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

Tangent Fractions Identities

Today's question comes from Yahoo Answers, and it includes trigonometric identities, fractions, and some hard work. Stay with me here, it will take some time.

Prove the identity:
(1 - tan x)/(1 + tan x) = cos 2x / (1 + sin 2x)

As you know, when working out an identity, you need to pick a side and work out until you reach the other side. Since the left side is more complex, I'll work from there to the simple right side. Here we go:

(1 - tan x)/(1 + tan x)

Using the tangent quotient (tan x = sin x / cos x), we can write the fraction as:
(1 - sin x / cos x)(1 + sin x / cos x)
(cos x / cos x - sin x / cos x)/(cos x / cos x + sin x / cos x)
((cos x - sin x)/cos x)/((cos x + sin x)/cos x)

Notice that both denominators are cos x, so they can be cancelled:
(cos x - sin x)/(cos x + sin x)

Multiply both parts of the fraction by (cos x + sin x)
(cos x - sin x)(cos x + sin x)/(cos x + sin x)²

Distribute the parentheses:
(cos² x - sin² x)/(cos²2 + 2sinxcosx + sin² x)

As you should know, cos² x - sin² x = cos 2x and cos² x + sin² x = 1, so:
cos 2x / (1 + 2sinxcosx)

As you should also know, 2sinxcosx = sin 2x, so:
cos 2x / (1 + sin 2x)

Q.E.D.

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