Trig Functions
Sine Wave
A sine wave starts at zero when $\theta=0$, rises to its maximum, drops through zero to its minimum and returns to zero, repeating this cycle. The vertical distance from zero to the peak is the amplitude. The number of peaks per unit time is the frequency (symbol $\nu$, "nu"). The horizontal distance between two peaks is the wavelength (symbol $\lambda$, "lambda").
Cosine Wave
A cosine wave is like a sine wave except it starts at its maximum, drops through zero to its minimum, then returns through zero to its maximum, repeating again and again. ($\cos\theta$ is $\sin\theta$ shifted by a quarter cycle.)
Combined Wave
Imagine a sine wave of amplitude $a$ and a cosine wave of amplitude $b$ adding together: $y=a\sin\theta+b\cos\theta$. We can rewrite this as a single shifted sine wave.
From the addition formula, $r\sin(\theta + \phi) = r\sin\theta\cos\phi + r\cos\theta\sin\phi$.
Comparing with $y = a\sin\theta + b\cos\theta$, let $a = r\cos\phi$ and $b = r\sin\phi$. Then $a^2 + b^2 = r^2(\cos^2\phi + \sin^2\phi) = r^2$, so
$$r=\sqrt{a^2 + b^2}, \qquad \tan\phi = \dfrac{b}{a}, \qquad \phi = \tan^{-1}\!\left(\tfrac{b}{a}\right)$$
Choose $\phi$ for the correct quadrant (your calculator's $\tan^{-1}$ only returns values between $-\pi/2$ and $\pi/2$):
$a > 0,\ b > 0$: $0 < \phi < \pi/2$
$a < 0,\ b > 0$: $\pi/2 < \phi < \pi$
$a < 0,\ b < 0$: $\pi < \phi < 3\pi/2$
$a > 0,\ b < 0$: $3\pi/2 < \phi < 2\pi$
Example 3b: Find the amplitude and phase of $y=-4\sin\theta+3\cos\theta$.
Comparing with $r\sin(\theta+\phi)$: $r\cos\phi=-4$ and $r\sin\phi=3$, so $r=\sqrt{(-4)^2+3^2}=5$.
$\tan\phi = 3/(-4)$, so $\phi=\tan^{-1}(3/-4)$. The calculator gives $-0.64$ rad ($-36.87^\circ$); since $a<0, b>0$ we add $\pi$ to get $\phi=2.50$ rad ($143.13^\circ$).
So $y=-4\sin\theta+3\cos\theta = 5\sin(\theta+2.50)$.
Equations of the Form $a\sin\theta+b\cos\theta = c$
To solve $a\sin\theta+b\cos\theta = c$, transform the left side to $r\sin(\theta+\phi)$ and set it equal to $c$:
$r\sin(\theta+\phi) = c \;\Rightarrow\; \sin(\theta+\phi) = c/r \;\Rightarrow\; \theta = \sin^{-1}(c/r) - \phi$
There is a second solution. The two solutions in $0 \le \theta \le 2\pi$ are
$$\theta_1 = \sin^{-1}(c/r)-\phi, \qquad \theta_2 = \pi-\sin^{-1}(c/r)-\phi$$
Add $2\pi$ to any negative result to bring it into range.
Example 4: Solve $4\sin\theta+3\cos\theta = 2$ for $0 \le \theta \le 2\pi$.
$r=\sqrt{3^2+4^2}=5$ and $\tan\phi=3/4$, so $\phi=0.64$ rad. Then $5\sin(\theta+0.64)=2$, so $\sin(\theta+0.64)=2/5$ and $\theta+0.64=\sin^{-1}(0.4)=0.41$.
$\theta=0.41-0.64=-0.23$; adding $2\pi$ gives $\theta_1=6.05$ rad. The other solution is $\theta_2=\pi-0.41-0.64=2.09$ rad.
Combined Wave Turning Points
For $y= a\sin\theta + b\cos\theta + c$, the sine term varies between $+1$ and $-1$, so:
Maximum $=\sqrt{a^2+b^2}+c$, at $\theta+\phi=\pi/2$, i.e. $\theta_{max}=\pi/2-\phi$
Minimum $=-\sqrt{a^2+b^2}+c$, at $\theta+\phi=3\pi/2$, i.e. $\theta_{min}=3\pi/2-\phi$
(In general maxima are at $\theta=(2n+\tfrac12)\pi-\phi$ and minima at $\theta=(2n+\tfrac32)\pi-\phi$; give values in $0 \le \theta \le 2\pi$.)
Example 5: Find the first maximum and minimum of $y= 4\sin\theta+3\cos\theta + 3$.
$r=\sqrt{3^2+4^2}=5$ and $\phi=\tan^{-1}(3/4)=0.64$ rad.
Maximum $=5+3=8$ at $\theta_{max}=\pi/2-0.64=0.93$ rad. Minimum $=-5+3=-2$ at $\theta_{min}=3\pi/2-0.64=4.07$ rad.