Question #dcedf

Given: #co2theta = tan2theta " "# #theta: theta in 0"<"theta"<"pi#

Required: The solution to #=>r(theta)=co2theta - tan2theta#
over #theta: theta in(0, pi )#

Solution Strategy:
a) Use trigonometric identity - #tan2theta= (sin2theta)/(cos2theta)#
b) Solve the rational function resulting from a) by any means necessary...

a) Substituting #tantheta= sintheta/costheta# we write:
#cos2theta = (sin2theta)/(cos2theta) #
#cos^(2)2theta = sin2theta# now this leads to
1) #r(theta)=cos^(2)2theta -sin2theta=0# find the roots
substitute #sin2theta = sqrt[1-cos^(2)2theta]#
2) #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 #theta:theta (0,pi)# we have 2 solutions
#theta_1 ~~ .333 and theta_2~~1.238#

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

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

Good luck!

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