cos(11pi/12) using Sum and difference identities

Question

could someone explain finding cos(11pi/12) using Sum and difference Identities.

it must use one of these:

 

cos(α+β)=cos(α)cos(β)−sin(α)sin(β)

 

or

 

cos(α+β)=cos(α)cos(β)+sin(α)sin(β)

 

Im a bit hung up figuring out what (α)and(β)shouldbethough

Answer 1

cos(11π/12) = cos(2π/12 + 9π/12)

 

                   = cos(π/6 + 3π/4)

 

                   = cos(π/6)cos(3π/4) - sin(π/6)sin(3π/4)

 

                   = (√3/2)(-√2/2) - (1/2)(√2/2)

 

                    = (-√6 - √2)/4

Answer 2

11π / 12

8π/12 + 3π/12

2π/3 + π/4

Answer 3

The key here is to break up 11pi/12 into a sum or difference of fractions. Knowing this is our intermediate goal we want to find common frations of pi (e.g. pi/2, pi/3, pi/4 and pi/6).

 

Let's begin with the 11pi/12. Break 11 into the largest factor of 12 >>> 6. This gives us 6pi/12 + 5pi/12 and the first identity can be written with 6pi/12 being simplified to pi/2 (90 degrees).

 

So we have cos(11pi/12)=cos(pi/2 + 5pi/12) = cos(pi/2)cos(5pi/12)−sin(pi/2)sin(5pi/12)

 

cos(pi/2) = 0 so the first term zero's out and sin(pi/2) is 1 leaving -sin(5pi/12).

 

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Note that part 1 can also be determined by:

cos(11pi/12) = -sin(5pi/12) based on two other identities that cos(x) = sin(pi/2 - x) = - sin(x - pi/2)

 

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Deal with complexity by finding the structure or pattern.

 

In this case the next factor of 12 to work with is 3 leaving 2 behind. 5pi/12 = 3pi/12 + 2pi/12 = pi/4 + pi/6

 

So we have -sin(5pi/12)=-sin(pi/4 + pi/6) = -[sin(pi/4)cos(pi/6)+cos(pi/4)sin(pi/6)]

 

=-[sqrt(2)/2 *sqrt(3)/2 + sqrt(2)/2*1/2]

 

=-[sqrt(6)/4 + sqrt(2)/4]

 

=-0.25[sqrt(6) + sqrt(2)]

 

Hope this helps!