Proving $\operatorname{cos}(x+y)=\operatorname{cos}(x)\operatorname{cos}(y)-\operatorname{sin}(x)\operatorname{sin}(y)$ using differentiation

Question

While proving $\operatorname{cos}(x+y)=\operatorname{cos}(x)\operatorname{cos}(y)-\operatorname{sin}(x)\operatorname{sin}(y)$

by this $$\operatorname{sin}(x+y)=\operatorname{sin}(x)\operatorname{cos}(y)+\operatorname{cos}(x)\operatorname{sin}(y) \\ \text{differentiating both sides w.r.t } x \\ \operatorname{cos}(x+y) \left(1+\frac{dy}{dx}\right)=(\operatorname{cos}(x)\operatorname{cos}(y)-\operatorname{sin}(x)\operatorname{sin}(y))\left(1+\frac{dy}{dx}\right)\\ \text{for $\frac{dy}{dx} \neq -1$}\\\operatorname{cos}(x+y)=\operatorname{cos}(x)\operatorname{cos}(y)-\operatorname{sin}(x)\operatorname{sin}(y) $$ Now I am confused what happens when$\frac{dy}{dx} = -1$

Answer 1

If all you want to do is derive $$\cos(x+y)=\cos(x)\cos(y)-\sin(x)\sin(y)$$ from $$\sin(x+y)=\sin(x)\cos(y)+\cos(x)\sin(y)$$ then you actually don't really care about the case where $\frac{dy}{dx}=-1$ - the implication only uses any single evaluation of this general derivative. You could get the same result by differentiating treating $x$ as constant or $y$ as constant or taking $y=x+c$. The only case that wouldn't work is if you tried to derive this taking $x+y$ to be constant.

A bit more justified would be to take a total derivative of this expression; you would get: $$d\sin(x+y) = \cos(x+y)\,dx + \cos(x+y)\,dy$$ and $$d(\sin(x)\cos(y)+\cos(x)\sin(y))=\cos(x)\cos(y)\,dx-\sin(x)\sin(y)\,dy-\sin(x)\sin(y)\,dx+\cos(x)\cos(y)\,dy$$ which tells you how the expression must change no matter which direction you move in - geometrically, this tells you about the tangent plane to the bivariate function $\sin(x+y)$. Since these tangent planes are equal, you can actually just set the coefficient of $dx$ (or, equally well, $dy$) on either to be equal to get: $$\cos(x+y)=\cos(x)\cos(y)-\sin(x)\sin(y).$$ In more elementary terms, this is just the derivative of the given expression with respect to $x$ taking $y$ to be constant - but it's a bit more insightful to see that this is just half of an expression where the change in the given terms is written as a weighted sum of the change in $x$ and $y$.

Answer 2

To elaborate on my comment about taking a partial derivative. You are assuming that $y = y(x)$. Your expression is of the form $$f(x,y)(1 + y'(x)) = g(x,y)(1 + y'(x))$$ and your are trying to conclude the equality $$f(x,y) = g(x,y), \ \forall (x,y)\in \mathbb{R}^2 $$ But you choose $y$ to be an arbitrary function of $x$. So Suppose $$y'(x_0) = -1$$ Then of course you cannot claim that $$f(x_0,y(x_0)) = g(x_0,y(x_0))$$ So simply pick a new function $y_2$ with $y_2'(x_0) \neq 1$ and $y_2(x_0) = y(x_0)$. Then you have the desired equality $$f(x_0,y(x_0)) = g(x_0,y(x_0))$$ In particular if $y$ is constant with respect to $x$ then this is just the partial derivative since $y'\equiv 0$ . If you choose a "bad" $y= y(x)$ to begin with then your proof will not work.

Answer 3

After differentiation, you can proceed as follows $$\cos(x+y)\left(1+\frac{dy}{dx}\right)=\left(\cos(x)\cos(y)-\sin(x)\sin(y)\right)\left(1+\frac{dy}{dx}\right)$$ $$\cos(x+y)\left(1+\frac{dy}{dx}\right)-\left(\cos(x)\cos(y)-\sin(x)\sin(y)\right)\left(1+\frac{dy}{dx}\right)=0$$ $$\left(\cos(x+y)-\left(\cos(x)\cos(y)-\sin(x)\sin(y)\right)\right)\left(1+\frac{dy}{dx}\right)=0$$ $$\text{If}\ \ 1+\frac{dy}{dx}=0\quad \implies \quad \cos(x+y)-\left(\cos(x)\cos(y)-\sin(x)\sin(y)\right)\ne0 $$ $$\text{If}\ \ \frac{dy}{dx}=-1\quad \implies \cos(x+y)\ne\cos(x)\cos(y)-\sin(x)\sin(y) $$