Study Companion: Programme F.9 — Trigonometry
You have met \sin, \cos and \tan as buttons on a calculator and as "SOH CAH TOA", and you have probably used a^2+b^2=c^2. Programme F.9 makes them forced rather than memorised. Two facts about triangles do all the work: similar triangles have their sides in the same ratios, so a ratio of sides depends only on the angle; and Pythagoras' theorem, which ties the three sides together. Every exact value, every identity and even the compound-angle formulas in this chapter are those two facts, read carefully off a drawing.
The rules, procedures and drill are in the companion reference sheet, which follows the book's topic order. This file tells the story in 6 units instead, because the programme's many topics are really 6 ideas.
Scope, as in the book: every angle here is an angle of a right-angled triangle, so it lies between 0^\circ and 90^\circ. Programme F.11 extends \sin and \cos to every angle by rotation and shows that nothing proved here has to be unlearned.
How to use this companion
Each unit is an argument: a real situation and a question, the attempt a sensible person would make first, the picture that settles it, a check with numbers, and only then the rule. Read it once front to back. The Narration spine at the end of each unit summarises what the video shows.
Unit 01: An Angle Is an Amount of Turning
The problem
A drive belt runs round a pulley of radius 0.2 m. The pulley turns through a quarter of a revolution, 90^\circ. How much belt passes over it? More generally: is there a way of measuring angle that turns "how far round" into "how far along" with no conversion factor at all?
First attempt
The arc is "radius times angle", so try s=r\theta=0.2\times90=18 m. That is 18 metres of belt for a quarter turn of a pulley whose whole rim is about 1.26 m long. The formula is not wrong; the unit is. A degree is \frac1{360} of a turn, a size chosen by the Babylonians, and nothing about it is tied to the radius. Multiplying a length by a count of degrees gives a number with no geometric meaning.
The picture
An angle is produced by turning: a line pivots about one end and sweeps round. The book names the landmarks: a full turn is a full angle, 360^\circ; half of it is a straight angle, 180^\circ; a quarter is a right angle, 90^\circ. An angle smaller than a right angle is acute, one between 90^\circ and 180^\circ is obtuse.
Now measure the turning with the radius itself. Take the radius, a straight piece of length r, and bend it onto the rim, starting where the line began:
The angle it covers is one radian: the turn that carries the tip one radius along the rim. Keep laying radii end to end round the rim. The rim is 2\pi r long, so exactly 2\pi\approx6.283 radii fit round a full turn:
Because the angle is now counted in radii of arc, the arc for any angle is just that count times the radius, and s=r\theta holds with no stray factor.
Check it with numbers
A quarter turn is \frac14\times2\pi=\frac\pi2=1.5708 rad. The belt that passes is
a quarter of the 1.2566 m rim, as it must be. The failed attempt was off by a factor of 18/0.3142\approx57.3, and that factor is exactly how many degrees make one radian:
The rule it gives
Common angles are best left as multiples of \pi: 30^\circ=\frac\pi6, 45^\circ=\frac\pi4, 90^\circ=\frac\pi2, 120^\circ=\frac{2\pi}3, 270^\circ=\frac{3\pi}2. In the other direction, 2.34\text{ rad}=2.34\times\frac{180}\pi=134.1^\circ.
No unit means radians. The book writes \sin2 for the sine of 2 radians and keeps the degree sign whenever degrees are meant. A calculator has to be told which: in radian mode \sin2=0.9093, in degree mode it returns \sin2^\circ=0.0349. Check the mode before every calculation.
Degrees, minutes and seconds (DMS). A degree is split into 60 minutes and a minute into 60 seconds: 1^\circ=60', 1'=60''. That is place value in base 60 (the idea of F01 Unit 01), so converting to decimal degrees is reading each place at its weight:
and in radians 63.21\times\frac\pi{180}=1.1032 rad. Going the other way, the fractional part of a degree times 60 gives minutes, and the fractional part of the minutes times 60 gives seconds:
Worked example
Convert 45^\circ36'18'' to radians, to 4 dp.
Where this shows up
A winch drum of radius 0.3 m turns through 40^\circ. In radians that is 40\times\frac\pi{180}=0.6981 rad, so the cable wound in is s=r\theta=0.3\times0.6981=0.2094 m. The same relation divided by time gives the cable speed: a drum turning at 2 rad/s winds cable at v=r\omega=0.3\times2=0.6 m/s. That clean v=r\omega, with no \frac\pi{180} anywhere, is why every rotating-machine formula in engineering uses radians.
Narration spine. A pulley and a belt; the naive s=r\theta with degrees gives an absurd 18 m and is struck out. A line turns about its end; the radius bends onto the rim to mark one radian, and radii keep being laid round until 2\pi of them close the circle. A tracker sweeps the angle while its value in radians is read as arc divided by radius. The quarter turn gives 0.314 m of belt, and the lost factor 57.3 is revealed as one radian in degrees.
Unit 02: Similar Triangles Make the Ratios Well Defined
The problem
A surveyor lays a 50 m line up a uniform slope that rises at 12^\circ to the horizontal. How much height does the line gain, and how far does it reach horizontally? A number for "how steep" must exist, since the slope is uniform, but where does it come from?
First attempt
Height seems to grow with the angle, so share it out proportionally: at 90^\circ the line would rise its full 50 m, so at 12^\circ it rises \frac{12}{90}\times50=6.67 m. Draw the triangle to scale and measure it, though, and the rise is about 10.4 m. The proportional guess fails because equal steps of angle do not give equal steps of height: the tip of the line moves along a circle, climbing steeply at first and hardly at all near the top.
The picture
A scale drawing did give the right answer, and it is worth asking why a drawing on paper can answer a question about a hillside. Any two right-angled triangles with the same angle \theta have the same three angles (the third is 90^\circ-\theta in both), so they are similar: one is the other enlarged. Enlargement multiplies every side by the same factor k, and a ratio of two sides does not notice k:
So within one triangle, \frac{\text{opposite}}{\text{hypotenuse}} is fixed by the angle alone. The book's own example: triangles with AB=2, AC=5, BC=4 cm and a similar one with A'B'=3 cm have scale factor \frac32, so A'C'=7.5 cm and B'C'=6 cm, and every ratio of two sides is the same in both triangles.
Watch the triangle grow with its angle held at 12^\circ: the sides all change, the ratios never do. That fixed ratio is the number the question needs.
Check it with numbers
For 12^\circ the fixed ratios are \frac{\text{opp}}{\text{hyp}}=0.2079 and \frac{\text{adj}}{\text{hyp}}=0.9781. The survey line is the hypotenuse, so
The rise matches the scale drawing, not the 6.67 m guess.
The rule it gives
In a right-angled triangle with an acute angle \theta, name the sides relative to \theta: the hypotenuse faces the right angle, the opposite side faces \theta, the adjacent side runs from \theta to the right angle. The trigonometric ratios are
They are well defined, depending on \theta and not on the size of the triangle, because of similarity. A calculator stores them: \sin27^\circ=0.4540, \cos84^\circ=0.1045, \tan43^\circ=0.9325.
Turning each fraction upside down gives the reciprocal ratios:
There are no calculator buttons for them; use the reciprocal key: \cot12^\circ=\frac1{0.21256}=4.7046, \sec37^\circ=1.2521, \operatorname{cosec}71^\circ=1.0576. They earn their place when the unknown is the hypotenuse: the answer is then known side times reciprocal ratio, with no division to rearrange.
Worked example
A strut is to reach 5 m up a wall, meeting the ground at 43^\circ. How long is it? The strut is the hypotenuse and 5 m is opposite the angle:
Where this shows up
A 3 m ladder leaning at 56^\circ to the ground reaches 3\sin56^\circ=3\times0.8290=2.49 m up the wall; its foot stands 3\cos56^\circ=1.68 m out. Every force on an inclined member is resolved the same way: a 500 N pull along a cable at 12^\circ to the horizontal has a vertical component 500\sin12^\circ=104.0 N and a horizontal component 500\cos12^\circ=489.1 N, the survey numbers scaled by ten.
Narration spine. A 50 m survey line at 12^\circ; the proportional guess of 6.67 m is drawn against the true triangle and struck out. A right triangle with its angle locked at 12^\circ grows under a tracker while live readouts show its sides changing and its side ratios standing still. The ratios get their names, the survey line's rise and run come out as 10.40 m and 48.91 m on the same triangle, and flipping each ratio gives cosec, sec and cot, used at once on a strut.
Unit 03: Pythagoras, Its Converse, and the Fundamental Identity
The problem
A builder setting out the corner of a slab pegs 3 m along one edge and 4 m along the other, then adjusts until the diagonal between the pegs measures exactly 5 m, and declares the corner square. Why does 5 m guarantee a right angle? And for a roof whose rafter spans 4.8 m horizontally while rising 1.4 m, how long is the rafter?
First attempt
The rafter goes across 4.8 and up 1.4, so try 4.8+1.4=6.2 m. But that is the length of the path that goes along and then up; the rafter cuts the corner and must be shorter. Adding lengths along two directions at right angles does not give the length of the straight line between the ends.
The picture
Take four copies of any right-angled triangle, legs a and b, hypotenuse c, and place them inside a square of side a+b, one in each corner. They leave a tilted square of side c in the middle:
Now slide the four triangles, without turning them, into two rectangles along the diagonal of the big square. The uncovered area becomes a square of side a and a square of side b. The big square is the same, and the four triangles are the same, so the uncovered area is the same in both arrangements:
That is a proof, not an illustration: it holds for every right-angled triangle, because nothing in the rearrangement depended on the particular a and b.
The builder needs the theorem the other way round: the squares add up, so is the angle right? That is the converse, and it needs its own argument. Suppose a triangle has sides a, b, c with a^2+b^2=c^2. Build a genuine right angle with legs a and b; by Pythagoras its third side is \sqrt{a^2+b^2}=c. The two triangles have the same three sides, so they are the same shape (congruent), and the first triangle's angle is right too.
Check it with numbers
The builder's triangle: 3^2+4^2=9+16=25=5^2, so the corner is square. The rafter: c^2=4.8^2+1.4^2=23.04+1.96=25, so c=5.0 m, well short of the 6.2 m guess.
The test also detects a triangle that is not right-angled: 7,24,25 gives 49+576=625=25^2 (right-angled), while 5,11,12 gives 25+121=146\ne144 (not right-angled).
The rule it gives
Pythagoras' theorem: the square on the hypotenuse of a right-angled triangle is equal to the sum of the squares on the other two sides, a^2+b^2=c^2. Its converse, the book's test: if the squares on the two shorter sides add to the square on the longest, the triangle is right-angled; if they do not, it is not.
To find a side, rearrange: with hypotenuse 8 and one side 3, the other is \sqrt{64-9}=\sqrt{55}=7.416.
The fundamental identity. Divide Pythagoras by c^2, in a triangle where \theta has adjacent side a and opposite side b:
(\sin^2\theta means (\sin\theta)^2.) It is written with \equiv because it is an identity: true for every angle \theta, not an equation to be solved for particular ones. Here "every angle" means every acute angle, the only kind a right-angled triangle has; F.11 shows it survives for all angles.
Dividing the same identity by \cos^2\theta, and then by \sin^2\theta, gives two more identities:
The identity is also a right-angle test in ratio form. For sides 8,12,10 the would-be \cos and \sin are \frac8{12} and \frac{10}{12}, and \frac{64}{144}+\frac{100}{144}=\frac{164}{144}\ne1: not right-angled.
Worked example
Is a triangle with sides 9, 40, 41 right-angled? The longest side is 41: 9^2+40^2=81+1600=1681=41^2, so yes, with the right angle opposite the 41 side. In ratio form, \left(\frac9{41}\right)^2+\left(\frac{40}{41}\right)^2=\frac{1681}{1681}=1.
Where this shows up
Setting out a building by "3-4-5" is still done on site with a tape, and it scales: 6, 8, 10 m is used for large slabs, since the diagonal error is easier to read on a longer tape. A 0.1 m error in a 10 m diagonal means c^2=98.01 or 102.01 instead of 100, a corner off square by about 1.2^\circ, which a builder can see in a brick course.
Narration spine. A builder's 3-4-5 corner and a roof rafter; the guess 4.8+1.4=6.2 m is struck because the straight line cuts the corner. Four copies of the triangle sit in a square around a tilted square c^2, then slide without turning to leave a^2 and b^2: the uncovered area never changed. The numbers 3, 4, 5 go into the picture, the converse is argued by building a second triangle, and dividing by c^2 turns the theorem into \cos^2\theta+\sin^2\theta\equiv1 beside the same triangle.
Unit 04: Two Triangles Give Exact Values
The problem
A calculator says \sin30^\circ=0.5 exactly but \sin27^\circ=0.45399\ldots, a decimal that never ends. It also says \sin60^\circ=0.8660254\ldots: is that another endless decimal, or an exact number in disguise? A tent whose front is an equilateral triangle has a centre pole \sqrt3 m tall. How long are its sloping sides, exactly?
First attempt
Sixty degrees is twice thirty, so perhaps \sin60^\circ=2\sin30^\circ=1. But a sine of 1 would need the opposite side as long as the hypotenuse, a triangle with no adjacent side at all, and the calculator gives 0.866. As in Unit 02, a ratio of sides does not scale with the angle.
The picture
Two shapes whose angles are known exactly produce right-angled triangles whose sides can be found by Pythagoras.
Halve a square of side 1 along its diagonal. Each half is a right-angled isosceles triangle with angles 45^\circ, 45^\circ, 90^\circ and hypotenuse \sqrt{1^2+1^2}=\sqrt2.
Halve an equilateral triangle of side 2 by its altitude. The altitude bisects the base, so each half has hypotenuse 2, short side 1, angles 30^\circ at the top and 60^\circ at the base, and height \sqrt{2^2-1^2}=\sqrt3.
Read the ratios straight off the sides.
Check it with numbers
From the half equilateral, \sin60^\circ=\frac{\sqrt3}2=0.8660, exactly the calculator's decimal, so 0.8660254\ldots is \frac{\sqrt3}2 in disguise. From the half square, \sin45^\circ=\frac1{\sqrt2}=0.7071. And 2\sin30^\circ=1\ne0.8660, confirming the doubling guess was wrong.
The rule it gives
| \theta | \sin\theta | \cos\theta | \tan\theta |
|---|---|---|---|
| 30^\circ=\frac\pi6 | \frac12 | \frac{\sqrt3}2 | \frac1{\sqrt3} |
| 45^\circ=\frac\pi4 | \frac1{\sqrt2} | \frac1{\sqrt2} | 1 |
| 60^\circ=\frac\pi3 | \frac{\sqrt3}2 | \frac12 | \sqrt3 |
Nothing here needs memorising: sketch the two half-shapes and read the table off them. Two patterns fall out of the drawing. The 30^\circ and 60^\circ angles sit in the same triangle, so the side opposite one is adjacent to the other: \sin30^\circ=\cos60^\circ and \sin60^\circ=\cos30^\circ. And \tan60^\circ=\sqrt3=\frac1{\tan30^\circ}, the same two sides read in the other order. (\frac1{\sqrt2} is also written \frac{\sqrt2}2, and \frac1{\sqrt3} as \frac{\sqrt3}3.)
Worked example
The tent. Its front is equilateral, so each half is the half-equilateral triangle, with the pole as the side opposite the 60^\circ base angle:
exactly, with no rounding anywhere.
Where this shows up
A timber prop braced at 45^\circ with its foot 3.4 m from the wall is the hypotenuse of a half square: \frac{3.4}{L}=\cos45^\circ=\frac1{\sqrt2}, so L=3.4\sqrt2=4.81 m. A roof pitched at 30^\circ over a half-span of 4 m has rafters \frac4{\cos30^\circ}=\frac8{\sqrt3}=4.62 m long and a ridge 4\tan30^\circ=\frac4{\sqrt3}=2.31 m above the eaves. Exact values let a designer keep the \sqrt3 until the very last line, so rounding happens once.
Narration spine. \sin30^\circ is a clean 0.5 but \sin60^\circ looks endless; the doubling guess \sin60^\circ=1 is struck. A unit square is cut along its diagonal to give 1,1,\sqrt2; an equilateral triangle of side 2 is cut by its altitude, and Pythagoras produces \sqrt3 on screen. The exact ratios are read off each half, \sqrt3/2 is matched to the calculator's 0.8660, and the tent's side comes out as exactly 2 m.
Unit 05: Verifying Identities
The problem
Simplifying a formula before coding it, an engineer meets \frac{1-\cos^2\theta}{\sin\theta\cos\theta} and suspects it is just \tan\theta. Is it, for every angle? Checking every angle one by one is impossible, so what would count as settling it?
First attempt
Try an angle. At \theta=30^\circ:
It works, so is it proved? Try the same test on a different claim, 2\sin\theta\cos\theta\equiv\tan\theta. At \theta=45^\circ both sides are 1, a perfect match. At \theta=30^\circ, though, the left side is 2\cdot\frac12\cdot\frac{\sqrt3}2=0.866 and the right is 0.577. The claim is false, and a lucky angle made it look true. A numerical check can refute an identity, but it can never prove one.
The picture
The honest route here is algebraic, and the picture is the triangle that supplies its one fact. In the right-angled triangle, 1-\cos^2\theta is not a new quantity: it is \frac{c^2-a^2}{c^2}=\frac{b^2}{c^2}=\sin^2\theta by Pythagoras. That is the fundamental identity of Unit 03, rearranged, and it is the move that unlocks most identities in this chapter.
Every line is equal to the one above it for every \theta (every acute \theta; \sin\theta and \cos\theta are never zero there, so the cancelling is allowed). A chain of identities from the left side to the right side is a proof.
Check it with numbers
Put \theta=30^\circ through the chain: \frac14\big/\frac{\sqrt3}4, then \frac14\big/\frac{\sqrt3}4 again (since \sin^230^\circ=\frac14), then \frac12\big/\frac{\sqrt3}2, then \frac1{\sqrt3}=0.5774. Every line gives the same number, as it must when each step is an identity.
The rule it gives
To verify an identity (the book's term):
- Start from one side, usually the more complicated, and transform it step by step until it becomes the other side.
- Rewrite \tan, \sec, \operatorname{cosec} and \cot in terms of \sin and \cos.
- Add fractions over a common denominator.
- Use \cos^2\theta+\sin^2\theta\equiv1 (or one of its two relatives) to remove a 1-\cos^2\theta, a 1-\sin^2\theta or a \sin^2\theta+\cos^2\theta.
Sometimes it is quicker first to do the same thing to both sides, as the book does when it multiplies both sides of \frac{1+\sin\theta}{\cos\theta}\equiv\frac{\cos\theta}{1-\sin\theta} by \cos\theta(1-\sin\theta) to get 1-\sin^2\theta\equiv\cos^2\theta. That is legitimate only because the step can be undone: the multiplier is never zero for an acute angle, so dividing by it takes you straight back. Squaring both sides cannot be undone (-1\ne1, but their squares are equal), so it is never a valid step in a proof of an identity.
Worked example
Verify \frac1{1-\cos\theta}+\frac1{1+\cos\theta}\equiv2\operatorname{cosec}^2\theta.
The same pattern gives \tan\theta+\cot\theta\equiv\sec\theta\operatorname{cosec}\theta: over a common denominator the numerator is \sin^2\theta+\cos^2\theta=1.
Where this shows up
A ramp rises at angle \theta over a horizontal run d. Its sloping length is L=d\sec\theta and its rise is h=d\tan\theta. A spreadsheet computing L^2-h^2 for many ramps can be replaced by one number, since \sec^2\theta-\tan^2\theta\equiv1 gives L^2-h^2=d^2 whatever the angle. For d=6 m and \theta=20^\circ: L=6.385 m, h=2.184 m, and L^2-h^2=40.77-4.77=36.00=6^2 m^2. The identity turned a trigonometric calculation into none at all.
Narration spine. The claim \frac{1-\cos^2\theta}{\sin\theta\cos\theta}\equiv\tan\theta passes a test at 30^\circ; then a false claim passes a test at 45^\circ and fails at 30^\circ, and is struck: spot checks refute but never prove. Beside a triangle whose angle a tracker sweeps, live readouts of the true identity's two sides stay equal while the false one's drift apart. The proof chain is built one line at a time, each line transformed out of the one above.
Unit 06: Compound Angles from Stacked Triangles
The problem
A robot arm has a joint at 45^\circ above the horizontal and a second link, 0.4 m long, that turns a further 30^\circ at the elbow. The second link therefore points at 75^\circ to the horizontal. How far does it reach horizontally, exactly? The exact-value table of Unit 04 stops at 30^\circ, 45^\circ and 60^\circ; is \cos75^\circ out of reach?
First attempt
75^\circ=45^\circ+30^\circ, so try \cos75^\circ=\cos45^\circ+\cos30^\circ=0.7071+0.8660=1.5731. That is impossible: a cosine is adjacent over hypotenuse, and no side is longer than the hypotenuse, so a cosine is at most 1. Cosine does not share out over a sum of angles, any more than the square root does over a sum of numbers.
The picture
Build the angle \theta+\varphi out of two right-angled triangles stacked on each other, the book's own construction. Give the outer line AB length 1.
- In triangle AEB (angle \varphi at A, hypotenuse 1): AE=\cos\varphi and BE=\sin\varphi.
- In triangle AFE (angle \theta at A, hypotenuse AE): AF=AE\cos\theta=\cos\theta\cos\varphi.
- The angle at B in the small triangle BDE is also \theta: triangles AXC and BXE both have a right angle and share the vertically opposite angle at X, so their third angles match. In triangle BDE the side DE is opposite that angle and BE is the hypotenuse, so DE=BE\sin\theta=\sin\theta\sin\varphi.
Now read \cos(\theta+\varphi) off the figure. It is \frac{AC}{AB}=AC, and AC is AF with a piece CF cut off the end. CF is the same length as DE (opposite sides of the rectangle CFED), so
The same figure gives the sine as well. BC=BD+DC, where BD=BE\cos\theta=\cos\theta\sin\varphi and DC=EF=AE\sin\theta=\sin\theta\cos\varphi:
(The figure needs \theta+\varphi acute to be drawn as shown. F.11 shows the formulas hold for all angles.)
Check it with numbers
Put \theta=45^\circ, \varphi=30^\circ into the same figure:
which is exactly what the calculator gives for \cos75^\circ, and nowhere near the 1.5731 of the first attempt.
The rule it gives
Identities for compound angles:
Differences, without any negative angles. Let \alpha=\theta-\varphi with \theta>\varphi, so that \theta=\alpha+\varphi and the sum formulas give \cos\theta=\cos\alpha\cos\varphi-\sin\alpha\sin\varphi and \sin\theta=\sin\alpha\cos\varphi+\cos\alpha\sin\varphi. Multiply the first by \cos\varphi, the second by \sin\varphi, and add: the \sin\alpha terms cancel and \cos^2\varphi+\sin^2\varphi=1 leaves \cos\alpha. Doing the same with a subtraction isolates \sin\alpha:
Tangents. Divide \sin(\theta+\varphi) by \cos(\theta+\varphi) and then divide top and bottom by \cos\theta\cos\varphi:
Double angles: put \varphi=\theta, then use the fundamental identity for the other two forms of \cos2\theta:
Products of ratios: add or subtract the sum and difference formulas, and the cross terms cancel:
Sums and differences of ratios: read the products backwards. Write A=\frac{\theta+\varphi}2 and B=\frac{\theta-\varphi}2, so that A+B=\theta and A-B=\varphi; then 2\sin A\cos B=\sin\theta+\sin\varphi, and so on:
Note the -2 in the last one: it comes from the minus sign in 2\sin A\sin B=\cos(A-B)-\cos(A+B), read the other way round. For example, \sin5\theta+\sin3\theta\equiv2\sin4\theta\cos\theta.
Beyond the book: rearranging the two forms of \cos2\theta gives \sin^2\theta\equiv\frac{1-\cos2\theta}2 and \cos^2\theta\equiv\frac{1+\cos2\theta}2, which F.13 uses to integrate \sin^2 and \cos^2.
Worked example
Find \cos15^\circ exactly. 15^\circ=60^\circ-45^\circ:
Where this shows up
The robot arm: its first link, 0.5 m long at 45^\circ, and the second, 0.4 m long at 45^\circ+30^\circ, put the tool at
A robot controller computes exactly these sums, and it uses the compound-angle formulas to update \cos and \sin of the new angle from the old ones without recomputing them from scratch.
Narration spine. A two-link arm points its second link at 45^\circ+30^\circ; the guess \cos45^\circ+\cos30^\circ=1.573 is struck because a cosine cannot exceed 1. The stacked-triangle figure is built one line at a time, the equal angle \theta at B is justified at the crossing point, and the lengths AF=\cos\theta\cos\varphi and DE=\sin\theta\sin\varphi are coloured on the figure. AC=AF-DE is read off as \cos(\theta+\varphi), the numbers 0.6124-0.3536=0.2588 go into the same figure, and BC=BD+DC gives the sine formula beside it.