logb1 = 0
because b0 = 1

logee2x = 2x
because e2x = e2x

logbbx = x
because bx = bx

elogee2x = e2x
because logee2x = logee2x

logex = logx / loge

the inverse of y=logbx is by=x

3 points that always exist in a log
( 1/b , -1 )
( 1 , 0 )
( b , 1 )

When b > 1:
Logarithm When 0 < b < 1

For today there is another question from Yahoo Answers, also about trigonometric identities. This time it’s not about simple ones like before, but a more complex one, like the ones I sometimes answer on super math tips. Here it is:

Prove:
(cos 4x + cos 2x) / (sin 4x + sin 2x) = cot 3x

At first glance this seems irritating and impossible, but with some creative thinking it’s very possible.

Now, what do 2, 3, and 4 have in common? Of course, 3 is the average, and we’re going to use that property heavily.

Since we want to have cot 3x at the end, let’s change all sines and cosines to 3x + x or 3x - x and hope they cancel:
(cos (3x + x) + cos (3x - x)) / (sin (3x + x) + sin(3x - x)) = cot 3x.

Well, they don’t cancel so fast, so we’ll have to expand all expressions (remember what’s sin (A + B)?) and then hope that we can cancel out some terms. Let’s start (I put the arguments in parentheses so you can see them better):
(cos (3x + x) + cos (3x - x)) / (sin (3x + x) + sin(3x - x)) = cot 3x
Let’s start from the numerator:
cos(3x)cos(x) - sin(3x)sin(x) + cos(3x)cos(x) + sin(3x)sin(x)
2cos(3x)cos(x)

Now, the denominator:
sin(3x)cos(x) + cos(3x)sin(x) + sin(3x)cos(x) - cos(3x)sin(x)
2sin(3x)cos(x)

Now divide those two terms and see what happens:
2cos(3x)cos(x) / 2sin(3x)cos(x)
The 2 and cos(x) cancel:
cos(3x) / sin(3x) = 1/(sin(3x) / cos(3x)) = 1/tan(3x) = cot (3x)

Job complete.

Besides solving these questions, here is a challenge for you: try to solve the previous question (with the towns) using calculus. I haven’t solved it myself yet (except for the geometry part, which is easy).

The first one to solve (on comments or email) will be written in the hall of fame on the blog’s list (which I’ll create) with a link of his choice and the right to post here any question.

Go on to solving!
Pranav

How the fundamental and Pythagorean identities were derived, along with a few identity problems.

Trigonometric identities.

Identities
Pythagorean Identites:

start out with the unit circle
x2 + y2 = 12

x = cosθ and y = sinθ

cos2θ + sin2θ = 1 (1 of 3 pythagorean identities)

divide by cos2θ
cos2θ/cos2θ + sin2/cos2θ = 1/cos2θ
1 + sin2θ/cos2θ = 1/cos2θ

tan2θ + 1 = sec2θ (2 of 3 pythagorean identities)

divide unit circle equation by sin2θ
cos2θ/sin2θ + sin2θ/sin2θ = 1/sin2θ
cos2θ/sin2θ + 1 = 1/sin2θ
cot2θ + 1 = csc2θ (3 of 3 pythagorean identities)

Fundamental Identities
tanθ = sinθ/cosθ

cotθ = cosθ/sinθ

cotθ = 1/tanθ

cscθ = 1/sinθ

secθ = 1/cosθ

cos2θ + sin2θ = 1

tan2θ + 1 = sec2θ

cot2θ + 1 = csc2θ

Using Identities to generate new equations

start with the pythagorean identity:
cos2θ + sin2θ = 1
subtact sin2θ from unit circle equation and take square root
cosθ = ±√(1 - sin2θ)

subtact cos2θ from unit circle equation and take square root
sinθ = ±√(1- cos2θ)
divide by cos2θ from unit circle equation and take square root
cos2θ + sin2θ = 1
cos2θ/cos2θ + sin2θ/cos2θ = 1/cos2θ
1 + tan2θ = sec2θ
±√(1 + tan2θ) = secθ
secθ = ±√(1 + tan2θ)
divide by sin2θ from unit circle equation and take square root
cos2θ + sin2θ = 1
cos2θ/sin2θ + sin2θ/sin2θ = 1/sin2θ
cot2θ + 1 = csc2θ
±√(cot2θ + 1) = cscθ
cscθ = ±√(cot2θ + 1)