cos(2?)=1/v2 and 180 degrees<?<270 degrees

Question

Find the value for cos(θ) if the following conditions hold: cos(2θ)=1/√2 and 180^o<θ<270^o

a) -(√1+(√2))

b) -(√2+(√2))/4

c) -(√2(√2))/2

d) -(√2(√2))

 

Answer 1

 answer withdrawn because I misread the question

 

 

 

          

Answer 2

Given 180º < θ < 270º, which is angle in Quadrant III.  For cos(θ) in this quadrant, the value is a negative number.

 

Now 360º < 2θ < 540º => 0º < 2θ < 180º  Since cos(2θ) > 0, 2θ must be in Quadrant I and has a generalize form of 2θ + 360n, n is any integer.  This also means that half this angle is θ + 180n, which lies in Quadrant III when n is an odd integer.  This would make cos(θ) a negative number.

 

From the half-angle formula, cos(θ) = -√((1 + cos(2θ))/2), where the correct sign has been used:

 

When cos(2θ) = 1/√(2) = √(2)/2, then cos(θ) = -√((1 +√(2)/2)/2)

 

                                                                        = -√(2 + √2)/2 (after simplification) ≈ -0.924

 

Also, θ = 202.5º

 

This value is not the same as the given answers, so I am assuming that there is a misprint in the answers.

Answer 3

Leslie,

 

If the cos(2θ) = 1/√2, then

 

cos^-1 (1/√2) = 2θ

 

Rationalizing 1/√2 gives us √2/2.  This is something we recognize from the unit circle.

cos 45^o = √2/2 

 

45^o = 2θ

 

We're not finished yet because we need θ, not 2θ and we need our answer in the between 180^o and 270^o

 

Cos is positive in quadrants I and IV.

So, 2θ is also equal to 315^o

 

Our general solution is 

2θ = 45^o + 360^on and

2θ = 315o + 360^on, where n = number of times around the unit circle

 

Dividing by 2

θ = 22.5^o + 180^on  and

θ = 162.5^o + 180^on

 

When n = 1

θ = 202.5^o and
θ = 342.5^o

But the 2nd solution is outside the given range so θ = 202.5^o

 

Cos 202.5 = -0.924

 

 

Answer 4

Use identity:  cos(2x) = 2(cosx)^2 -1 = 1/ sqrt(2)

 

Your answer will be negative since the angle condition places it in the 3rd quadrant.