What exactly is going on here? (Trigonometric functions) The exercise below confuses me so much and we are going to have these on our exams. I always thought that the way to solve such a problem was by combining the two functions $f_1 + f_2$ and then solving for A, omega and phi by using trigonometric identities so that the sum of the two functions can be written on the standard form $f_2(t)=A*cos(ω*t+ω)$. However, in this case that wouldn't make much sense either, since the function that needs to be changed to the standard form is called $f_2$, and that function has already been given. If the exercise was about combining the two functions (summing the two functions), then we would have created a new function named for instance $f_3$, which would have consisted of $f_1 + f_2$.
Could someone provide me with the name of the theory that they are using to obtain the solution below?

 A: There's no deep theory that's being used here.  It's simply using the trig identity $\sin(\theta)=\cos\left (\theta - \frac{\pi}{2}\right)$ and taking $\theta = \frac{\pi}{2}t+\frac{\pi}{4}$.  
The identity comes from the fact that sine and cosine are 90 degrees apart.  In radians this is $\frac{\pi}{2}$.  The problem doesn't ask you to combine the functions.
A: 
A function $f(x)$ is co-function of a function $g(x)$ if $f(A) = g(B)$ whenever $A$ and $B$ are complementary angles. 

The co-function identities show the relationship between sine, cosine, tangent, cotangent, secant and cosecant. The value of a trigonometric function of an angle equals the value of the co-function of the complement. 
Recall from geometry that a complement is defined as two angles whose sum is $90°$.
In Degree
$$sin(90° - x) = cos x \quad \text{and} \quad cos(90° - x) = sin x$$
$$tan(90° - x) = cot x \quad \text{and} \quad cot(90° - x) = tan x$$
$$sec(90° - x) = csc x \quad \text{and} \quad csc(90° - x) = sec x$$
In Radians $\pi=180°\implies \frac{\pi}{2}=90°$
$$sin(\frac{\pi}{2} - x) = cos x \quad \text{and} \quad cos(\frac{\pi}{2} - x) = sin x$$
$$tan(\frac{\pi}{2} - x) = cot x \quad \text{and} \quad cot(\frac{\pi}{2} - x) = tan x$$
$$sec(\frac{\pi}{2} - x) = csc x \quad \text{and} \quad csc(\frac{\pi}{2} - x) = sec x$$
with these you also have to remember the table 

