Find the exact value of cos(11p/8)

Question

A) √(2+√2)/4

B) √(2-√2)/4

C) - √(2-√2)/4

D) -√(2+√2)/4

 

 

 

 

2) 1 + cos (12x)=

 

a)2cos^2(12x)

b) 2sin^2(12x)

c)2cos^2(6x)

d) 2sin^2(6x)

 

Answer 1

Think about that angle 11π/8. That's exactly half the value of 11π/4. So, 11π/8 = (1/2)(11π/4). That means you can use the Half-Angle Formulas for θ = 11π/4 and θ/2 = 11π/8:

 

cosθ = 2cos²(θ/2) - 1

cos(11π/4) = 2cos²(11π/8) - 1

 

Take a step back and figure out what the cosine of 11π/4 is. This angle is equivalent to 11π/4 - 2π = 3π/4, and, looking at a unit circle, you'll be able to tell that 3π/4 has a cosine of -√(2)/2. Plug that value in and continue solving for cos(11π/8):

 

-√(2)/2 = 2cos²(11π/8) - 1

-√(2)/2 + 1 = 2cos²(11π/8)

-√(2)/2 + 2/2 = 2cos²(11π/8)

(-√(2) + 2)/2 = 2cos²(11π/8)

(-√(2) + 2)/4 = cos²(11π/8)

(2 - √(2))/4 = cos²(11π/8)

±√[(2 - √2)/4] = cos(11π/8)

 

11π/8 is in the third quadrant (between π and 3π/2), so cosine is negative:

 

-√[(2 - √2)/4] = cos(11π/8)

 

 

 

For the second question, let θ = 12 and let θ/2 = 6. Then:

 

cosθ = 2cos²(θ/2) - 1

cos(12x) = 2cos²(6x) - 1

cos(12x) + 1 = 2cos²(6x)

1 + cos(12x) = 2cos²(6x)