Cosine Simplification

Yahoo Answers is definitely the place everyone goes to find an answer for a math question. Otherwise, there is no explanation for the kinds of questions you can find there. Today's question is about simplification of a trigonometric expression:

Simplify: (cos 3x)(1 - cos 2x + cos 4x - cos 6x)

To solve that, we will need some major identities: the sum of cosines, the difference of cosines, the sum of sines, and the product of sines and cosines. With these formulas, this question can be easily solved.

Let's start with the formulas:
cos α + cos β = 2cos((α + β)/2)cos((α - β)/2)
cos α - cos β = -2sin((α + β)/2)sin((α - β)/2)
sin α + sin β = 2sin((α + β)/2)cos((α - β)/2)
sinαcosβ = 1/2(sin(α + β)sin(α - β))

Now, to avoid complications, let's aim for the average. We can turn 1 into cos 0, so we have the cosines of 0 and 6x and the cosines of 2x and 4x. In both formulas they are divided by two, so they should probably cancel each other.

cos 0 - cos 6x = -2sin((6x + 0) / 2)sin(6x - 0)/2 = -2sin² 3x
cos 4x - cos 2x = -2sin((4x + 2x)/2)sin((4x - 2x)/2) = -2sin 3x * sinx

In conclusion, the problem has just become:
(cos 3x)(-2sin² 3x - 2sin 3x * sin x)

Distribute:
-2sin 3x * sin 3x * cos 3x - 2sin 3x * cos 3x * sinx

Transform into a sum of a cosine and a sine (I picked α and β to be 3x, because there will be a sine of zero, which is zero):
-2 * 1/2 * (sin(3x + 3x) + sin(3x - 3x)) * sin 3x - 2 * 1/2 * (sin(3x + 3x) + sin(3x - 3x)) * sin x
-sin 6x * sin 3x - sin 6x * sin x

We are getting really close now, just hold on:
-(sin 6x)(sin 3x + sin x)
-(sin 6x)(2sin((3x + x)/2)cos((3x - x)/2)
-2sin 6x * sin 2x * cos x

And that's it.

Nadav

nadavs

Elliptic Triangles

As promised, today I have a question involving ellipses. In my opinion, it's a great question.

C is a point on the ellipse x²/16 + y²/9 = 1. A and B are the foci of the ellipse. Prove that angle ∠ACB cannot be 90°.

To do that, we first need to find the foci of the ellipse. As you remember, in an ellipse, a² = b² + c². From the equation of the ellipse, a² = 16 and b² = 9. This means:
c² = a² - b² = 16 - 9 = 7
c = √7

Now that we know c, we know one side of triangle ABC. The side is AB and it equals 2√7. Now we only have to find the other sides and show that ∠ACB cannot equal 90°.

Let's call one side, AC, x. By the definition of an ellipse, the sum of the distances from a point on the ellipse to the foci is a constant, 2a. In this case, 2a = 2√16 = 8. This means the third side, BC, is 8 - x.

We have three sides of the triangle and we want to find an angle. To do that, we can use the law of cosines, which involves all three sides and an angle. Since we want ∠ACB, we will use the side opposing it, AB.

The law of cosines states that:
a² = b² + c² - 2bc * cos∠A
(a, b, c, and ∠A are not related to the ellipse)

Let's plug our known values in there:
(2√7)² = x² + (8 - x)² - 2x(8 - x)cos∠ACB

We want to know what happens when ∠ACB = 90°, which means cos∠ACB = 0. Let's do all the squaring and eliminate the last part, because it's zero:
28 = x² + 64 - 16x + x²
2x² - 16x + 36 = 0
x² - 8x + 18 = 0

When trying to solve this through the quadratic formula, the expression inside the square root equals:
8² - 4*18 = 64 - 72 = -8

There is no real number for x which makes the law of cosines correct for 90°, so angle ∠ACB cannot equal 90° in this ellipse.

There are even cooler things on Super Math Tips, go check them out.

Hope you liked it,
Nadav

nadavs

Trigonometry on the Table

Today I have a very simple question from Yahoo Answers. Hopefully tomorrow I'll be able to find something a little more interesting:

Prove that tan² x + 1 = sec² x

Every trigonometry student must remember this identity, but here is the reason why the trig students must know this identity:

As you all should know, sin² x + cos² x = 1 (if you don't know that, blame your math teacher). Now all we have to do is use some more trigonometric identities and rules of exponents, and we're all set:

tan x = sin x/cos x
sec x = 1/cos x
(a/b)ⁿ = aⁿ/bⁿ

Now we take the basic trigonometric identity and divide it by cos² x:
sin² x / cos² x + cos² x / cos² x = 1 / cos² x
(sin x / cos x)² + 1 = (1/cos x)²
tan² x + 1 = sec² x

Simple, nice, and easy.

Bring in more questions!
Nadav

nadavs