Trigonometry: A Crash Review

Easily review trigonometry as you work at a computer!

Note: There are several kludge notations lurking in this crash review:

• SQRT for "square root" (correctly notated for 2, 3, 5, and 6)
• pi for π (correctly notated outside of popup text for pictures)

Trigonometry Survival 101 -- basic material [not coordinate geometry]

Trigonometry Survival 201 -- useful material, but not in your standard Trigonometry course

Formula Lists

Self-quizzes

A Vastly Incomplete Selection of Web Resources

Their names are:

• sine [sin]
• 
• cosine [cos]
• 
• tangent [tan]
• cotangent [cot]
• secant [sec]
• cosecant [csc]

Incidentally, there are three angle units listed on my HP-48SX: degrees, radians, and gradients. [Gradients are not commonly used in the U.S.A.] To rebuild the conversion formulae, remember:

1 circle = 360 degrees = 2π radians = 400 gradients.

• We inherited degrees from Babylonia. There are 60 minutes in a degree, and 60 seconds in a minute. [When dealing with the Earth's rotation in astronomy, the Earth rotates "about" 1 angular degree in 4 solar-day minutes. The Earth rotates 1 degree in 4 sidereal-day minutes.] Also, a nautical mile is "about" 1 arc-minute on the Earth's surface. This would be exact if the Earth was a perfect sphere. A common notation for, say, 3 degrees 4 minutes 5 seconds is 3°4'5".
• Radians are the "natural" angle unit when dealing with "stratosphere" math. The defining formulae for sin and cos are much simpler then. Radians also behave better if you have to deal with calculus [deriviatives and integrals].
• I don't know much about the history of gradients.

For the record: Triangles are constructed by picking three points, then joining each pair by a straight line segment. We assume we can neglect curvature of the surface. That is, we assume the geometry is Euclidean. [You will not get the same kind of triangles on a sphere as on a flat desk or piece of paper. The sphere is visibly curved, the desk and the flat piece of paper are not supposed to be curved.] I will occasionally use the fact the geometry is Euclidean.

The sum of the three (interior) angles of a triangle is half a circle, i.e. 180° or π radians or 200 gradients. I will sometimes denote this by 2 in the rest of this crash review; this is not standard notation. An implicit exercise will be translating 2 to angular measure [degrees, radians, and/or gradients.].

If we pick a point on a straight line, the two parts of the straight line formed by taking out the point form an angle of half a circle, i.e. 180° i.e. π radians i.e. 200 gradients. In general, I will omit "interior" when talking about angles of a triangle.

We say a triangle, that contains a right angle, is a right triangle. If a triangle has two sides of equal length and one side of different length, we call it an isosceles triangle. If a triangle has three equal sides, we say it is an equilateral triangle.

Note:

• For an isosceles triangle, the two angles opposite the sides with equal length (i.e., the two angles whose bounding sides have one of the two sides of equal length, and the side with different length) are equal.
• For an equilateral triangle, all three angles are equal, and equal 60° i.e. π/3 radians i.e. 200/3 gradients.

• (arc length)=RX.

the generic right triangle, and the unit circle.

Here, A, H, and O are acronyms for Adjacent [to angle], Hypotenuse, and Opposite [from angle], respectively.

First of all, all physical angles have some size. We cannot visualize an angle with negative physical size. They do not exist in anything sufficiently similar to the physical space we live in.

However, (especially when dealing with the unit circle), it is often convenient to measure angles in a specific direction: counterclockwise. [I'm not going to digress into "stratosphere" math now, but that is dictating the convention here.]

Then, a negative sign means we are measuring the angle "unconventionally", i.e. clockwise.

This will simplify the use of some of the trigonometric identities we are going to look at.

Given a right triangle, the trig function values for the two acute angles [angles smaller than a right angle] can be computed without knowing the angles. I prefer to remember the formulae this way [X is an angle]:

• sin(X) = O/H; csc(X) = H/O
• cos(X) = A/H; sec(X) = H/A
• tan(X) = O/A; cot(X) = A/O

If H=1, then O and A are simply cos(X) and sin(X).

A mnemonic for the formulae for sin(X), cos(X), and tan(X) is the (fictitious) Indian Chief Soh-cah-toa, who had no problems with this part of trigonometry. [I got this from Mr. Coole, a long time ago -- I was in grade school then.]

The way I wrote the formulae, above, emphasizes the following identities:

• sin(X)csc(X)=1
• cos(X)sec(X)=1
• tan(X)cot(X)=1

Another fact, of some use, is the Pythagorean theorem: H²=A²+O². If we relabel A as a, O as b, and H as c, we get the familiar form of the Pythagorean theorem: c²=a²+b². This only works for right triangles. The generalization of the Pythagorean theorem to non-right triangles is called the law of cosines.

A common strategy is to memorize how to compute tan(X), cot(X), sec(X), and csc(X) in terms of sin(X) and cos(X), and then reduce everything to this. This is not necessarily the least painful way to do a trig problem, but it is often more important to start the problem, than figure out how to do it elegantly. [EXERCISE: derive this from the A, H, O formulation for acute angles. The formulae work for arbitrary angles.]

• tan(X)=sin(X)/cos(X)
• cot(X)=cos(X)/sin(X)
• sec(X)=1/cos(X)
• csc(X)=1/sin(X)

First of all, we say two angles are complementary if they add up to a right angle. That is, for an angle X, we say -X is complementary to X.

For instance, the angle in the upper right-hand corner is complementary to, i.e. the complement of, the marked angle in the lower left-hand corner.

A related piece of terminology is supplementary angles: two angles are said to be supplementary if they add up to half a circle, i.e. 180° i.e. π radians i.e. 200 gradients. That is, 2 - X is supplementary to the angle X. To get an intuition for this, draw a unit circle with the horizontal axis.. Draw an arbitrary radius from the center to somewhere on the unit circle. The two angles formed between the radius, and the horizontal axis, are supplementary.

Now, let's look at the trig function-computation rules again:

• sin(X) = O/H; csc(X) = H/O
• cos(X) = A/H; sec(X) = H/A
• tan(X) = O/A; cot(X) = A/O

What you should get are the cofunction identities:

• sin(-X)=cos(X)
• cos(-X)=sin(X)
• tan(-X)=cot(X)
• cot(-X)=tan(X)
• sec(-X)=csc(X)
• csc(-X)=sec(X)

These identities do work for arbitrary angles. If one side is undefined, both sides are undefined.

The reference triangles are right triangles that are "easy to construct". They provide easily-memorized values for the angles with measure 30°, 45°, and 60°, i.e. π/6, π/4, and π/3 radians, i.e. 100/3, 50, and 200/3 gradients.

There are two reference triangles. They [in the U.S.A.] are known by their degree names:

• 45-45-90 [constructed from square]
• 30-60-90 [constructed from equilateral triangle]

When we put A=1=O and H= into the generic right triangle, we get:

• sin(45°) = /2
• cos(45°) = /2
• tan(45°) = 1
• cot(45°) = 1
• sec(45°) =
• csc(45°) =

The 30-60-90 triangle is constructed from an equilateral triangle with sides of length 2. We put one side on the horizontal axis, and bisect the angle opposite this side. [Note that an equilateral triangle has three vertex angles, all of which are 60°.]

We now have a hypotenuse of length 2, one leg of length 1 [from bisecting the horizontal side], and one side with length [from the bisecting line].

We get:

• sin(30°)=cos(60°)=½
• cos(30°)=sin(60°)=/2
• tan(30°)=cot(60°)=1/=/3
• cot(30°)=tan(60°)=
• sec(30°)=csc(60°)=2/=(2)/3
• csc(30°)=sec(60°)=2

The unit circle provides a picture on which to memorize reference values of the trig functions.

Think of it as the set of possible endpoints for a length 1 hypotenuse H, with one endpoint of H fixed at the origin. We can construct generic right triangles with hypotenuse 1 in it. Pick a point on the circumference, draw a line segment from it to the origin, and then draw a perpendicular line segment down to the x-axis.

Notice that the coordinates of the vertex, in Cartesian coordinates, is (cos(X), sin(X)), where X is the central angle. The horizontal side (on the x-axis) is A, and the vertical side (parallel to the y-axis) is O. The radius (length 1) is H. The slope of the hypotenuse H is tan(X).

However, observe the four quadrants. Our example triangle has its hypotenuse in the upper-right quadrant [both coordinates positive]. Horizontal and vertical hypotenuses create line segments rather than triangles. Also, at least one of the coordinates go negative in the other three quadrants [upper-left, lower-left, lower-right].

We proceed by assuming that the coordinates of the vertex, in Cartesian coordinates, is (cos(X), sin(X)), regardless of where the vertex is on the unit circle. This immediately leads us to the Pythagorean identities:

• sin²(X)+cos²(X)=1 [standard; if you memorize only one, learn this one]
• tan²(X)+1=sec²(X) [divide by cos²(X)]
• 1+cot²(X)=csc²(X) [divide by sin²(X)]

Incidentally, the computation of the slope of the hypotenuse H via tan(X) also works for arbitrary angles. [Recall that an undefined slope corresponds to a vertical line].

The next point is that the signs of the various trigonometric functions are controlled by the quadrant the function is evaluated in. We number the quadrants I through IV [Roman numerals] counterclockwise, as follows:

Now, the trigonometric function signs are controlled by the quadrants as follows:

• Quadrant I:
  • sin(X), csc(X) positive
  • cos(X), sec(X) positive
  • tan(X), cot(X) positive
• Quadrant II:
  • sin(X), csc(X) positive
  • cos(X), sec(X) negative
  • tan(X), cot(X) negative
• Quadrant III:
  • sin(X), csc(X) negative
  • cos(X), sec(X) negative
  • tan(X), cot(X) positive
• Quadrant IV:
  • sin(X), csc(X) negative
  • cos(X), sec(X) positive
  • tan(X), cot(X) negative

• Boundary between Quadrants I, IV: X is 0° i.e. 0 radians i.e. 0 gradients; point (1,0)
  • sin(0) = 0
  • cos(0) = 1
  • tan(0) = 0
  • cot(0) = undefined
  • sec(0) = 1
  • csc(0) = undefined
• Boundary between Quadrants I, II: i.e. 90° i.e. π/2 radians i.e. 100 gradients
  • sin() = 1
  • cos() = 0
  • tan() = undefined
  • cot() = 0
  • sec() = undefined
  • csc() = 1
• Boundary between Quadrants II, III: 2 i.e. 180° i.e. π radians i.e. 200 gradients
  • sin(2 ) = 0
  • cos(2 ) = -1
  • tan(2 ) = 0
  • cot(2 ) = undefined
  • sec(2 ) = -1
  • csc(2 ) = undefined
• Boundary between Quadrants III, IV: 3 i.e. 270° i.e. 3π/2 radians i.e. 300 gradients
  • sin(3 ) = -1
  • cos(3 ) = 0
  • tan(3 ) = undefined
  • cot(3 ) = 0
  • sec(3 ) = undefined
  • csc(3 ) = -1

What happens if we reflect the unit circle about the line y=x? We are swapping the x and y coordinates. That is, (letting X be the central angle), we are swapping sin(X) and cos(X). We are also physically reflecting the central angle X to the angle -X. So, we find that (here are the cofunction identities again):

• sin(-X)=cos(X)
• cos(-X)=sin(X)
• tan(-X)=cot(X)
• cot(-X)=tan(X)
• sec(-X)=csc(X)
• csc(-X)=sec(X)

Now, let us consider reflecting the unit circle about the x-axis:

Note that the x-coordinate [cos(X)] is unaffected, while the y-coordinate [sin(X)] is negated. The resulting point is the point we get from rotating through the angle -X from the angle 0. This gives the following formulae:

• sin(-X)=-sin(X) "sin(X) is an odd function"
• cos(-X)=cos(X) "cos(X) is an even function"
• tan(-X)=-tan(X) "tan(X) is an odd function"
• cot(-X)=-cot(X) "cot(X) is an odd function"
• sec(-X)=sec(X) "sec(X) is an even function"
• csc(-X)=-csc(X) "csc(X) is an odd function"

These formulae classify the trigonometric functions into even functions and odd functions. As you may recall from College Algebra, even functions are those functions whose value is unchanged by negating the argument, and odd functions are those functions whose value is negated by negating the argument.

It is no coincidence that the slope of the x-axis is 0. The central angle of the x-axis with the x-axis is clearly 0 [degrees, radians, gradients, it matters not which unit]. 0 is also exactly half of 0.

Now, let us consider reflecting the unit circle about the y-axis:

Note that the y-coordinate [sin(X)] is unaffected, while the x-coordinate [cos(X)] is negated. The resulting point is the point we get from rotating through the angle -X from the angle 0. This gives the following formulae:

• sin(2 -X)=sin(X)
• cos(2 -X)=-cos(X)
• tan(2 -X)=-tan(X)
• cot(2 -X)=-cot(X)
• sec(2 -X)=-sec(X)
• csc(2 -X)=csc(X)

It is no coincidence that the slope of the y-axis is undefined. The central angle of the y-axis with the x-axis is clearly . This is also exactly half of 2 .

Now, we can directly compute the reference values of cos and sin for 45° i.e. π/4 radians i.e. 50 gradients, i.e. ½, without using the 45-45-90 reference triangle. 45° is the angle that is its own complementary angle. Thus,

• cos(45°) = sin(90° - 45°) = sin(45°)

• 2cos²(45°) = 1

• cos(45°) = 1/ = /2

[Just like an equilateral triangle, an equilateral hexagon has equal lengths for all of its sides. In general, an equilateral polygon (regardless of how many sides it has) has equal lengths for all of its sides.]

Note that the inscribed hexagon can be considered as the 'splicing together' of six equilateral triangles.

[If we draw the line segments from the vertices to the center of the inscribed hexagon, since the sides have the same length, the angles as viewed from the center will have the same size. 360/6° = 60°. (The prior two sentences use Euclidean geometry). We know that the triangle we just created is isosceles, (two of its sides are radii), so the two angles opposite the radii have the same size, and add up to 120°: they are both 60°. All three angles are equal. Thus all three sides of the triangle have equal length. (The very last sentence also uses Euclidean geometry)]

Now, observe that the horizontal side "above the x-axis" is bisected by the y-axis. Thus, the length of this side in the first quadrant of the unit circle is ½ [the triangle is equilateral, so the side being bisected has the length of the radius, i.e. is length 1]. That is, cos(60°)=½.

[The y-axis has an angle of 90° i.e. π/2 radians i.e. 100 gradients. The angle created by drawing line segments from the vertices of this side, to the center, starts at 60° and ends at 120°. So, the y-axis is 90° - 60° = 30° into the above angle, which means the y-axis bisects this angle.]

From the Pythagorean identity, we find

• (½)²+sin²(60°)=1, i.e.
• ¼+sin²(60°)=1

• sin(60°) = /2

The unit circle has a circumference of 2π. "Thus", all trig functions will have the same value when evaluated 2π radians apart. We say that all trig functions have a period. [In "stratosphere" math, this period is arrived at in a very different way.]

The trig function period identities are [X is an angle, n is a positive integer]:

• sin(X±n[circle]) = sin(X)
• cos(X±n[circle]) = cos(X)
• tan(X±[½] n[circle]) = tan(X)
• cot(X±[½] n[circle]) = cot(X)
• sec(X±n[circle]) = sec(X)
• csc(X±n[circle]) = csc(X)

To formally demonstrate the n part, I would use natural induction, which should be buried somewhere in College Algebra. If you don't recall this term clearly, don't worry about it. However, I'm only going to explain it for n=1.

We read these from the unit circle immediately:

• sin(X+[circle]) = sin(X)
• cos(X+[circle]) = cos(X)

• sec(X+[circle]) = sec(X)
• csc(X+[circle]) = csc(X)

• tan(X+[½][circle])
• = sin(X+[½][circle])/cos(X+[½][circle])
• = [-sin(X)]/[-cos(X)]
• = sin(X)/cos(X) = tan(X)

There are two basic formulae, and one "impractical" one.

"½ base times height"

• (Area)=(½)(length of base)(length of height)

This formula is easily visualized for a right triangle [a rectangle with a line segment between two opposite vertices gives two right triangles, both clearly with half the original area since they are congruent]. It behaves reasonably for line segments [it gives zero area, which is correct for a line segment; either base or height is zero for a line segment.]

EXERCISE: learn to use this formula by applying it to the reference triangles. The 45-45-90 triangle with hypotenuse should have area ½, and the 30-60-90 triangle with hypotenuse length 2 should have area /2.

This formula has useful analogies in higher dimensions. The volume of either a pyramid or a cone, for instance, is given by:

• (Volume)=(1/3)(area of base)(length of height)

• (n-d hypervolume)=(1/n)(n-1 hypervolume of base)(length of height)

• (hypervolume)=(¼)(volume of base)(length of height)

"½ of the product of the lengths of two sides, and the sine of the included angle"

• (Area)=[½](length of side 1)(length of side 2)sin(angle between sides 1, 2)
  • (Area)=[½]absin(C)
  • (Area)=[½]acsin(B)
  • (Area)=[½]bcsin(A)

We get:

• (length of height)=(length of side 2)sin(angle between sides 1,2)

EXERCISE: Learn to use this formula by applying it to the reference triangles. [The correct areas are the same as before.]

EXERCISE: Now, learn to use this formula by applying it to the equilateral triangle with all sides length 2. This triangle is essentially two copies of the 30-60-90 triangle referred to earlier, so its area is twice as large — .

I do not know how to generalize the above formula to n-dimensional Euclidean geometry.

Heron's area formula for a triangle:

• (Area)² = s(s-a)(s-b)(s-c) where s=(a+b+c)/2 and a, b, c are the lengths of the sides. [Take the positive square root to get area.]

EXERCISE: Learn to use this formula by applying it to the reference triangles, and also to the equilateral triangle with all sides length 2.

The mathematician Cartan generalized this formula to n-dimensional Euclidean geometry, using matrix determinants [this is the determinant of a certain 3x3 matrix]. Cartan's generalization is definitely beyond the scope of this crash review.

• sin(A)/a = sin(B)/b = sin(C)/c

If the solved-for sin(angle) is strictly between 0 and 1, then some work is required to determine the actual angles. The inverse sine function on a calculator, or spreadsheet, is programmed to give an acute angle [strictly between 0 and in angular measure]. However, since the sine of an angle X is equal to the sine of its supplementary angle 2-X [sin(X)=sin(2-X)], the supplementary angle is also a viable choice.

[EXERCISE: Learn to use the sine law by explicitly writing out the equalities for the 45-45-90 triangle with hypotenuse , the 30-60-90 triangle with hypotenuse 2, and the equilateral triangle with all sides length 2.]

This is the generalization of the Pythagorean theorem to non-right triangles. I'm going to present it "deus ex machina".

• c²=a²+b²-2abcos(C)
• b²=a²+c²-2accos(B)
• a²=b²+c²-2bccos(A)

• C=
  • cos()=0. We get c²=a²+b², i.e. the Pythagorean theorem.
• C=0
  • cos(0)=1. We get:
  • c²=a²+b²-2ab, i.e.
  • c²=a²-2ab+b², i.e. [this is a standard reverse-squaring rule from College Algebra]
  • c²=(a-b)²
  • The last equation is physically correct: if C=0, then the triangle is really a line segment, and the side c physically has length |a-b|. This is what the algebra states:
  • c²=(a-b)² i.e.
  • |c|=|a-b| i.e. [c is guaranteed to be non-negative, since it represents a physical length]
  • c=|a-b|
• C=2
  • cos(2)=-1. We get:
  • c²=a²+b²+2ab, i.e.
  • c²=a²+2ab+b², i.e. [this is a standard reverse-squaring rule from College Algebra]
  • c²=(a+b)²
  • The last equation is physically correct: if C=2, then the triangle is really a line segment, and the side c physically has length a+b. This is what the algebra states:
  • c²=(a+b)² i.e.
  • |c|=|a+b| i.e. [a, b, c are all guaranteed to be non-negative, since they represent physical lengths]
  • c=a+b

I'm going to present these "deus ex machina". [The least painful method, to me, is 2x2 matrices -- inappropriate here.]

The basic angle sum formulae are [A,B are arbitrary angles here]:

• sin(A+B)=sin(A)cos(B)+sin(B)cos(A)
• cos(A+B)=cos(A)cos(B)-sin(A)sin(B)

• tan(A+B)=[sin(A)cos(B)+sin(B)cos(A)]/[cos(A)cos(B)-sin(A)sin(B)]
• =[tan(A)+tan(B)]/[1-tan(A)tan(B)] [we divided both top and bottom by cos(A)cos(B)]

• tan(A+B)=[tan(A)+tan(B)]/[1-tan(A)tan(B)]

• sin(A-B)=sin(A)cos(B)-sin(B)cos(A)
• cos(A-B)=cos(A)cos(B)+sin(A)sin(B)
• tan(A-B)=[tan(A)-tan(B)]/[1+tan(A)tan(B)]

• sin(2A)=2sin(A)cos(A)
• cos(2A)=cos²(A)-sin²(A)
• tan(2A)=2tan(A)/[1-tan²(A)]

Note: several entries in Trigonometry Survival 201 are/will be based on this.

The angle-halving formulae are easily derived from the Pythagorean Identity and the formula for cos(2A). In general, we need to know which quadrant the angle A/2 is in to decide on the correct sign.

Observe:

• 1+cos(2A)=[cos²(A)+sin²(A)]+[cos²(A)-sin²(A)]=2cos²(A)
• 1-cos(2A)=[cos²(A)+sin²(A)]-[cos²(A)-sin²(A)]=2sin²(A)

• cos(A)=±SQRT[(1+cos(2A))/2]
• sin(A)=±SQRT[(1-cos(2A))/2]

• cos(A/2)=±SQRT[(1+cos(A))/2]
• sin(A/2)=±SQRT[(1-cos(A))/2]

We also have [exercise: derive the following]:

• tan(A/2)=±SQRT[(1-cos(A))/(1+cos(A))]

• tan(A/2)=±|sin(A)|/(1+cos(A))=±sin(A)/(1+cos(A))
• tan(A/2)=±(1-cos(A))/|sin(A)|=±(1-cos(A))/sin(A)

30° = [½]60°, 45° = [½]90°, 60° = [½]120°, and 90° = [½]180°. For the tangent identities, both sides should be undefined for the 90° case.]

These formulae may be derived from the angle sum and difference formulae. We have:

• sin(A+B)+sin(A-B)=2sin(A)cos(B)
• sin(A+B)-sin(A-B)=2sin(B)cos(A)
• cos(A+B)+cos(A-B)=2cos(A)cos(B)
• cos(A+B)-cos(A-B)=2sin(A)sin(B)
• tan(A+B)+tan(A-B)=2(tan(A)[1+tan²(B)])/[1-tan²(A)tan ²(B)]
• tan(A+B)-tan(A-B)=2(tan(B)[1+tan²(A)])/[1-tan²(A)tan ²(B)]

EXERCISE: Derive these from the angle sum and difference formulae, as follows:

• Set X=A+B and Y=A-B .
• Substitute in the angle sum and difference formulae for the affected functions, and simplify.
• Solve for A and B in terms of X and Y, and then replace A, B with their expressions in X, Y.

EXERCISE: Numerically use the product identities for A=(X+Y)/2, B=(X-Y)/2 where X, Y are reference angles.

Wait! I'm not completely sure how to solve for A and B in terms of X and Y!

Until I get a College Algebra page going, here's a quick summary on how to solve systems of linear equations. (This domain can use several.) [While not all systems of linear equations are solvable, the one we want to is solvable.]

• Substitution
  • We hope that by cleverly adding multiples of pairs of equations, we can get equations with reasonably isolated variables. This works fairly well with two variables.
  • How this works in practice:
    • X=A+B, Y=A-B.
    • To solve for A, eliminate B from the resulting sum and then solve. Since 1+(-1)=0, we simply add both equations: X+Y=2A. Dividing by 2 yields (X+Y)/2=A.
    • To solve for B, eliminate A from the resulting sum and then solve. Since 1+(-1)*1=0, we subtract Y=A-B from X=A+B: X-Y=2B. Dividing by 2 yields (X-Y)/2=B.
• Gaussian Elimination (named after Gauss, the mathematician)
  • Taking variables from left to right (in our case, A, then B):
    • Pick a linear equation using the "leading variable". If this variable's coefficient is not 1, divide the equation by this coefficient. This equation is now the "topmost equation".
    • By adding a suitable multiple of the "topmost equation" to the other equations, remove A from the resulting equations.
    • We are now done with the "topmost equation". Set it aside.
    • When the only equations left have a multiple of a single variable equal to a constant, solve those variables. Then replace those variables in the equations that have been set aside. Repeat until all variables have been explicitly solved.
  • How this works in practice:
    • X=A+B, Y=A-B: A is "leading variable". Both X=A+B and Y=A-B have coefficient 1 for A.
    • However, the multiple of the "topmost equation" I am subtracting is 1[=1/1; numerator is from the equation I am subtracting from, denominator is from the "topmost equation". I arbitrarily choose my topmost equation to be Y=A-B.
    • We subtract Y=A-B (i.e. 1*[Y=A-B]) from X=A+B to get X-Y=2B.
    • We solve for B: B=(X-Y)/2.
    • We then subsitute this into the "topmost equation" X=A+B, getting X=A+(X-Y)/2
    • Isolating A, we end up at (X+Y)/2=A.
• Cramer's Rule: This is theoretically interesting, since it directly informs you when the system of linear equations does not have a unique solution. However, it requires the introduction of even more terminology. I won't cover it in this refresher.

• Acute -- strictly between 0 and
  • strictly between 0 and 90°
  • strictly between 0 and π/2 radians
  • strictly between 0 and 100 gradients
• Right --
  • exactly 90°
  • exactly π/2 radians
  • exactly 100 gradients
• Obtuse -- strictly between and 2
  • strictly between 90 and 180°
  • strictly between π/2 and π radians
  • strictly between 100 and 200 gradients
• Straight Line i.e half a circle -- 2
  • 180°
  • π radians
  • 200 gradients

For angles X, Y:

• complementary: the sum of the two angles is
  • X+Y = 90°
  • X+Y = π/2 radians
  • X+Y = 100 gradients
• supplementary: the sum of the two angles is 2 i.e half a circle.
  • X+Y = 180°
  • X+Y = π radians
  • X+Y = 200 gradients

For an angle X:

• sin(-X)=cos(X)
  • sin(90° -X)=cos(X)
  • sin(π/2 radians-X)=cos(X)
  • sin(100 gradients-X)=cos(X)
• cos(-X)=sin(X)
  • cos(90° -X)=sin(X)
  • cos(π/2 radians-X)=sin(X)
  • cos(100 gradients-X)=sin(X)
• tan(-X)=cot(X)
  • tan(90° -X)=cot(X)
  • tan(π/2 radians-X)=cot(X)
  • tan(100 gradients-X)=cot(X)
• cot(-X)=tan(X)
  • cot(90° -X)=tan(X)
  • cot(π/2 radians-X)=tan(X)
  • cot(100 gradients-X)=tan(X)
• sec(-X)=csc(X)
  • sec(90° -X)=csc(X)
  • sec(π/2 radians-X)=csc(X)
  • sec(100 gradients-X)=csc(X)
• csc(-X)=sec(X)
  • csc(90° -X)=sec(X)
  • csc(π/2 radians-X)=sec(X)
  • csc(100 gradients-X)=sec(X)

For an angle X:

• sin(2-X)=sin(X)
  • sin(180° -X)=sin(X)
  • sin(π radians-X)=sin(X)
  • sin(200 gradients-X)=sin(X)
• cos(2-X)=-cos(X)
  • cos(180° -X)=-cos(X)
  • cos(π radians-X)=-cos(X)
  • cos(200 gradients-X)=-cos(X)
• tan(2-X)=-tan(X)
  • tan(180° -X)=-tan(X)
  • tan(π radians-X)=-tan(X)
  • tan(200 gradients-X)=-tan(X)
• cot(2-X)=-cot(X)
  • cot(180° -X)=-cot(X)
  • cot(π radians-X)=-cot(X)
  • cot(200 gradients-X)=-cot(X)
• sec(2-X)=-sec(X)
  • sec(180° -X)=-sec(X)
  • sec(π radians-X)=-sec(X)
  • sec(200 gradients-X)=-sec(X)
• csc(2-X)=csc(X)
  • csc(180° -X)=csc(X)
  • csc(π radians-X)=csc(X)
  • csc(200 gradients-X)=csc(X)

	sin	cos	tan	cot	sec	csc
0°	0	1	0	undefined	1	undefined
30°	½	/2	/3		(2)/3	2
45°	/2	/2	1	1
60°	/2	½		/3	2	(2)/3
90°	1	0	undefined	0	undefined	1

Exercise: convert the angles to radians. If you plan to use gradients, also convert the angles to gradients.

If you find the above table difficult to memorize for sin and cos, my ex-fianceé suggests this alternative table [it uses =1 and =2]:

	0°	30°	45°	60°	90°
sin	0/2	/2	/2	/2	/2
cos	/2	/2	/2	/2	0/2

If you find this alternative table convenient, you may want to check that the entries that look different, do algebraically reduce to those in the first table.

Exercise: verify that these tables are correct:

	sin	cos	tan	cot	sec	csc
15°	[-]/4	[+]/4	2-	2+	-	+
75°	[+]/4	[-]/4	2+	2-	+	-

Suggestions: Note that 15° is computable either with the half-angle or the difference-angle formula from the standard reference table. Use both of these methods. Also, 75° can be computed by the sum formula from the standard reference table; use this method. Once you are confident that these tables are correct, and if you want more practice with the angle sum, difference, halving, and doubling formulae, use these additional reference values in combination with the earlier ones [0°, 30°, 45°, 60°, and 90°, i.e. the standard table].

[Exercise: translate everything into radians. If you plan to use gradients, also translate everything into gradients.]

EXERCISE: Verify that these tables are correct:

	sin	cos	tan	cot
18°	[-1]/4	/4	[5-]/20	[+1]/4
36°	/4	[+1]/4	[-1]/4	[5+]/20
54°	[+1]/4	/4	[5+]/20	[-1]/4
72°	/4	[-1]/4	[+1]/4	[5-]/20

	sec	csc
18°	/5	+1
36°	-1	/5
54°	/5	-1
72°	+1	/5

Suggestions: First, let X be an angle such that 5X works out to be 0, 1, 2, 3, or 4 full circles.

[That is, we are directly interested in 72°, 144°, 216°, or 288°. I also included 0°, but that is a standard reference value. We will recover:

• 18° from 72° and the cofunction identities
• 36° from 144° and the supplementary angle identities
• 54° from 36° and the cofunction identities

Next, use the angle addition formulae to rewrite in terms of sin(X) and cos(X)

• sin(5X)=0
• cos(5X)-1=0 [a convenient rewrite of cos(5X)=1, for this problem]

These are the equations that describe an inscribed equilateral pentagon. They are solvable using nothing more than the quadratic formula(!), and what we already are supposed to know.

This would occur to me from sin(X) being an (algebraically) odd function, combined with the angle comments above:

sin(0°)=0, so we can factor sin(X) out of our expanded version of sin(5X)=0. Then, use the Pythagorean identity to replace cos²(X) with 1-sin²(X). Expand the results. This should give a quartic in sin(X) [oops], which is also a quadratic in sin²(X) [great]. Use the quadratic formula to solve for sin²(X). Then take square roots on the solutions we get for sin²(X); we need both the positive and negative roots. One of these root sets is for 72° and -72°, and the other one is for 144° and -144°.

Geometrically, the larger positive root goes with 72°, and the smaller positive root goes with 144°. Solve for cosine of 72° and 144° with the Pythagorean identity [be sure to use the quadrants to force the correct sign]. Then fill in the table as summarized initially.

Also: when completing the table, tan(18°) and tan(54°) [and their corresponding cofunction values, cot(72°) and cot(36°)] are technically difficult to algebraically compute directly from the sine and cosine values. [The denominator needs two stages to cancel out correctly.] I tried that three times in a row, and got three different answers, all of them wrong. The method I used for the table was to compute cot(18°) and cot(54°) respectively, and then take the multiplicative inverse algebraically to get tan(18°) and tan(54°).

This isn't really "fair", since it relies on the "stratosphere" math approach. The first mention of the basis for this, traditionally, is in the middle of a Calculus series [business or conventional].

However, it is very practical. (Especially if you plan to use trigonometry in an engineering course.) I'm going to present how the result is arrived at in just enough detail, that those readers who actually know the relevant material will be able to see that I'm doing it right.

The key point is: for an angle X with radian measure 0<X<1, we have

• sin(X)<X, and furthermore X-sin(X)<X³/6
• cos(X)>1-X², and furthermore cos(X)-1+X²<(X⁴)/24

For three decimal place accuracy (the least we are interested in, for practical purposes), we need to control the error to less than 5*10^-4 i.e. [½]10^-3. This error term is:

• for sin(X), X³/6
• for cos(X), (X⁴)/24

• sin(X)
  • X³/6<5*10^-4 i.e.
  • X³<30*10^-4 i.e.
  • X³<3*10^-3 i.e.
  • X<[3^1/3]/10 which is between 0.144 and 0.145
• cos(X)
  • (X⁴)/24<5*10^-4 i.e.
  • X⁴<120*10^-4 i.e. [yes, there's a reason I'm not reducing here!]
  • X<[120^1/4]/10 which is between 0.330 and 0.331

• when 0<X<0.144, sin(X)=X to three decimal places
• when 0<X<0.330, cos(X)=1-X² to three decimal places

Computing tan(X) and sec(X) from the results of this "instant trig table" will give reasonably accurate numerical results. Computing cot(X) and csc(X) will not work reasonably "near 0"; in general, you cannot get more significant digits out than you put in, and sin(X) will lose significant digits as X ends up near 0. This means you are dividing a large number by a small, imprecise number, which will not give reasonable numerical results. [The zero decimal places between the decimal point and the first non-zero digit are "lost significant digits". For example, sin(0.021)=0.021 to three decimal places -- but there are only two significant digits.]

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