For the expression cos (ωt + 0) the correct Pascal equivalent is

1.	For the expression cos (ωt + 0) the correct Pascal equivalent is
A.	BETA * JCOS (OMEGA * T + THETA) / SQRT (SQR (BETA) + SQR (OMEGA))
B.	BETA COS (OMEGA * T + THETA) / SQRT [SQR (BETA) + SQR (OMEGA)]
C.	BETA COS (OMEGA T + THETA) / SQRT [SQR (BETA) + SQR (OMEGA)]
D.	None of the above
Answer» B. BETA COS (OMEGA * T + THETA) / SQRT [SQR (BETA) + SQR (OMEGA)]

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For the expression cos (ωt + 0) the correct Pascal equivalent is

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