The addition formulas for cos(a+b) and sin(a+b) rely on a unique algebraic mechanism: the distributive property applied to complex exponentials. We start from Euler’s identity, and everything else follows from the addition or subtraction of two equalities. Mastering this mechanism makes it unnecessary to memorize the dozens of variants that textbooks pile up.
Derivation by complex exponential of the formulas cos a cos b and sin a sin b
The most direct way to rediscover the addition formulas is through exponential notation. Knowing that ei(a+b) = eia . eib, we expand the product of the two factors:
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(cos a + i sin a)(cos b + i sin b) = cos a cos b – sin a sin b + i(sin a cos b + cos a sin b).
The real part directly gives cos(a+b) = cos a cos b – sin a sin b. The imaginary part provides sin(a+b) = sin a cos b + cos a sin b. No figures, no auxiliary geometric reasoning are necessary.
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For those preparing for the baccalaureate or a first exam, remembering this single approach allows one to rediscover in a few lines the formulas for cos a cos b and sin a sin b without relying on a formula sheet.
Subtraction formulas cos(a-b) and sin(a-b): sign substitution
The evenness of cosine (cos(-x) = cos x) and the oddness of sine (sin(-x) = -sin x) are sufficient to deduce the subtraction formulas. We replace b with -b in the previous expressions.
cos(a-b) = cos a cos b + sin a sin b. The sign between the two terms reverses compared to cos(a+b). Symmetrically, sin(a-b) = sin a cos b – sin b cos a.
This change of sign is the only difference. A good reflex is to remember that cosine “likes” products of the same nature (cos.cos and sin.sin), while sine mixes the two functions (sin.cos and cos.sin).
Product-sum transformation: isolating cos a cos b and sin a sin b
The addition formulas also serve in reverse. By adding cos(a+b) and cos(a-b), we eliminate the sine term:
cos(a+b) + cos(a-b) = 2 cos a cos b.
From this, we derive cos a cos b = 1/2 [cos(a-b) + cos(a+b)]. This linearization relation is ubiquitous in integral calculus and signal processing.
By subtracting cos(a+b) from cos(a-b), we isolate the product of the sines:
sin a sin b = 1/2 [cos(a-b) – cos(a+b)].
For the cross product, we proceed similarly with sin(a+b) and sin(a-b):
sin a cos b = 1/2 [sin(a+b) + sin(a-b)].
These three identities form the trio of product-sum formulas. Their utility far exceeds the high school curriculum: they come into play whenever a trigonometric product needs to be integrated or simplified in a Fourier series.
Why linearization is more useful than duplication
The duplication formulas (cos 2a, sin 2a) are just a special case where b = a. We recommend not memorizing them separately. Setting b = a in cos(a+b) immediately gives cos(2a) = cos²a – sin²a, then the variants 2cos²a – 1 and 1 – 2sin²a via the fundamental identity cos²a + sin²a = 1.
Linearization, on the other hand, does not appear as naturally if one only knows duplication. It is therefore better to anchor the addition and product-sum formulas and treat duplication as a corollary.
Remarkable angles and quick verification of trigonometric formulas
Recent baccalaureate and diploma exams mobilize remarkable values (0, pi/6, pi/4, pi/3, pi/2) directly related to the addition formulas. Verifying a formula on a pair of known angles is the quickest way to detect a sign error.
Let’s take cos(pi/3 + pi/6). The formula gives:
cos(pi/3)cos(pi/6) – sin(pi/3)sin(pi/6) = (1/2)(sqrt(3) / 2) – (sqrt(3) / 2)(1/2) = 0.
The expected result is cos(pi/2) = 0. The consistency is immediate. If the sign had been reversed, we would have obtained sqrt(3) / 2, which would have indicated the error.
Here are the most useful checks for each formula:
- cos(a+b) with a = b = pi/4: should give cos(pi/2) = 0, i.e., cos²(pi/4) – sin²(pi/4) = 1/2 – 1/2 = 0.
- sin(a+b) with a = pi/6, b = pi/3: should give sin(pi/2) = 1, i.e., (1/2)(1/2) + (sqrt(3) / 2)(sqrt(3) / 2) = 1/4 + 3/4 = 1.
- sin a sin b with a = b = pi/4: should give 1/2, i.e., 1/2 [cos(0) – cos(pi/2)] = 1/2 [1 – 0] = 1/2.
Testing a formula on remarkable angles takes less than thirty seconds and eliminates the majority of sign errors in exams.
Memorization strategy for the baccalaureate: two formulas, not twelve
We assert without reservation: remembering cos(a+b) and sin(a+b) is enough to reconstruct all other identities. Substituting b with -b gives the subtraction formulas. The sum and difference of the addition formulas yield the product-sum. The case b = a gives duplication. The case b = pi/2 gives the co-function formulas.
The frequent mistake is to memorize each variant independently, which leads to sign confusion under pressure. It is better to automate the derivation than to store a dozen fragile formulas.
- Source formula 1: cos(a+b) = cos a cos b – sin a sin b (remember the minus sign).
- Source formula 2: sin(a+b) = sin a cos b + cos a sin b (remember the plus sign and the sin/cos alternation).
- Derived formulas: obtained by substitution, addition, or subtraction of the two source formulas.
On the interval [0, pi] required by the first program, the sin and cos functions take sufficiently varied values so that verification by remarkable angles remains always possible. Two memorized formulas, one mastered derivation method: this is the most favorable effort/result ratio in trigonometry.