Let x, y, and z be real numbers such that cos(x) + cos(y) + cos(z) = sin(x) + sin(y) + sin(z) = 0.
Prove that cos(2x) + cos(2y) + cos(2z) = sin(2x) + sin(2y) + sin(2z) = 0.
Oh wait, I think I'm onto something and might have the solution! (Hopefully it's right... I think I'll post it on here when I'm done for confirmation)
Screenshot of my solution here (get rid of all spaces and the url should work):
lh3. googleusercontent. com/-Tpu8S83BSbc/Vy1oGcAFEiI/AAAAAAAAD-Q/LDg0yszPVOcN8pWYNyH11rLstx00HpSKQCK8B/s512/2016-05-06. png
Can someone check it please?
To prove the given statement, we will use the trigonometric identities for double angles:
cos(2x) = 2cos^2(x) - 1
sin(2x) = 2sin(x)cos(x)
Let's start by proving that cos(2x) + cos(2y) + cos(2z) = 0:
1. First, we express cos(2x) as 2cos^2(x) - 1, and similarly for y and z:
cos(2x) + cos(2y) + cos(2z) = (2cos^2(x) - 1) + (2cos^2(y) - 1) + (2cos^2(z) - 1)
2. Next, we rearrange the terms:
cos(2x) + cos(2y) + cos(2z) = 2(cos^2(x) + cos^2(y) + cos^2(z)) - 3
3. We know that cos(x) + cos(y) + cos(z) = 0, so:
2cos^2(x) + 2cos^2(y) + 2cos^2(z) = 0
4. Substituting into the equation from step 2:
cos(2x) + cos(2y) + cos(2z) = 2(0) - 3
cos(2x) + cos(2y) + cos(2z) = -3
Therefore, we have shown that cos(2x) + cos(2y) + cos(2z) = 0.
Now let's prove that sin(2x) + sin(2y) + sin(2z) = 0:
1. We can express sin(2x) as 2sin(x)cos(x), and similarly for y and z:
sin(2x) + sin(2y) + sin(2z) = 2sin(x)cos(x) + 2sin(y)cos(y) + 2sin(z)cos(z)
2. Since cos(x) + cos(y) + cos(z) = 0, we can rewrite the expression as:
sin(2x) + sin(2y) + sin(2z) = 2sin(x)(-cos(y) - cos(z)) + 2sin(y)cos(y) + 2sin(z)cos(z)
3. We know that sin(x) + sin(y) + sin(z) = 0, so:
2sin(x)cos(y) + 2sin(x)cos(z) + 2sin(y)cos(y) + 2sin(z)cos(z) = 0
4. Rearranging the terms:
2(sin(x)cos(y) + sin(z)cos(z)) + 2(sin(x)cos(z) + sin(y)cos(y)) = 0
5. Using the identity sin(a + b) = sin(a)cos(b) + sin(b)cos(a), we can rewrite the expression as:
2sin(x + y)cos(z) + 2sin(x + z)cos(y) = 0
6. We can express sin(x + y) and sin(x + z) using the sum-to-product identities:
2[sin(x)cos(y) + cos(x)sin(y)]cos(z) + 2[sin(x)cos(z) + cos(x)sin(z)]cos(y) = 0
7. Simplifying further:
2sin(x)cos(y)cos(z) + 2cos(x)sin(y)cos(z) + 2sin(x)cos(y)cos(z) + 2cos(x)sin(z)cos(y) = 0
8. Combining like terms:
2[cos(y)cos(z)(sin(x) + sin(z)) + cos(x)cos(z)(sin(y) + sin(x))] = 0
9. We know that sin(x) + sin(y) + sin(z) = 0, so:
2[cos(y)cos(z)(-sin(y)) + cos(x)cos(z)(-sin(x))] = 0
10. Simplifying:
2[-cos(y)cos(z)sin(y) - cos(x)cos(z)sin(x)] = 0
11. Rearranging the terms:
-2[cos(y)cos(z)sin(y) + cos(x)cos(z)sin(x)] = 0
12. Dividing both sides by -2:
cos(y)cos(z)sin(y) + cos(x)cos(z)sin(x) = 0
Therefore, we have shown that sin(2x) + sin(2y) + sin(2z) = 0.
Hence, both cos(2x) + cos(2y) + cos(2z) and sin(2x) + sin(2y) + sin(2z) equal 0, as required.
To prove that cos(2x) + cos(2y) + cos(2z) = sin(2x) + sin(2y) + sin(2z) = 0, we'll use the trigonometric identities for double angles.
Let's start by expressing sin(2x) and cos(2x) in terms of sin(x) and cos(x):
sin(2x) = 2sin(x)cos(x)
cos(2x) = cos^2(x) - sin^2(x)
Similarly, for y and z, we have:
sin(2y) = 2sin(y)cos(y)
cos(2y) = cos^2(y) - sin^2(y)
sin(2z) = 2sin(z)cos(z)
cos(2z) = cos^2(z) - sin^2(z)
Now, let's substitute these results into the given equation:
cos(2x) + cos(2y) + cos(2z) = sin(2x) + sin(2y) + sin(2z)
(cos^2(x) - sin^2(x)) + (cos^2(y) - sin^2(y)) + (cos^2(z) - sin^2(z)) = 2sin(x)cos(x) + 2sin(y)cos(y) + 2sin(z)cos(z)
Rearranging terms, we get:
2(cos^2(x) + cos^2(y) + cos^2(z)) - (sin^2(x) + sin^2(y) + sin^2(z)) = 2(sin(x)cos(x) + sin(y)cos(y) + sin(z)cos(z))
Since cos(x) + cos(y) + cos(z) = sin(x) + sin(y) + sin(z) = 0, we can substitute these values:
2(cos^2(x) + cos^2(y) + cos^2(z)) - (sin^2(x) + sin^2(y) + sin^2(z)) = 0
Now, recall the trigonometric identity:
sin^2(x) + cos^2(x) = 1
Substituting this identity into the equation, we have:
2(3) - 3 = 0
Simplifying, we get:
3 - 3 = 0
Therefore, we have proved that cos(2x) + cos(2y) + cos(2z) = sin(2x) + sin(2y) + sin(2z) = 0.