Prove the trigonometric identity?

Question

cos π/5 + cos 3π/5 = 1/2

This one is very troubling...

^_^

Answer 1

One way to prove this is by knowing something about the golden triangle. This is a 36-72-72 degree triangle, whose sides are the ratio of Φ : Φ : 1, where Φ is the golden ratio, (1+√5)/2. This number has the property of Φ^2 = Φ+1, which is useful for simplifying. If you draw the perpendicular bisector of the triangle on the 36-degree angle, you get two right triangles, and can show that cos(72) = sin(18) = 1 / 2Φ, and cos(18) = sin(72) = √(4Φ^2 - 1) / 2Φ = √(4Φ+3) / 2Φ.

So, if you don't mind me converting the radians to degrees here...

cos(36) + cos(108) =
cos(2*18) + cos(180 - 72) =
[ cos^2(18) - sin^2(18) ] - cos(72) =
[√(4Φ+3) / 2Φ]^2 - (1 / 2Φ)^2 - [1 / 2Φ] =
[ (4Φ+3 - 1) / 4Φ^2 ] - [1 / 2Φ] =
[ (4Φ+2) / 4Φ^2 ] - [2Φ / 4Φ^2] =
(4Φ + 2 - 2Φ) / 4Φ^2 =
(2Φ + 2) / 4Φ^2 =
(Φ + 1) / 2Φ^2 =
(Φ^2) / 2Φ^2 =
1/2

Or, if you wanted to just simplify this into terms of π/5:
cos(π/5) + cos(3π/5) =
cos(π/5) + cos(π/5 + 2π/5) =
cos(π/5) + cos(π/5)cos(2π/5) - sin(π/5)sin(2π/5)
cos(π/5) + cos(π/5)cos(2π/5) - sin(π/5)*2sin(π/5)cos(π/5)
cos(π/5) [ 1 + cos(2π/5) - 2sin(π/5)sin(π/5) ]
cos(π/5) [ 1 + cos(2π/5) - 2sin^2(π/5) ]
cos(π/5) [ 1 + cos(2π/5) - 2[1 - cos^2(π/5)] ]
cos(π/5) [ cos(2π/5) - 1 + 2cos^2(π/5)] ]
cos(π/5) [ (2cos^2(π/5)-1) - 1 + 2cos^2(π/5)] ]
cos(π/5) [ 4cos^2(π/5) - 2 ]
2 cos(π/5) [ 2cos^2(π/5) - 1 ]
Then if you knew cos(π/5) = Φ/2, you could plug it in and simplify

Answer 2

The solution is fairly simple.
The identity to be proved is cos(36) + cos (108) = 1/2, or cos(36) - sin (18) = 1/2.
Let A = 18 degrees, then we want to prove
cos(2A) - sin A = 1/2.
Starting from sin (36) = cos (54), or sin (2A) = cos (3A) = cos (2A + A) = cos(2A) cos A - sin(2A) sin A = cos(2A) cos A - 2*(sin A)^2* cos A,
or
2*sin A cos A = cos(2A) cos A - 2*(sin A)^2* cos A
Dividing cos A on both sides, (noting cos(18) not = 0),
we get
2*sin A = cos(2A) - 2*(sin A)^2
= cos (2A) - (1 - cos (2A))
= 2* cos (2A) -1
So
cos (2A) - sin A = 1/2.
It is just what we want to prove.

Answer 3

You could use the identity cos2theta = cos^2 theata - sin^2 theta....you don't have a two but you have a 1/5...I think you should then be able to evaluate whats. left.

Answer 4

very easy if you use roots of unity.

Denote:
A = e^i*pi/5,
B = e^3i*pi/5
C = e^5i*pi/5 = -1
D = e^7i*pi/5
E = e^9i*pi/5

We want to find Re(A+B)which is equivalent to the given expression.

Note A,B,C,D,E are equally spaced about the unit circle in the complex plane so Re(A+B+C+D+E) = 0

Also, the points are symmetric about the real axis (x-axis) so Re(A)=Re(E) and Re(B) = Re(D). Obviously Re(C) = -1.

So,
Re(A+B+C+D+E)=0 => Re(2A+2B) = 1 => Re(A+B) = 1/2.

I'm almost certain this is the way it was meant to be solved. Most tricky problems involving sums of cosines or sines can be solved by converting to the complex plane (typical of USAMO problems).

Click Best Answer if this was helpful. Thanks!