Find the angle at which the curves sin y=ae^-x and cos
Find the angle at which the curves $\sin y=ae^{-x}$ and $\cos y=be^{-x}$ intersect.
Explanation of Method
Two curves that turn out to be orthogonal at every crossing, regardless of the parameters.
Step 1: Slopes by implicit differentiation
\[ \cos y\cdot y'=-ae^{-x} \;\Longrightarrow\; k_{1}=\frac{-ae^{-x}}{\cos y}, \qquad -\sin y\cdot y'=-be^{-x} \;\Longrightarrow\; k_{2}=\frac{be^{-x}}{\sin y}. \tag{1,2} \]
Step 2: Evaluate at a common point
At any common point $(x_0,y_0)$: $\sin y_0=ae^{-x_0}$, $\cos y_0=be^{-x_0}$. Substituting, \[k_{1}=\frac{-ae^{-x_0}}{be^{-x_0}}=-\frac ab, \qquad k_{2}=\frac{be^{-x_0}}{ae^{-x_0}}=\frac ba. \tag{3}\]
Step 3: Product of slopes
\[ k_{1}k_{2} = \left(-\frac ab\right)\!\left(\frac ba\right) = -1. \tag{4} \]
Result
Final answer:
\[\boxed{\,\theta = 90°\,}\]
the two curves are orthogonal at every intersection point, for all $a,b$.
Remark
Remark. Squaring and adding the two defining equations gives $(a^2+b^2)e^{-2x_0}=1$, so an intersection point exists at $x_0=\tfrac12\ln(a^2+b^2)$ (with $y_0$ then determined up to the usual periodicity of $\sin,\cos$).