Expert: Sherry Wallin - 7/10/2010

Question
I have a very difficult problem I am trying to solve and keep hitting a brick wall.  If you can tell me if I made a wrong step or what I should do next to solve it, I would appreciate it.

The problem is:
Sin(2Θ)+cosΘ=1  (trying to solve for Θ)

This is how I have worked it so far:
Sin(2Θ)+cosΘ=1
(2SinΘ)(cosΘ)+cosΘ=1
(2SinΘ)+1=(1/cosΘ)
2SinΘ=(1/cosΘ)-1
2SinΘ=(1-cosΘ)/cosΘ

Thanks!

Answer
JanaLee~
    Were you told to prove the identity or were you just suppose to find the value of theta that makes the statement true? By inspection when theta is zero you have a solution.
The sin 2*0 + cos 0 = 0 +1 = 1 which is what you want, right??

Math Prof

BTW, there was nothing wrong with what you were doing, you just weren't going anywhere that would help you.

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Ok so you want to solve for theta rather than find theta by inspection. This can be done in the following way:

sin(2Θ)+cosΘ=1
2sinΘcosΘ+cosΘ=1
cosΘ(2sinΘ+1)=1
2sinΘ+1=1/cosΘ
2sinΘ=1/cosΘ-1
2sinΘ=(1-cosΘ)/cosΘ     **so far nothing different than what you did, but NOW...
2sqrt(1-cos^2Θ)=(1-cosΘ)/cosΘ  where all I did was solve cos^2Θ+sin^2Θ=1 for sinΘ
4(1-cos^2Θ)=(1-2cosΘ+cos^2Θ)/cos^2Θ   I squared both sides
4-4cos^2Θ=1/cos^2Θ-2/cosΘ+1   I just simplified by expanding
0=4cos^2Θ+1/cos^2Θ-2/cosΘ+1-4   moved everything to the right
0=4cos^2Θ+1/cos^2Θ-2/cosΘ-3   just simplified
0=4cos^4Θ+1-2cosΘ-3cos^2Θ    multiplied by cos^2Θ
0=4cos^4Θ-3cos^2Θ-2cosΘ+1  put things in descending order
0=4x^4-3x^2-2x+1   let cosΘ=x and substituted
0=(x-1)(4x^3+4x^2+x-1)   factored using the rational roots theorem and synthetic division
x-1=0 and 4x^3+4x^2+x-1=0

I used a polynomial root finder found online at:
http://www.solvemymath.com/online_math_calculator/algebra_combinatorics/polynomi...

and found that 4x^3+4x^2+x-1 = 0 has a triple root of .348
x-1=0 -> x=1
Now substitute back into cosΘ=x for x=1 then  cosΘ=1 and then take the cosine inverse
of both sides getting Θ=(cos^(-1))(1) which gives us that Θ=0 degrees

again use the cosine inverse on the other value of x=.348 and you find that
cosΘ=.348 and Θ=(cos^(-1)(.348))=69.634965 degrees

check it in the original equation:
sin(2*69.634965)+cos(69.634965)=.652496+.348=1.000496 where 496 millionths is insignificant
and
sin(2*0)+cos(0) = 0 + 1 = 1

Math Prof