i = 2 cos(ωt) (Given); i^2 = { (i1 sin(ωt) + i2 cos(ωt))^2 }
Understand the Problem
The image presents a mathematical derivation involving sine and cosine functions, likely in the context of electrical engineering or physics. The calculations seem to derive a parameter based on these trigonometric functions.
Answer
The relationship from squaring gives $4 \cos^2(\omega t) = i_1^2 \sin^2(\omega t) + 2i_1i_2 \sin(\omega t) \cos(\omega t) + i_2^2 \cos^2(\omega t)$.
Answer for screen readers
The result combines two equal forms of $i^2$ leading to a relationship between $i_1$, $i_2$, and constants derived from $4 \cos^2(\omega t)$.
Steps to Solve

Understand the given expression The expression is $i = 2 \cos(\omega t)$. We need to square this and relate it to another expression involving sine and cosine.

Square the current equation Start with $$ i^2 = (2 \cos(\omega t))^2 $$ This expands to: $$ i^2 = 4 \cos^2(\omega t) $$

Expand the other side From the problem, you have: $$ i^2 = (i_1 \sin(\omega t) + i_2 \cos(\omega t))^2 $$ Now expand this using the formula $(a+b)^2 = a^2 + 2ab + b^2$: $$ i^2 = i_1^2 \sin^2(\omega t) + 2i_1i_2 \sin(\omega t) \cos(\omega t) + i_2^2 \cos^2(\omega t) $$

Set the two expressions equal Now equate both expressions for $i^2$: $$ 4 \cos^2(\omega t) = i_1^2 \sin^2(\omega t) + 2i_1i_2 \sin(\omega t) \cos(\omega t) + i_2^2 \cos^2(\omega t) $$

Organize and solve To solve for parameters, separate terms involving $\sin^2(\omega t)$ and $\cos^2(\omega t)$. Use $ \sin^2(\omega t) + \cos^2(\omega t) = 1 $ to simplify terms.
The result combines two equal forms of $i^2$ leading to a relationship between $i_1$, $i_2$, and constants derived from $4 \cos^2(\omega t)$.
More Information
This problem involves squaring trigonometric functions and exploring relationships between sine and cosine components, which often appears in electrical engineering problems related to alternating current.
Tips
 Confusing the expansion of $(a+b)^2$; remember to include all terms.
 Failing to simplify expressions and mix terms correctly.