Question #dcedf

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Question #dcedf

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Answer 1 · 2016-04-07T17:20:38

$θ_{1} \approx .333 and θ_{2} \approx 1.238$ $theta_1 ~~ .333 and theta_2~~1.238$
Solution obtained Graphically... see below and explanation

enter image source here

Solution set on $(0, π) \Rightarrow (.333, 0) and (1.238, 0)$ $(0,pi) => (.333, 0) and (1.238,0)$

Explanation:

Given: $c o 2 θ = tan 2 θ$ $co2theta = tan2theta " "$ $θ : θ \in 0 < θ < π$ $theta: theta in 0"<"theta"<"pi$

Required: The solution to $\Rightarrow r (θ) = c o 2 θ - tan 2 θ$ $=>r(theta)=co2theta - tan2theta$
over $θ : θ \in (0, π)$ $theta: theta in(0, pi )$

Solution Strategy:
a) Use trigonometric identity - $tan 2 θ = \frac{sin 2 θ}{cos 2 θ}$ $tan2theta= (sin2theta)/(cos2theta)$
b) Solve the rational function resulting from a) by any means necessary...

a) Substituting $tan θ = \frac{sin θ}{cos θ}$ $tantheta= sintheta/costheta$ we write:
$cos 2 θ = \frac{sin 2 θ}{cos 2 θ}$ $cos2theta = (sin2theta)/(cos2theta)$
${cos}^{2} 2 θ = sin 2 θ$ $cos^(2)2theta = sin2theta$ now this leads to
1) $r (θ) = {cos}^{2} 2 θ - sin 2 θ = 0$ $r(theta)=cos^(2)2theta -sin2theta=0$ find the roots
substitute $sin 2 θ = \sqrt{1 - {cos}^{2} 2 θ}$ $sin2theta = sqrt[1-cos^(2)2theta]$
2) $r (θ) = {cos}^{2} 2 θ - \sqrt{1 - {cos}^{2} 2 θ}$ $r(theta)= cos^(2)2theta - sqrt[1-cos^(2)2theta]$

b) Solve 1) form directly or 2) form indirectly
I have chosen to so solve form 1) graphically see below:
Over $θ : θ (0, π)$ $theta:theta (0,pi)$ we have 2 solutions
$θ_{1} \approx .333 and θ_{2} \approx 1.238$ $theta_1 ~~ .333 and theta_2~~1.238$
enter image source here

You also cam graph, #r(theta)=cos^(2)2theta -sin2theta and locate the zeros, x intercept...

enter image source here

Get the same answer (.333, 0) and (1.238,0)

Good luck!

Answer 2 · 2016-04-07T17:46:01

$19^{\circ} 09 and 70^{\circ} 92$ $19^@09 and 70^@92$

Explanation:

$cos 2 t = \frac{sin 2 t}{cos 2 t}$ $cos 2t = (sin 2t)/(cos 2t)$
Cross multiply -->
${cos}^{2} 2 t - sin 2 t = 0$ $cos^2 2t - sin 2t = 0$
$(1 - {sin}^{2} 2 t) - sin 2 t = 0$ $(1 - sin^2 2t) - sin 2t = 0$ . --> Trig identity: ${cos}^{2} a + {sin}^{2} a = 1$ $cos^2a + sin^2 a = 1$
$- {sin}^{2} 2 t - sin 2 t + 1 = 0$ $- sin^2 2t - sin 2t + 1 = 0$
Solve this quadratic equation for sin 2t.
$D = d^{2} = b^{2} - 4 a c = 1 + 4 = 5$ $D = d^2 = b^2 - 4ac = 1 + 4 = 5$ --> $d = \pm \sqrt{5}$ $d = +- sqrt5$
There are 2 real roots:
$sin 2 t = - \frac{b}{2 a} \pm \frac{d}{2 a} = - \frac{1}{2} \pm \frac{\sqrt{5}}{2} = - (\frac{1}{2}) ((1 \pm \sqrt{5})$ $sin 2t = -b/(2a) +- d/(2a) = - 1/2 +- sqrt5/2 = -(1/2)((1 +- sqrt5)$
a. $sin 2 t = - \frac{1 + \sqrt{5}}{2} = - \frac{3.23}{2} = - 1.62$ $sin 2t = -(1 + sqrt5)/2 = - 3.23/2 = - 1.62$ (Rejected since < -1)
b. $sin 2 t = - \frac{1 - \sqrt{5}}{2} = 0.62$ $sin 2t = - (1 - sqrt5)/2 = 0.62$
sin 2t = 0.62. Calculator gives --> $2 t = {38.17}^{\circ} - \to t = 19^{\circ} 09$ $2t = 38.17^@ --> t = 19^@09$
Trig unit circle gives another arc 2t that has the same sin value
$2 t = 180 - 38.17 = 141^{\circ} 83 - \to x = 70^{\circ} 92$ $2t = 180 - 38.17 = 141^@83 --> x= 70^@92$
Answer for $(0, π) :$ $(0, pi):$
1 $9^{\circ} 09; 70^{\circ} 92$ $9^@09; 70^@92$

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