Miscellaneous · Q19: $\cos(A+B)$ by differentiating the formula for $\sin(A+B)$

Basics you need

Solution

1. Fix $B$ and let $A$ vary

   Choose any real number $B$ and keep it fixed. Writing $x$ in place of $A$⁠, the given fact says that for every real $x$⁠,

   $$\sin(x + B) = \sin x\cos B + \cos x\sin B.$$
   To use differentiation we need a variable. The formula is true for all $A$⁠, so we may treat $A$ as a variable $x$ while $B$ plays the role of a constant. Then $\cos B$ and $\sin B$ are fixed numbers.
   The sine sum formula (Class XI, Trigonometric Functions). Basics: trigonometric identities

2. Why we may differentiate both sides

   Let $F(x) = \sin(x + B)$ and $G(x) = \sin x\cos B + \cos x\sin B$⁠. Step 1 says $F(x) = G(x)$ for every real $x$⁠, so $F$ and $G$ are the same function. Hence

   $$F'(x) = G'(x) \quad \text{for every real } x.$$
   Differentiating "both sides of an equation" is only safe when the equation is an identity in the variable, as here. (By contrast, $x^2 = 4$ holds only at $x = \pm 2$⁠, and differentiating it to get $2x = 0$ would be nonsense.)
   Basic 4.

3. Differentiate the left side with the chain rule

   Outer function $\sin u$⁠, inner function $u = x + B$ with $\frac{du}{dx} = 1$ (since $B$ is constant):

   $$F'(x) = \cos(x + B)\cdot 1 = \cos(x + B).$$
   The chain rule multiplies the outer derivative $\cos u$ by the inner derivative. Here the inner derivative is 1, so the result is simply $\cos(x+B)$⁠.
   Chain rule, Theorem 4, Section 5.3.1 (Basic 2). Lesson: chain rule

4. Differentiate the right side term by term

   $$\begin{aligned} G'(x) &= \cos B\cdot\frac{d}{dx}\sin x \\ &\quad + \sin B\cdot\frac{d}{dx}\cos x \\ &= \cos B\cos x + \sin B\,(-\sin x) \\ &= \cos x\cos B - \sin x\sin B. \end{aligned}$$
   $\cos B$ and $\sin B$ are constants, so they stay as factors. The derivative of $\sin x$ is $\cos x$ and the derivative of $\cos x$ is $-\sin x$⁠: that minus sign is the minus sign in the cosine formula.
   Sum and constant multiple rules, Section 5.3; Table 5.3 (Basics 1, 3). Basics: derivative rules

5. Equate, and put $x = A$

   By Step 2, $F'(x) = G'(x)$⁠, so for every real $x$

   $$\cos(x + B) = \cos x\cos B - \sin x\sin B.$$

   Taking $x = A$⁠, and recalling that $B$ was arbitrary,

   $$\cos(A + B) = \cos A\cos B - \sin A\sin B$$

   for all real $A$ and $B$⁠.

   The identity of Step 5 holds for every $x$ and for whichever $B$ we fixed in Step 1, so it holds for all pairs $A$⁠, $B$⁠. That is the sum formula for cosines.
   Basic 4. Basics: trigonometric identities

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