Three derivatives to know by heart, with $x$ in radians: $\dfrac{d}{dx}\,\sin x = \cos x$, the derivative of sine is cosine; $\dfrac{d}{dx}\,\cos x = -\sin x$, the derivative of cosine is minus sine; $\dfrac{d}{dx}\,\tan x = \sec^2 x$, the derivative of tangent is secant squared, where $\sec x = \dfrac{1}{\cos x}$ (read aloud secant) so that $\sec^2 x$ is $\dfrac{1}{\cos^2 x}$. The minus on cosine's derivative is the first trap in the lesson's name, and the picture settles it: $\cos x$ starts at its peak of 1 and FALLS, so its slope just after 0 is negative, and $-\sin x$ is negative there. Sine starts at 0 and rises at slope exactly 1, which is $\cos 0$. Tangent comes from the quotient rule on $\dfrac{\sin x}{\cos x}$: $\dfrac{\cos^2 x + \sin^2 x}{\cos^2 x} = \dfrac{1}{\cos^2 x}$.

The second trap is degrees. These formulas hold ONLY in radians: a full turn is $2\pi$, with $\pi = 3.14159\ldots$, never 3. In degrees, every slope picks up a factor of $\dfrac{\pi}{180}$, because a degree is a smaller step than a radian; a calculator left in degree mode reads $\dfrac{\pi}{6}$ as 0.523599°, a sliver of an angle whose cosine is nearly 1, and reports nonsense. Keep $\pi$ as $\pi$: the exam's points are $\dfrac{\pi}{6}, \dfrac{\pi}{4}, \dfrac{\pi}{3}, \dfrac{\pi}{2}, \pi$ and their multiples, where sine and cosine are exact.