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
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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.
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Square the current equation Start with $$ i^2 = (2 \cos(\omega t))^2 $$ This expands to: $$ i^2 = 4 \cos^2(\omega t) $$
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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 $(a+b)^2 = a^2 + 2ab + b^2$: $$ i^2 = i_1^2 \sin^2(\omega t) + 2i_1i_2 \sin(\omega t) \cos(\omega t) + i_2^2 \cos^2(\omega t) $$
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Set the two expressions equal Now equate both expressions for $i^2$: $$ 4 \cos^2(\omega t) = i_1^2 \sin^2(\omega t) + 2i_1i_2 \sin(\omega t) \cos(\omega t) + i_2^2 \cos^2(\omega t) $$
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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.
