6.5 - Principal Values of the Inverse Trigonometric Functions
Nicole B
Ms. Blackwell
ICM(H)
Melanie was taking a electrical engineering class at North Carolina State University this past fall. During her class she learned that the average amount of power (P) of an electrical circuit with alternating current is determined by the equation P=VICosf, where V represents the voltage in the circuit, I represents the current in the circuit, and f represents the measure of the phase angle. Soon after Melanie completed the class her home was having some electrical difficulties. Melanie knew that her home circuit had a voltage of 120 volts, it produced 7.5 watts of power, and the measure of the phase angle was 84.6º. What is the current of her home circuit?
ANSWERS:
| STEP
1: First you
know that to figure out the amount of current in Melanie's home you
would use the equation P=VICosf.
From there you
would plug the correct variables into their right spot.
7.5=(120)(I)(Cos84.6) *The capital C in Cos is used to distinguish the function because it has restricted domains versus the average trigonometric function. In this case you know that you can not have a negative amperes, so I>0. STEP 2: Divide by 120*Cos84.6, to separate the I in order to solve for it. 7.5/((120)(Cos84.6))= I/((120)(Cos84.6)) I=7.5/(120*Cos84.6) STEP 3: Solve for I. I=0.66 amperes |
|
ALSO TAUGHT IN THIS SECTION!!
Step 1: Let q = Arcsin1 STEP2: Sinq = 1 STEP3: Solve for q, and we know that when sinq = 1 than q is equal to 90º. STEP 4: Therefore, Arcsin1 = 90º. |
Helpful Web Sites
http://mathworld.wolfram.com/InverseTrigonometricFunctions.html
QUOTE:
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