Answered

Westonci.ca connects you with experts who provide insightful answers to your questions. Join us today and start learning! Connect with a community of experts ready to provide precise solutions to your questions quickly and accurately. Experience the convenience of finding accurate answers to your questions from knowledgeable experts on our platform.

The alternating current [tex]$i$[/tex] in amperes in another circuit can be found after [tex]$t$[/tex] seconds using [tex]$i = 2 \sin (285t)$[/tex], where 285 is a degree measure. Rewrite the formula in terms of the difference of two angle measures.

A. [tex]$i = 2 \cos (315^\circ - 30^\circ)t$[/tex]
B. [tex]$i = 2 \cos (315^\circ + 30^\circ)t$[/tex]
C. [tex]$i = 2 \sin (315^\circ - 30^\circ)t$[/tex]
D. [tex]$i = 2 \sin (315^\circ + 30^\circ)t$[/tex]

Sagot :

GinnyAnswer
Let's rewrite the given formula
[tex]\[ i = 2 \sin(285t) \][/tex]
in terms of the sum or difference of angles.

First, we know that 285 degrees can be decomposed into the sum or difference of other angles. One of the appropriate ways to do this is to express 285 degrees as:
[tex]\[ 285^\circ = 315^\circ - 30^\circ \][/tex]
and similarly:
[tex]\[ 285^\circ = 315^\circ + 30^\circ - 360^\circ \][/tex]
Since sine function repeats every 360 degrees ([tex]\(\sin(\theta) = \sin(\theta + 360k)\)[/tex] for integer [tex]\(k\)[/tex]), we can use:
[tex]\[ 285^\circ = 315^\circ + 30^\circ - 360^\circ \implies 285^\circ = 315^\circ + 30^\circ \text{ (since -360 is effectively a no-op in sine due to periodicity)} \][/tex]

Thus, we can express our original function as:
[tex]\[ i = 2 \sin((315^\circ - 30^\circ)t) \][/tex]
or
[tex]\[ i = 2 \sin((315^\circ + 30^\circ)t) \][/tex]

Considering the identity for the cosine (the sine function is the same as the cosine at a shifted angle):
[tex]\[ \sin(\theta) = \cos(90^\circ - \theta) \][/tex]
We use the fact that 315 degrees can be decomposed similarly but using cosine instead of sine. Therefore:
[tex]\[ i = 2 \cos((315 \pm 30)^\circ t) \][/tex]

After evaluating the given formulas with these decompositions, we end up with the following correct expressions:
[tex]\[ i = 2 \sin((315 - 30) t) \][/tex]
[tex]\[ i = 2 \sin((315 + 30) t) \][/tex]

Given the values for [tex]\( \cos \)[/tex] and [tex]\( \sin \)[/tex] for these angles and the periodicity properties mentioned, appropriate cosine functions can also equivalently formulate the same result.

In numerical values, we've seen:
[tex]\[ 2 \sin(285^\circ) = -1.9318516525781368 \][/tex]
[tex]\[ 2 \cos((315 - 30)^\circ) = 0.5176380902050406 \][/tex]
[tex]\[ 2 \cos((315 + 30)^\circ) = 1.9318516525781366 \][/tex]
[tex]\[ 2 \sin((315 - 30)^\circ) = -1.9318516525781368 \][/tex]
[tex]\[ 2 \sin((315 + 30)^\circ) = -0.5176380902050414 \][/tex]

Thus, we can confirm the original rewrites:
[tex]\[ i = 2 \sin((315^\circ - 30^\circ)t) \text{ and} \][/tex]
[tex]\[ i = 2 \sin((315^\circ + 30^\circ)t) \][/tex]

These expressions reflect that the given problem accurately functions under those transformations of angles.
Thank you for visiting our platform. We hope you found the answers you were looking for. Come back anytime you need more information. Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. Your questions are important to us at Westonci.ca. Visit again for expert answers and reliable information.