Question

Prove cos (pi /7) + cos(3pi /7) + cos(5pi/7) = 1/2
( no calculator)​

Answer 1

To prove that cos (pi/7) + cos(3pi/7) + cos(5pi/7) = 1/2, we can use the identity:

cos(a) + cos(b) = 2 cos[(a+b)/2] cos[(a-b)/2]

Let's apply this identity to the first two terms:

cos(pi/7) + cos(3pi/7) = 2 cos[(pi/7 + 3pi/7)/2] cos[(3pi/7 - pi/7)/2]

cos(pi/7) + cos(3pi/7) = 2 cos(2pi/7) cos(pi/7/2)

Now, let's apply the identity again to the second and third terms:

cos(3pi/7) + cos(5pi/7) = 2 cos[(3pi/7 + 5pi/7)/2] cos[(5pi/7 - 3pi/7)/2]

cos(3pi/7) + cos(5pi/7) = 2 cos(4pi/7) cos(pi/7/2)

Simplifying each term, we get:

cos(pi/7) + cos(3pi/7) = 2 cos(2pi/7) cos(pi/7/2) = 2 cos(2pi/7) sin(pi/14)

cos(3pi/7) + cos(5pi/7) = 2 cos(4pi/7) cos(pi/7/2) = 2 cos(2pi/7) cos(pi/14)

Adding these two equations together, we get:

cos(pi/7) + cos(3pi/7) + cos(3pi/7) + cos(5pi/7) = 2 cos(2pi/7) [sin(pi/14) + cos(pi/14)]

cos(pi/7) + cos(3pi/7) + cos(3pi/7) + cos(5pi/7) = 2 cos(2pi/7) [(sin(pi/14) + cos(pi/14)) / 2]

Using the Pythagorean identity sin^2(x) + cos^2(x) = 1, we can find that:

sin(pi/14) = sqrt[(1 - cos^2(pi/14)]

cos(pi/14) = sqrt[(1 - sin^2(pi/14)]

Substituting these into our equation, we get:

cos(pi/7) + cos(3pi/7) + cos(3pi/7) + cos(5pi/7) = 2 cos(2pi/7) [sqrt((1-cos(pi/14))/2) + sqrt((1+cos(pi/14))/2)]

cos(pi/7) + cos(3pi/7) + cos(3pi/7) + cos(5pi/7) = 2 cos(2pi/7) [sqrt((2sin^2(pi/28))/2) + sqrt((2cos^2(pi/28))/2)]

cos(pi/7) + cos(3pi/7) + cos(3pi/7) + cos(5pi/7) = 2 cos(2pi/7) [sin(pi/28) + cos(pi/28)]

cos(pi/7) + cos(3pi/7) + cos(3pi/7) + cos(5pi/7) = 2 cos(2pi/7) cos(pi/14)

Using the identity cos(2x) = 1 - 2sin^2

Answer 2

Explanation:

Let's start by using the trigonometric identity for the sum of two cosines:

cos(a) + cos(b) = 2 cos((a+b)/2) cos((a-b)/2)

We can apply this identity to the first two terms:

cos(pi/7) + cos(3pi/7) = 2 cos((pi/7 + 3pi/7)/2) cos((3pi/7 - pi/7)/2)

Simplifying the angles in the equation above:

cos(pi/7) + cos(3pi/7) = 2 cos(2pi/7) cos(pi/7/2)

Now we can apply the identity again to the second and third terms:

cos(3pi/7) + cos(5pi/7) = 2 cos((3pi/7 + 5pi/7)/2) cos((5pi/7 - 3pi/7)/2)

Simplifying the angles again:

cos(3pi/7) + cos(5pi/7) = 2 cos(4pi/7) cos(pi/7/2)

Notice that cos(4pi/7) is the same as cos(3pi/7 + pi/7) and we can use the identity again:

cos(4pi/7) = cos(3pi/7 + pi/7) = cos(3pi/7) cos(pi/7) - sin(3pi/7) sin(pi/7)

Substituting the expression for cos(4pi/7) and adding the two previous equations:

cos(pi/7) + cos(3pi/7) + cos(5pi/7) = 2 cos(pi/7/2) [ cos(2pi/7) + cos(3pi/7) cos(pi/7) - sin(3pi/7) sin(pi/7) ]

Now we can use the identity cos(2a) = cos^2(a) - sin^2(a) to simplify the expression inside the brackets:

cos(pi/7) + cos(3pi/7) + cos(5pi/7) = 2 cos(pi/7/2) [ cos(3pi/7) cos(pi/7) - sin(3pi/7) sin(pi/7) + cos^2(3pi/7) - sin^2(3pi/7) ]

The terms in the brackets can be simplified using the identity cos(pi/7) = cos(6pi/7) and sin(pi/7) = sin(6pi/7):

cos(pi/7) + cos(3pi/7) + cos(5pi/7) = 2 cos(pi/7/2) [ cos(3pi/7) cos(6pi/7) - sin(3pi/7) sin(6pi/7) + cos^2(3pi/7) - sin^2(3pi/7) ]

cos(pi/7) + cos(3pi/7) + cos(5pi/7) = 2 cos(pi/7/2) [ cos(3pi/7 - 6pi/7) + cos^2(3pi/7) - sin^2(3pi/7) ]

cos(pi/7) + cos(3pi/7) + cos(5pi/7) = 2 cos(pi/7/2) [ -cos(3pi/7) + cos^2(3pi/7) - sin^2(3pi/7) ]

Now we can use the identity cos^2(a)