Trigonometry

Functions

Sine
Cosine
Tangent
Hyperbolic Functions

Solving something of the form

A \sin x + B \cos x = C

Express $g (t) = \cos (5 t) - \sin (5 t)$ as $g (t) = A \cos (5 t + β)$
Separate the second one by trig identity
Compare that with the first equation
Then you get $A \cos β = 1$ , $- A \sin β = - 1$
Square the two and add

\begin{aligned} A^{2} \cos^{2} β + A^{2} \sin^{2} β & = 1^{2} + 1^{2} \\ A & = \sqrt{2} (∵ A > 0) \\ \cos β & = \frac{1}{\sqrt{2}} \\ \sin β & = \frac{1}{\sqrt{2}} \\ β & = \frac{π}{4} \\ g (t) & = \sqrt{2} \cos (5 t + \frac{π}{4}) \end{aligned}

we can say $A > 0$ because $A$ represents amplitude (allowing $A < 0$ yields an equivalent result anyway)

Integral of $\sec θ$ :

\begin{aligned} \int \sec θ d θ & = \int \frac{\sec θ (\sec θ + \tan θ)}{\sec θ + \tan θ} d x \\ u & = \sec θ + \tan θ \\ d u & = \tan θ \sec θ + \sec^{2} θ d x \\ \int \sec θ d θ & = \int \frac{1}{u} d u \\ = \ln | u | \\ = \ln | \sec θ + \tan θ | \end{aligned}