Runnel Zhang

The Derivation of Trigonometric Sum and Difference Formulas from Advanced Standpoints

Overview

The origin of this note was during class when I wanted to verify the tangent (tan) sum formula directly using the Reciprocal Basis from tensor calculus. I initially tried using the scalar triple product (mixed product), but that didn't work. Later, I attempted orthogonal decomposition, but the reciprocal basis of a unit orthogonal basis is the basis itself, which rendered the specific properties of reciprocal bases useless.

Although simply "grabbing" the result using rotation matrices is a very reasonable approach, I had an obsession with finding a "direct proof for tan." Therefore, I organized various derivation paths ranging from elementary geometry to tensor operations, and finally to higher-level mathematical viewpoints.

N-20230220: View via here, or visit DOI:10.6084/M9.FIGSHARE.22268962

N-20230223: View via here, or visit DOI:10.6084/M9.FIGSHARE.22268965

I. Approaches via Geometry & Vectors

1. Proof of the Sine Formula via Cross Product

This is the most intuitive method. We restrict the angle range to (0,π)(0, \pi). Let OM⃗=(cos⁡β,sin⁡β)\vec{OM} = (\cos\beta, \sin\beta) and ON⃗=(cos⁡α,sin⁡α)\vec{ON} = (\cos\alpha, \sin\alpha). Using the geometric definition of the cross product's modulus: ∣ON⃗×OM⃗∣=∣ON⃗∣∣OM⃗∣sin⁡(β−α)=sin⁡(β−α)|\vec{ON} \times \vec{OM}| = |\vec{ON}||\vec{OM}|\sin(\beta-\alpha) = \sin(\beta-\alpha) On the other hand, using the coordinate form (determinant): ON⃗×OM⃗=∣cos⁡αsin⁡αcos⁡βsin⁡β∣k⃗=(sin⁡βcos⁡α−sin⁡αcos⁡β)k⃗\vec{ON} \times \vec{OM} = \begin{vmatrix} \cos\alpha & \sin\alpha \\ \cos\beta & \sin\beta \end{vmatrix} \vec{k} = (\sin\beta\cos\alpha - \sin\alpha\cos\beta)\vec{k} Comparing both expressions (noting direction and sign), we directly obtain: sin⁡(β−α)=sin⁡βcos⁡α−sin⁡αcos⁡β\sin(\beta-\alpha) = \sin\beta\cos\alpha - \sin\alpha\cos\beta

2. The Failed Attempt via Projection Vectors

I attempted to find a shortcut using vector projection. Calculating the squared modulus of the projection vector NH⃗\vec{NH} resulted in 1−cos⁡2(β−α)=sin⁡2(β−α)1 - \cos^2(\beta-\alpha) = \sin^2(\beta-\alpha). This circled back to the sin⁡/cos⁡\sin/\cos form and did not achieve my goal of "deriving tan directly." This path is a dead end.

II. Perspectives from Algebra & Analysis

3. Rotation Matrices

It is perhaps better to say that this is the most reasonable method, and my personal favorite. Even before encountering textbooks, I preferred thinking in terms of matrices. Algebraically, "rotating by α+β\alpha+\beta" is equivalent to the composition of linear transformations: first rotating by α\alpha, then by β\beta. R(α+β)=R(β)⋅R(α)R(\alpha+\beta) = R(\beta) \cdot R(\alpha) [cos⁡(α+β)−sin⁡(α+β)sin⁡(α+β)cos⁡(α+β)]=[cos⁡β−sin⁡βsin⁡βcos⁡β]⋅[cos⁡α−sin⁡αsin⁡αcos⁡α]\begin{bmatrix} \cos(\alpha+\beta) & -\sin(\alpha+\beta) \\ \sin(\alpha+\beta) & \cos(\alpha+\beta) \end{bmatrix} = \begin{bmatrix} \cos\beta & -\sin\beta \\ \sin\beta & \cos\beta \end{bmatrix} \cdot \begin{bmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{bmatrix} Directly expanding the matrix multiplication yields the sum formulas for sine and cosine instantly.

4. Euler's Formula

When dealing with trigonometric formulas, using complex numbers is generally the most convenient (Trivial) approach. ei(α+β)=eiα⋅eiβe^{i(\alpha+\beta)} = e^{i\alpha} \cdot e^{i\beta} cos⁡(α+β)+isin⁡(α+β)=(cos⁡α+isin⁡α)(cos⁡β+isin⁡β)\cos(\alpha+\beta) + i\sin(\alpha+\beta) = (\cos\alpha + i\sin\alpha)(\cos\beta + i\sin\beta) Simply expanding the real and imaginary parts suffices. While this proof is extremely elegant, it still targets sin⁡\sin and cos⁡\cos.

III. The Tensor Challenge: Direct Proof of the Tangent Formula

To satisfy my obsession with "proving tan directly," I attempted to introduce concepts of non-orthogonal bases and the Reciprocal Basis.

5. Reciprocal Basis Attempt

Idea: Let OM⃗,ON⃗\vec{OM}, \vec{ON} be the basis {gi}\{g_i\}. The metric tensor matrix is G=[cos⁡αcos⁡βsin⁡αsin⁡β]G = \begin{bmatrix} \cos\alpha & \cos\beta \\ \sin\alpha & \sin\beta \end{bmatrix}. Using Gaussian elimination to find the inverse yields the reciprocal basis {gi}\{g^i\}.

Process: In the note from the 20th, I derived a determinant involving tan⁡(α−β)\tan(\alpha-\beta), but there was a sign error. In the note from the 23rd, I corrected the calculation, constructed a vector z⃗=(sin⁡(α−β),0,cos⁡(α−β))\vec{z} = (\sin(\alpha-\beta), 0, \cos(\alpha-\beta)), and calculated the following determinant: 1/tan⁡(α−β)=∣101tan⁡(α−β)−tan⁡β10tan⁡α−10∣1/\tan(\alpha-\beta) = \begin{vmatrix} 1 & 0 & \frac{1}{\tan(\alpha-\beta)} \\ -\tan\beta & 1 & 0 \\ \tan\alpha & -1 & 0 \end{vmatrix} After expansion and simplification, I finally obtained: 1tan⁡(α−β)=1+tan⁡αtan⁡βtan⁡α−tan⁡β\frac{1}{\tan(\alpha-\beta)} = \frac{1 + \tan\alpha\tan\beta}{\tan\alpha - \tan\beta} This verifies the tangent difference formula. Although the manual calculation was somewhat tedious, it proves the feasibility of handling trigonometric functions using the properties of covariant and contravariant vectors.

IV. Supplements from Higher Viewpoints

After organizing the notes above, I realized that these methods actually imply deeper mathematical structures. Here are a few higher-level perspectives I have added:

6. Analytical Perspective: ODE Uniqueness

Disregarding geometry, sin⁡\sin and cos⁡\cos are essentially linearly independent solutions to the differential equation y′′+y=0y'' + y = 0. Define f(x)=sin⁡(x+a)f(x) = \sin(x+a). It is easy to prove that f(x)f(x) is also a solution to this equation. According to the Uniqueness Theorem for solutions of second-order linear ODEs, f(x)f(x) must be a linear combination of sin⁡x\sin x and cos⁡x\cos x: sin⁡(x+a)=C1sin⁡x+C2cos⁡x\sin(x+a) = C_1 \sin x + C_2 \cos x By substituting x=0x=0 and x′(0)x'(0) to determine the coefficients, one can derive the formula purely analytically. This is a "hardcore" analytical proof.

7. The Ultimate Geometric Perspective: Ptolemy's Theorem

This is a "dimensional strike" (a powerful simplification) in geometry. Construct a cyclic quadrilateral on a unit circle. According to the theorem—"the product of the diagonals equals the sum of the products of the opposite sides"—the side lengths correspond directly to chord lengths (i.e., sine values). sin⁡(α+β)=sin⁡αcos⁡β+cos⁡αsin⁡β\sin(\alpha+\beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta This requires no complex auxiliary lines, relying directly on the profound properties of Euclidean geometry.

8. Algebraic Perspective: Lie Groups & The Exponential Map

Returning to my favorite "Rotation Matrix Method," elevating it to the level of the Lie Group SO(2)SO(2) makes everything transparent. Elements of the 2D rotation group SO(2)SO(2) can be represented via the exponential map R(θ)=eθXR(\theta) = e^{\theta X}, where XX is the generator (skew-symmetric matrix). Using the homomorphism property of the exponential function: e(α+β)X=eαX⋅eβXe^{(\alpha+\beta)X} = e^{\alpha X} \cdot e^{\beta X} This indicates that the trigonometric addition formula is essentially the manifestation of group addition under the exponential map. This also explains why Euler's formula (U(1)U(1) group) and the Rotation Matrix (SO(2)SO(2) group) approaches are so formally similar—they are isomorphic.

Conclusion: From initially drawing circles and finding segments, to deriving matrices and tensors, and finally understanding the underlying Group Theory structure. So-called "formulas" are merely projections of the same mathematical entity across different spaces. Although it was a long detour, and the reciprocal basis calculation was laborious, the process of exploration was perhaps more interesting than the result itself.