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lyudmila [28]
3 years ago
7

Find the exact values of sin 2u, cos 2u, and tan 2u using the double-angle formulas. sec u = −2, π/2 < u < π

Mathematics
1 answer:
arlik [135]3 years ago
6 0
\bf sec(\theta)=\cfrac{hypotenuse}{adjacent}&#10;\\\\\\&#10; sec(u)=-2\iff sec(u)=-\cfrac{2}{1}\cfrac{\leftarrow hypotenuse}{\leftarrow adjacent}

now... we have -2/1   but... which one is the negative?   well
we know the angle is "<span> π/2 < u < π", that simply means, the 2nd quadrant
well, in the 2nd quadrant, the "x" or adjacent side is negative, so the adjacent side is then -1

now, let us use the pythagorean theorem, to get the opposite side

</span>\bf c^2=a^2+b^2\implies \pm\sqrt{c^2-a^2}=b\implies \pm\sqrt{(2)^2-(-1)^2}=b&#10;\\\\\\&#10;\pm\sqrt{3}=b
<span>
now..which one though?  well, recall, we're in the 2nd quadrant, in the 2nd quadrant "y" or the opposite side, is positive, so is </span>√(3) then

now... let's plug in those values in the double-angle identities then

\bf \textit{Double Angle Identities}&#10;\\ \quad \\&#10;sin(2\theta)=2sin(\theta)cos(\theta)&#10;\\ \quad \\\\&#10;cos(2\theta)=&#10;\begin{cases}&#10;cos^2(\theta)-sin^2(\theta)\\&#10;\boxed{1-2sin^2(\theta)}\\&#10;2cos^2(\theta)-1&#10;\end{cases}&#10;\\ \\\\&#10;tan(2\theta)=\cfrac{2tan(\theta)}{1-tan^2(\theta)}\\\\&#10;-----------------------------\\\\

\bf sin(2u)=2\left( \cfrac{\sqrt{3}}{2} \right)\left(  \cfrac{-1}{2}\right)&#10;\\\\\\&#10;cos(2u)=1-2\left( \cfrac{\sqrt{3}}{2} \right)^2&#10;\\\\\\&#10;tan(2u)=\cfrac{2\left( \frac{\sqrt{3}}{-1} \right)}{1-\left( \frac{\sqrt{3}}{-1} \right)^2}

simplify away
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Answer:

The quantity of salt at time t is m_{salt} = (60)\cdot (30 - 29.833\cdot e^{-\frac{t}{10} }), where t is measured in minutes.

Step-by-step explanation:

The law of mass conservation for control volume indicates that:

\dot m_{in} - \dot m_{out} = \left(\frac{dm}{dt} \right)_{CV}

Where mass flow is the product of salt concentration and water volume flow.

The model of the tank according to the statement is:

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Where:

c - The salt concentration in the tank, as well at the exit of the tank, measured in \frac{pd}{gal}.

\frac{dc}{dt} - Concentration rate of change in the tank, measured in \frac{pd}{min}.

V - Volume of the tank, measured in gallons.

The following first-order linear non-homogeneous differential equation is found:

V \cdot \frac{dc}{dt} + 6\cdot c = 3

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\frac{dc}{dt} + \frac{1}{10}\cdot c = 3

This equation is solved as follows:

e^{\frac{t}{10} }\cdot \left(\frac{dc}{dt} +\frac{1}{10} \cdot c \right) = 3 \cdot e^{\frac{t}{10} }

\frac{d}{dt}\left(e^{\frac{t}{10}}\cdot c\right) = 3\cdot e^{\frac{t}{10} }

e^{\frac{t}{10} }\cdot c = 3 \cdot \int {e^{\frac{t}{10} }} \, dt

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c = 30 + C\cdot e^{-\frac{t}{10} }

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Now, the integration constant is:

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The solution of the differential equation is:

c(t) = 30 - 29.833\cdot e^{-\frac{t}{10} }

Now, the quantity of salt at time t is:

m_{salt} = V_{tank}\cdot c(t)

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