halves to 1, the apex to 30°.

Here is one half, turned to lie flat. By Pythagoras $2^2-1^2=3$2² − 1² = 3, so that side is $\sqrt{3}$√3.

Nine values from two shapes. Forget one and you can cut the shape again.

Sine and Cosine of Complementary Angles #

One leg faces one corner and lies beside the other.

Sine and cosine trade places at complementary angles because one triangle names both

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The two angles beside the right angle add to 90°, so one is 90° minus the other.

The gold side of 3 faces x°. That same side lies beside the corner (90 − x)°.

Both ratios divide that same 3 by that same 5, so both come to the same number.

This is the cofunction relationship: each ratio is the other at the complement.

Check it against the exact values: sin 30° and cos 60° are both one half.

Solving Right Triangles #

One side and one angle give you the rest.

One side and one angle are enough to find every other part of a right triangle

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We know the hypotenuse is 10 and the angle is 30°. The opposite side is missing.

Pick the ratio that links the side you know to the side you want.

$\sin 30^{\circ }=\frac{1}{2}$sin 30° = 1/2, so the opposite side is 5.

Finding an Angle from a Ratio #

Sine, cosine and tangent, run backwards.

An inverse trig ratio runs the ratio backwards and hands you the angle

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Both sides are known this time, and the angle is not. The problem runs the other way.

The ratio is known. It is the angle inside it that is not.

${\tan }^{-1}$tan⁻¹ undoes the tangent: give it a ratio and it returns the angle that made it.

Equal sides give a tangent of 1, and ${\tan }^{-1}1$tan⁻¹ 1 is 45°. Worth knowing by heart.

Elevation and Depression #

Angles measured up or down from level.

An angle of elevation looks up from the horizontal and depression looks down

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Standing back from a tower, the angle up from the horizontal is the elevation.

From 40 meters away the sight line rises at 45°. Tangent links height to distance.

tan 45° = 1, so the height matches the distance: 40 meters.

Looking down from the cliff top, the angle below the horizontal is the depression.

The horizontal at the eye runs parallel to the sea, so the boat looks up at 45°.

tan 45° = 1, so the boat lies 30 meters out from a cliff 30 meters high.

Sine and Cosine of Angles Past 90 Degrees #

The unit circle keeps going where a triangle stops.

Past 90 degrees the sine matches the acute angle at 180 minus it and the cosine turns negative

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On a circle of radius 1 the distance across is the cosine and the height is the sine.

Keep turning past 90° and the point crosses to the left of the center.

The two points sit at the same height, so sin 120° is exactly sin 60°.

The distance across is mirrored, not copied: cos 120° is −½ where cos 60° is ½.

Compare any obtuse angle with 180 minus itself: same sine, opposite cosine.

The Graphs of Sine and Cosine #

The unit circle's height, unrolled into a wave.

Unrolled from the circle, sine and cosine draw one wave 90 degrees apart

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Turn the radius and its tip has a height — the sine of the angle turned through.

Unroll the turn, one cell per 90°: the heights draw a wave from 1 down to −1.

The distance across draws the same wave slid 90° along — cosine, starting from 1.

A full turn brings the point home, so the wave repeats itself every 360°.

The Graph of Tangent #

The curve that escapes to infinity.

Tangent divides sine by cosine, so it breaks at 90 degrees and repeats every 180

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Tangent is a division: the sine on top and the cosine underneath.

One cell is 45°. cos 90° is 0, and dividing by 0 has no answer — the graph breaks there.

Past 90° the sine stays positive but the cosine turns negative — so tan is negative.

Half a turn flips sine and cosine together, so tangent repeats every 180°, not 360°.

Amplitude and the Principal Axis #

Read both off the highest and lowest values.

The principal axis sits halfway between the highest and lowest values, and the amplitude is the distance from it to either extreme

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One cell is 90°. The plain sine wave climbs to 1 and falls to −1.

y = 2 sin x + 3 swings about the dashed line y = 3 instead of about the x-axis.

It reaches 5 at the crest and drops to 1 in the trough.

Halfway between the two is the principal axis; half the gap is the amplitude.

So a is the amplitude, and d is the height the principal axis sits at.

The Period and Phase Shift of a Wave #

One number sets the period, one shifts the wave.

The number multiplying x divides the period and a number subtracted from x slides the wave along

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y = sin 2x fits two whole waves where the dashed y = sin x fits one.

The 2 inside the bracket makes x count twice as fast, so one wave fits in half a turn.

In general the period is 360° divided by whatever multiplies x.

y = sin(x − 90°) is the very same wave, slid 90° to the right.

Four numbers, four jobs: amplitude a, period $360/b$360 / b, phase shift c, level d.

Solving Simple Trigonometric Equations #

One sine value names two angles.

One sine value names two angles in a turn, x and 180 − x; cosine pairs x with 360 − x

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This is the ratio question run backwards: which angle has a sine of one half?

But the height ½ is hit twice in a turn — at 30°, and again at 180 − 30 = 150°.

Every sine equation gives such a pair: x, and 180 − x at the same height.

Cosine pairs differently: the same cosine returns at 360 − x, so 60° pairs with 300°.

The Principal Values of the Inverse Functions #

One agreed window each; symmetry does the rest.

Each inverse function returns one agreed window of angles; symmetry supplies the rest

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$\sin \theta =0.5$sin θ = 0.5 names many angles: 30°, 150°, and again every turn. The inverse picks one.

Each inverse keeps one window: ${\sin }^{-1}$sin⁻¹ and ${\tan }^{-1}$tan⁻¹ center on zero, ${\cos }^{-1}$cos⁻¹ runs 0° to 180°.

The inverse hands back the principal angle; symmetry supplies the rest of them.

Quadratic Equations in Sine, Cosine or Tangent #

Substitute a letter and factor as usual.

An equation quadratic in one ratio becomes an ordinary quadratic under a substitution

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Write s for sin x and the equation is one you already know how to factor.

Factor it and read off the roots: s = −½ and s = 1.

Put sin x back. The cut at −½ meets the wave twice, at 210° and 330°.

The other root, sin x = 1, sits at the peak: the full turn gives 90°, 210° and 330°.

The same move on a cosine equation, and this time one root cannot be a cosine.

Reject the root outside −1 to 1, then solve what is left: 60° and 300°.

Trigonometric Equations with a Multiple Angle #

Stretch the range before you solve it.

An equation in a multiple of the angle needs the range stretched before the answers are shrunk back

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Call the whole angle u. The equation is now simple — but it is solved in u, not x.

If x runs from 0° to 360°, then u = 2x runs from 0° all the way to 720°.

Over those two turns the cut at ½ meets the wave four times, not twice.

Halve each answer to come back to x, and all four land inside 0° to 360°.

A shift inside moves the window too: double the range, then slide it back.

Solve for the whole angle, then undo the shift and the doubling last of all.

Modeling Tides and Daylight with a Sine Function #

Read the level, the amplitude and the period from data.

A quantity that rises and falls on a fixed cycle is modeled by a sine whose numbers come from the data

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The graph shows harbor depth against hours: it climbs to 8 m, falls to 2 m, repeats.

The level it swings about is 5 m, and it travels 3 m either side of that.

One cycle every 12 hours puts 360 ÷ 12 = 30 inside the bracket, and the model is built.

Put a time in and the model gives the depth: at t = 1 hour it is 6.5 m.

Run it backwards and the model becomes an equation: 6.5 m at t = 1 and t = 5.

Daylight hours follow the same shape: level 12, amplitude 2.5, period 12 months.

Area with Sine #

Two sides and the angle between them.

Two sides and the angle between them give the area of any triangle

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Half base times height needs a height, and it is neither of the sides you know.

Drop the height h onto a. It makes a right triangle whose hypotenuse is b.

In that right triangle $\sin C=h/b$sin C = h / b, so the height itself is b sin C.

Substituting gives Area = ½ a b sin C, and no height is needed.

The Sine Rule #

Each side over the sine of its opposite angle.

In any triangle $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$a/sin A = b/sin B = c/sin C, the same value for every side

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Every side has one angle facing it from across the triangle.

Drop the height h from A. It splits the triangle into two right triangles.

One height, two right triangles: h is c sin B on one and b sin C on the other.

Divide both sides by sin B sin C and each side sits over the angle facing it.

Know one side, the angle facing it, and one more angle: the rule then finds b.

The Cosine Rule #

Pythagoras, with a correction for the angle.

The cosine rule is Pythagoras with a correction term for an angle that is not 90 degrees

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With a right angle, $c^2=a^2+b^2$c² = a² + b².

No right angle now. Drop the height: it is b sin C, as the area rule showed.

Beside C the foot cuts off b cos C, so the rest of side a is a − b cos C.

Apply Pythagoras to the right triangle holding c and multiply out: that is the rule.

At 90° cos C = 0, so the correction term vanishes and Pythagoras is left.

Finding Angles with the Sine Rule #

The rule runs backwards to name an angle.

Rearranged for a sine, the rule turns three known parts into the missing angle

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The sine rule links side-angle pairs — and it can find an angle, not just a side.

Side 8 faces 30° and side 16 faces B: three of the four parts are known.

Rearrange for sin B and work it out: 16 × ½ ÷ 8 = 1.

A sine of 1 can only be 90°. The special values hand back their angles the same way.

The Ambiguous Case of the Sine Rule #

One sine, two possible triangles.

One sine value names two candidate angles, and each must still fit inside 180

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One sine names two angles: 30° and 150° stand level, both with a sine of ½.

Take A = 20° with sin B = ½. Choosing B = 30° closes a triangle: C is 130°.

The same data, drawn again: B = 150° closes too, with a 10° sliver left over.

The check: each candidate must leave the third angle above zero. Here both pass.

A big enough given angle rules the obtuse candidate out: with A = 75°, only 30° fits.

Finding Angles with the Cosine Rule #

Three sides, and the angle comes out alone.

Three sides pin the cosine of any angle, and its sign settles acute or obtuse

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With all three sides known, the rule holds exactly one unknown: the cosine of C.

Rearranged, the cosine comes out alone — ready to take three sides at once.

The sides are 3, 5 and 7: the cosine comes out at exactly −½, so C is 120°.

No second candidate here: one cosine names one angle, and its sign says which kind.

Trigonometry in Three Dimensions #

Flat triangles hiding inside a box.

A box hides flat right triangles, and two rounds of Pythagoras walk its diagonals

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Two diagonals hide in a box: one across the floor, one through the air.

The floor is a flat 4 by 3 rectangle, so its diagonal is $\sqrt{16+9}=5$√(16 + 9) = 5.

A triangle stands on that diagonal: base 5, height 2 — the space diagonal is $\sqrt{29}$√29.

The diagonal’s angle with the floor lives in that flat triangle: $\tan \theta =2/5$tan θ = 2 / 5.

Exact Trigonometric Ratios with Surds #

The side comes out exact, not rounded.

Keep the answer in surd form and it stays exact — a decimal is already rounded

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The hypotenuse is 8 and the angle is 60°, so the opposite side is $8\sin 60^{\circ }=8\times \frac{\sqrt{3}}{2}=4\sqrt{3}$8 sin 60° = 8 × √3/2 = 4√3.

The height is $5\tan 30^{\circ }=\frac{5}{\sqrt{3}}$5 tan 30° = 5/√3. Rationalize it: multiply numerator and denominator by $\sqrt{3}$√3 to get $\frac{5\sqrt{3}}{3}$5√3/3.

Squaring shows the difference: ${\left(4\sqrt{3}\right)}^2=16\times 3=48$(4√3)² = 16 × 3 = 48 exactly, while the rounded $6.93^2=48.0249$6.93² = 48.0249.

Bearings with the Sine and Cosine Rules #

The journey’s triangle, solved from its bearings.

The bearings tell you the angle inside the triangle; the cosine rule does the rest

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A ship sails 8 km on bearing 040°, then 5 km on bearing 160°. How far is it back to A?

The heading changes by 160° − 40° = 120° at B, so the angle inside the triangle is 60°.

Two sides and the included angle: the cosine rule gives the third side, d = 7 km.