Double, Triple and Half Angles

Page 7 gave the sine, cosine and tangent of A + B. This page sets B equal to A. That is the entire mechanism, and there is nothing new to memorise here.

One note on letters first, so you are never guessing. On this page A is the angle you set, and 2A and 3A are multiples of it. The letter θ appears in one place only, the half angle step, where the standard statement is about θ/2 and θ is the whole angle being halved.

Take sin(A + B) = sin A cos B + cos A sin B and put B = A. The right side becomes sin A cos A + cos A sin A, which is two copies of the same product:

sin 2A = 2 sin A cos A

Take cos(A + B) = cos A cos B − sin A sin B and put B = A:

cos 2A = cos²A − sin²A

Take tan(A + B) = (tan A + tan B) / (1 − tan A tan B) and put B = A:

tan 2A = 2 tan A / (1 − tan²A)

Now the guess these formulas exist to kill. sin 2A is not 2 sin A. Sine is not a function that you can pull a multiplier out of.

Test it at A = 30 degrees. sin(30 degrees) is 0.500, so 2 sin(30 degrees) is 1.000. But sin(60 degrees) is 0.866 to three decimal places. The wrong answer is out by about 0.134, and at A = 90 degrees it is worse: 2 sin(90 degrees) is 2.000, and no sine is ever bigger than 1. The real formula agrees: 2 sin(30 degrees) cos(30 degrees) is 2 × 0.500 × 0.866, which is 0.866 to three decimal places.

The tangent version inherits a gap from its parent. At A = 45 degrees, tan A is 1, so 1 − tan²A is 0 and the formula divides by zero. That is correct rather than broken: 2A is then 90 degrees, and page 4 established that tangent has an asymptote there.

cos 2A = cos²A − sin²A has a practical problem. It mixes both functions, and in most calculations you have only one of them.

The first Pythagorean identity from page 6 fixes that. sin²A + cos²A = 1 lets you replace either square by 1 minus the other, and the choice of which one to replace is the only choice in the whole step.

cos 2A = cosine squared A − sine squared A Value 0.500.

Replace sin²A by 1 − cos²A. = cosine squared A − (1 − cosine squared A) Value 0.500.

Multiply out the bracket. Minus a minus is a plus. = 2 cosine squared A − 1 Value 0.500.

Go back to the first line and replace cos²A instead. cos 2A = (1 − sine squared A) − sine squared A Value 0.500.

Multiply out that bracket too. = 1 − 2 sine squared A Value 0.500.

So there are three forms, and they are one fact. The value column is where you check that, and at A = 30 degrees it reads 0.500 on every line, because 2A is 60 degrees and cos(60 degrees) is 0.5 exactly.

Which one do you reach for? The one that leaves you with the function you already have. If a problem has given you cos A, use 2cos²A − 1 and never mention sine. If it has given you sin A, use 1 − 2sin²A. Choosing badly costs you an extra substitution, nothing worse.

The sine form and the cosine form matter for a second reason, which is what they look like rearranged. Nothing new is being proved here: each line does the same thing to both sides, which is why both sides of the value column move together.

1 − 2 sine squared A = cos 2A Value 0.500 = 0.500.

Add 2sin²A to both sides and subtract cos 2A from both. 1 − cos 2A = 2 sine squared A Value 0.500 = 0.500.

Divide both sides by 2. sine squared A = (1 − cos 2A) / 2 Value 0.250 = 0.250.

2 cosine squared A − 1 = cos 2A Value 0.500 = 0.500.

Add 1 to both sides. 2 cosine squared A = 1 + cos 2A Value 1.500 = 1.500.

Divide both sides by 2. cosine squared A = (1 + cos 2A) / 2 Value 0.750 = 0.750.

Read left to right, those turn a squared function into a plain cosine of a doubled angle. That is how sin²x gets integrated: the square has no elementary antiderivative you can write down by inspection, while (1 − cos 2x)/2 integrates term by term. The same two rearrangements are the whole of the next step.

The problem here is the opposite of the last step. You know cos θ and you want the sine or cosine of half of θ.

The last step already solved 1 − 2sin²A = cos 2A for the square. Nothing further has to be done to it: the angle has to be given the name everyone else uses, and that is the whole of this step. A rename cannot move a number, which is what the value column is there to say.

sine squared A = (1 − cos 2A) / 2 Value 0.067.

Let θ = 2A, so A is θ/2. Write θ/2 wherever A appeared and θ wherever 2A appeared. sine squared (θ/2) = (1 − cos θ) / 2 Value 0.067.

The other form, 2cos²A − 1 = cos 2A, gives the partner the same way:

cosine squared A = (1 + cos 2A) / 2 Value 0.933.

Let θ = 2A, so A is θ/2. Write θ/2 wherever A appeared and θ wherever 2A appeared. cosine squared (θ/2) = (1 + cos θ) / 2 Value 0.933.

Both formulas give a square. Taking a square root introduces a sign, and the formula cannot supply it. sin(θ/2) is plus or minus the root of (1 − cos θ)/2, and which one it is depends on the quadrant One of the four regions the x and y axes cut the plane into, numbered anticlockwise from the top right. Quadrant 1 spans 0 to 90 degrees, quadrant 2 spans 90 to 180, quadrant 3 spans 180 to 270, quadrant 4 spans 270 to 360. The signs of the six functions in each follow from the signs of x and y there, which is what the CAST and ASTC mnemonics encode. that θ/2 lands in. The CAST The four quadrant sign pattern for sine, cosine and tangent, read anticlockwise from the bottom right: Cosine positive in quadrant 4, All three positive in quadrant 1, Sine positive in quadrant 2, Tangent positive in quadrant 3. It is a compression of the signs of the x and y coordinates in each quadrant, not an independent fact, and it is what supplies the missing sign after a half angle square root. procedure from page 5 answers that in two steps: find the quadrant of θ/2, then read off whether sine is positive there.

The sign is not a formality that a better formula could remove. The right side depends on θ only through cos θ, and cosine cannot tell θ from θ + 360 degrees.

Take θ = 40 degrees and θ = 400 degrees. cos(40 degrees) and cos(400 degrees) are the same number, 0.766 to three decimal places, so the formula returns the same square, 0.117 to three decimal places, and the same magnitude, 0.342 to three decimal places.

But half of 40 is 20 degrees, where sine is plus 0.342, and half of 400 is 200 degrees, where sine is minus 0.342. Two different answers from one input. The formula gives the size and the quadrant gives the sign, and neither is optional.

Work one through. Take θ = 300 degrees, where cos θ is 0.5 exactly. Then sin²(150 degrees) is (1 − 0.5)/2, which is 0.25, so the magnitude is 0.5. And cos²(150 degrees) is (1 + 0.5)/2, which is 0.75, so that magnitude is 0.866 to three decimal places.

Half of 300 is 150 degrees, which sits in quadrant 2, where sine is positive and cosine is negative. So sin(150 degrees) is 0.5 and cos(150 degrees) is minus 0.866 to three decimal places, which is the worked example from page 5, arrived at from a different direction.

Tangent has a half angle formula too, and it comes in three versions. Dividing the two formulas above gives tan(θ/2) = ±√((1 − cos θ)/(1 + cos θ)), which carries the same sign problem. Two other versions do not:

tan(θ/2) = sin θ / (1 + cos θ) = (1 − cos θ) / sin θ

Each of those is a single fraction of ordinary signed values, so the sign takes care of itself and no quadrant test is needed. Check at θ = 60 degrees: sin(60 degrees)/(1 + cos(60 degrees)) is 0.866/1.5, which is 0.577 to three decimal places, and that is tan(30 degrees). Hold on to the first of those two fractions. It comes back in the last step of this page as the number t.

Nothing stops you doing it again. There is no compound angle formula for three angles and none is needed, because 3A is 2A + A and the compound angle formulas take any two, including two that are multiples of the same one. Everything after that choice is substitution, and every substitution has to leave the value alone.

3A is 2A + A, and the compound angle formula takes any two angles. sin 3A = sin(2A + A) Value 1.000.

The compound angle formula, with 2A in place of A and A in place of B. = sin 2A cos A + cos 2A sin A Value 1.000.

Substitute sin 2A = 2 sin A cos A, and the sine form of cos 2A from step 2. = (2 sin A cos A) cos A + (1 − 2 sine squared A) sin A Value 1.000.

Multiply out. = 2 sin A cosine squared A + sin A − 2 sine cubed A Value 1.000.

One cosine is left, and it is a square, so cos²A = 1 − sin²A removes it. = 2 sin A (1 − sine squared A) + sin A − 2 sine cubed A Value 1.000.

Multiply out again. Now every term is a sine. = 2 sin A − 2 sine cubed A + sin A − 2 sine cubed A Value 1.000.

Reorder so like terms are next to each other. Nothing has changed but the order. = 2 sin A + sin A − 2 sine cubed A − 2 sine cubed A Value 1.000.

Collect. Two sin A terms make three, and two sin³A terms make four. = 3 sin A − 4 sine cubed A Value 1.000.

Cosine goes the same way, and the one decision worth making is which face of cos 2A to put in, so that the substitution at the end has one function to remove rather than two.

Again 3A is 2A + A, this time in the cosine formula. cos 3A = cos 2A cos A − sin 2A sin A Value 0.000.

Substitute the cosine form of cos 2A, so that no cosine has to be removed later, and sin 2A = 2 sin A cos A. = (2 cosine squared A − 1) cos A − (2 sin A cos A) sin A Value 0.000.

Multiply out. = 2 cosine cubed A − cos A − 2 sine squared A cos A Value 0.000.

One sine is left, and it is a square, so sin²A = 1 − cos²A removes it. = 2 cosine cubed A − cos A − 2(1 − cosine squared A) cos A Value 0.000.

Multiply out and collect. Two cos³A terms make four, and the two cos A terms make three. = 4 cosine cubed A − 3 cos A Value 0.000.

And tangent, from tan(2A + A):

tan 3A = (3 tan A − tan³A) / (1 − 3tan²A)

Look at the shape of the first two. Each is a polynomial An expression built from a variable using only addition, subtraction and multiplication by constants, such as 4x³ − 3x. The triple angle identities are polynomials in a single trigonometric function: sin 3A = 3s − 4s³ with s = sin A, and cos 3A = 4c³ − 3c with c = cos A. in a single function. Every appearance of the other function was removed by the Pythagorean identity, and the result contains one function name and nothing else. That is not something you could have predicted from the compound formula, and it is the reason these are worth writing down.

Check them where you can check by hand. At A = 30 degrees, 3A is 90 degrees and sin(90 degrees) is 1. The polynomial gives 3 × 0.5 minus 4 × 0.125, which is 1.5 minus 0.5, which is exactly 1.

cos(90 degrees) is 0. cos(30 degrees) is √3/2, so cos³(30 degrees) is 3√3/8, and 4 × 3√3/8 minus 3 × √3/2 is 3√3/2 minus 3√3/2, which is exactly 0.

The tangent formula at the same angle has 1 − 3tan²(30 degrees) = 1 − 3 × (1/3) = 0 on the bottom, which is the division by zero you should want, because tan(90 degrees) does not exist.

One honest limit on the pattern. cos nA is always a polynomial in cos A, for every whole number n. sin nA is a polynomial in sin A only when n is odd. sin 2A is the counterexample sitting at the top of this page: 2 sin A cos A keeps a cosine and cannot be written in sin A alone.

The cosine family has a name. The polynomial that sends cos A to cos nA is the Chebyshev polynomial The polynomial T_n defined by T_n(cos A) = cos nA, so that T₂(x) = 2x² − 1 and T₃(x) = 4x³ − 3x. The multiple angle formulas for cosine are exactly this family, which is why cos nA can always be written in cos A alone. They are used in numerical analysis to choose sample points for polynomial approximation, because they spread the error evenly. of the first kind, written T_n, so 2x² − 1 is T₂ and 4x³ − 3x is T₃. They are used well outside trigonometry, mostly for choosing where to sample a function when approximating it.

Page 4 read the period and the amplitude off the six curves at their natural scale. This is the first time the scale changes.

sin 2x reaches at x whatever sin x reaches at 2x. Every feature therefore happens at half the x it used to, which squeezes the whole picture towards the vertical axis by a factor of two. The period The smallest interval over which a periodic function repeats. Sine, cosine, secant and cosecant have period 2π (360 degrees). Tangent and cotangent have period π (180 degrees), because a half turn negates both coordinates and their quotient is unchanged. of sin x is 2π, or 360 degrees. The period of sin 2x is π, or 180 degrees. In general the period of sin kx is 2π/k, so sin 3x has period 2π/3, which is 120 degrees.

Count the zeros over one full turn, from 0 degrees up to but not including 360 degrees. sin x is zero twice, at 0 and 180. sin 2x is zero four times, at 0, 90, 180 and 270. sin 3x is zero six times, at 0, 60, 120, 180, 240 and 300. Doubling the angle doubles the count in a fixed window, because it fits two periods where one used to go.

Now the error this step exists to correct. Doubling the angle changes the period, not the height. The amplitude Half the distance between the minimum and maximum values of a wave, that is, how far it departs from its centre line. Sine and cosine have amplitude 1. Tangent, cotangent, secant and cosecant have no amplitude, because they are unbounded. of sin 2x is 1, exactly as for sin x, because the output of a sine is capped at 1 whatever you feed into it. Changing the height means multiplying the output instead: 2 sin x has amplitude 2 and period 2π, unchanged. Inside the brackets or outside the brackets, two different operations, and they do not affect the same measurement.

The formula agrees with the picture. sin 2x = 2 sin x cos x is a product of two things that never exceed 1 in size, so it never exceeds 1 in size either. The factor of 2 out the front does not raise the peak, because the two factors are never both at their maximum at the same x.

Trigonometric functions are not algebraic. You cannot factorise sin A, and an integral or an equation containing several different trigonometric functions has no mechanical method to attack it. One substitution removes that problem entirely.

Set t = tan(A/2). Then:

sin A = 2t / (1 + t²)

cos A = (1 − t²) / (1 + t²)

tan A = 2t / (1 − t²)

The derivation is two lines of this page's own material, and the second of the two is the move page 7 already made on the tangent formula: divide the top and the bottom by a square, and every sine turns into a tangent. Dividing by 1 first is what puts a denominator there to divide.

The double angle formula, with A/2 in place of A. sin A = 2 sin(A/2) cos(A/2) Value 0.500.

Divide by sin²(A/2) + cos²(A/2), which is 1, so nothing changes. Watch the value column. = 2 sin(A/2) cos(A/2) / (sine squared (A/2) + cosine squared (A/2)) Value 0.500.

Now divide top and bottom by cos²(A/2), the same move as the tangent formula on page 7. Each small fraction below prints its own value, and t is tan(A/2). = (2 sin(A/2) cos(A/2) / cosine squared (A/2)) / (sine squared (A/2) / cosine squared (A/2) + cosine squared (A/2) / cosine squared (A/2)) 2 sin(A/2) cos(A/2) / cosine squared (A/2) becomes 2t. 2 sin(A/2) cos(A/2) / cosine squared (A/2) is 0.536. sine squared (A/2) / cosine squared (A/2) becomes t². sine squared (A/2) / cosine squared (A/2) is 0.072. cosine squared (A/2) / cosine squared (A/2) becomes 1. cosine squared (A/2) / cosine squared (A/2) is 1.000. Value 0.500.

Each one is a tangent or a 1, and t is tan(A/2). sin A = 2t / (1 + t squared) Value 0.500.

Cosine goes the same way from cos A = cos²(A/2) − sin²(A/2) over the same denominator of 1, giving (1 − t²)/(1 + t²). Dividing one result by the other gives the tangent formula.

Check them at A = 60 degrees. Then A/2 is 30 degrees and t is 0.577 to three decimal places, so t² is 1/3 exactly. sin A comes out as 2 × 0.577 over 4/3, which is 0.866 to three decimal places, matching sin(60 degrees). cos A comes out as (1 − 1/3) over (1 + 1/3), which is 0.5 exactly, matching cos(60 degrees). tan A comes out as 1.732 to three decimal places, matching √3.

Every one of the six functions is now a rational function A ratio of two polynomials, such as 2t/(1 + t²). The t-substitution t = tan(A/2) rewrites all six trigonometric functions as rational functions of t, which is what makes a trigonometric integral tractable by partial fractions and a trigonometric equation solvable as a polynomial one. of one variable, a ratio of two polynomials in t. That is the payoff, and it is a large one. A trigonometric integral becomes an integral of a rational function, which partial fractions handles by a fixed procedure. A trigonometric equation becomes a polynomial equation, which the usual root finding applies to.

The substitution is usually called the Weierstrass substitution The substitution t = tan(A/2), which turns sin A into 2t/(1 + t²), cos A into (1 − t²)/(1 + t²) and tan A into 2t/(1 − t²), so that any rational combination of trigonometric functions becomes a rational function of t. Geometrically t is the slope of the line from (−1, 0) through the point (cos A, sin A) on the unit circle. It is undefined when A is an odd multiple of 180 degrees, which is the single point (−1, 0). Euler used it in his Institutionum calculi integralis of 1768, well before the name attached to Weierstrass., though Euler had used it in his *Institutionum calculi integralis* of 1768, decades before Weierstrass was born. The name is conventional rather than historical.

There is a picture behind it. On the unit circle, draw the line from (−1, 0) through the point (cos A, sin A). Its slope is sin A / (1 + cos A), and the last step showed that number is exactly tan(A/2). So t is a slope, and the three formulas above say that the coordinates of the point can be recovered from that slope alone.

That is where the exact rational points come from. Give t a fraction and both coordinates come out as fractions, because the formulas contain nothing but multiplication, addition and division.

Take t = 1/2. Then cos A = (1 − 1/4)/(1 + 1/4) = 3/5 and sin A = 1/(5/4) = 4/5, so the point is (0.6, 0.8) and the triangle underneath it is the 3-4-5 triangle from page 1. That angle A is 53.130 degrees to three decimal places.

Run t over every fraction and you land on every rational point of the circle exactly once, with (−1, 0) the single point left out. Clearing the denominators at a point where neither coordinate is zero gives a Pythagorean triple Three whole numbers a, b, c with a² + b² = c², so they are the side lengths of a right triangle with no rounding anywhere, such as 3, 4, 5. Feeding a fraction into the t-substitution produces one: t = 1/2 gives the point (3/5, 4/5) on the unit circle, which is the 3-4-5 triangle. Babylonian tablet Plimpton 322 lists such triples., and every primitive triple arises this way, from some t between 0 and 1.

What the correspondence does not do is deal each triple out once. 7-24-25 comes back at t = 7, at t = 4/3, at t = 3/4 and at t = 1/7, and again at the negatives of all four, because those eight values name eight different points that share one set of side lengths. Triples that are not primitive never arrive at all, since 6-8-10 and 3-4-5 name the same point, (0.6, 0.8). That is why this substitution turns up in number theory as often as in integration.

Where it fails, and it does fail somewhere. t = tan(A/2) is undefined when A/2 is an odd multiple of 90 degrees, which means A is an odd multiple of 180 degrees. In the picture, that is the point (−1, 0) itself, where there is no line from (−1, 0) through the point to take a slope of.

So the substitution covers the whole circle except one point. In an integral that means checking the excluded angle separately rather than assuming the answer extends to it.