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denis23 [38]
3 years ago
7

Prove the trig identity: tan^2 x cos^2 x = (sec^2 x -1)(1-sin^4 x)/1+sin^2 x

Mathematics
1 answer:
ella [17]3 years ago
5 0

Answer:

See steps below

Step-by-step explanation:

We need to work with each side of the equation at a time:

Left hand side:

Write all factors using the basic trig functions "sin" and "cos" exclusively:

tan^2(x)\,cos^2(x)=\frac{sin^2(x)}{cos^2(x)} cos^2(x)=sin^2(x)

now, let's work on the right side, having in mind the following identities:

a)  sec^2(x)-1=tan^2(x)=\frac{sin^2(x)}{cos^2(x)}

b) 1-sin^4(x) =(1-sin^2(x))\,(1+sin^2(x))

c) 1-sin^2(x)=cos^2(x)

Then replacing we get:

\frac{(sec^2(x)-1)\,(1-sin^4(x))}{1+sin^2(x)} =\frac{sin^2(x)(1+sin^2(x))(1-sin^2(x))}{cos^2(x)(1+sin^2(x)))} =\frac{sin^2(x)(1+sin^2(x))\,cos^2(x)}{cos^2(x)(1+sin^2(x)))}=sin^2(x)

Therefore, we have proved that the two expressions are equal.

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Meg plotted the graph below to show the relationship between the temperature of her city and the number of sweaters sold at a st
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The explanation about the graph is illustrated below.

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7x=2x-35<br><br><br> A. 7<br><br> B.5<br><br><br> C.-5<br><br><br> D.-7
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Hydrocodone bitartrate is used as a cough suppressant. After the drug is fully absorbed, the quantity of drug in the body decrea
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Answer:

a. dQ/dt = -kQ

b. Q = 9e^{-kt}

c. k = 0.178

d. Q = 1.063 mg

Step-by-step explanation:

a) Write a differential equation for the quantity Q of hydrocodone bitartrate in the body at time t, in hours, since the drug was fully absorbed.

Let Q be the quantity of drug left in the body.

Since the rate of decrease of the quantity of drug -dQ/dt is directly proportional to the quantity of drug left, Q then

-dQ/dt ∝ Q

-dQ/dt = kQ

dQ/dt = -kQ

This is the required differential equation.

b) Solve your differential equation, assuming that at the patient has just absorbed the full 9 mg dose of the drug.

with t = 0, Q(0) = 9 mg

dQ/dt = -kQ

separating the variables, we have

dQ/Q = -kdt

Integrating we have

∫dQ/Q = ∫-kdt

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when t = 0, Q = 9

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So, Q = 9e^{-kt}

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ln\frac{1}{2} = ln(e^{-3.9k})\\\\-ln2 = -3.9k\\k = -ln2/-3,9 \\k = -0.693/-3.9\\k = 0.178

d) How much of the 9 mg dose is still in the body after 12 hours?

Since k = 0.178,

Q = 9e^{-0.178t}

when t = 12 hours,

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