Question

1. what is the relationship of sine and cosine where one angle is Φ another is 90 and the third one is 90-Φ

2. prove that ㏑(x^n)/㏑(x)=n

13 points and brainliest to the best answer

1. MichaelMet
Step-by-step explanation:
When Ptolemy produced his table of chords of functions, discussed in the section on computing trigonometric functions, he needed ways of computing the trig functions for sums and differences of angles. His basic trig function was the chord of an angle while we use sines and cosines. When we convert his formulas to sines and cosines, we get the following four important identities.
You can use these identities without knowing why they’re true. If you’re interested in why, then keep reading, otherwise, skip on to the next page.
Ptolemy’s theorem
In order to prove his sum and difference forumlas, Ptolemy first proved what we now call Ptolemy’s theorem.
Ptolemy’s theorem: For a cyclic quadrilateral (that is, a quadrilateral inscribed in a circle), the product of the diagonals equals the sum of the products of the opposite sides.
If the cyclic quadrilateral is ABCD, then Ptolemy’s theorem is the equation
AC BD = AB CD + AD BC.
We won’t prove Ptolemy’s theorem here. The proof depends on properties of similar triangles and on the Pythagorean theorem. Instead, we’ll use Ptolemy’s theorem to derive the sum and difference formulas.
The sum formula for sines
We’ll follow Ptolemy’s proof, but modify it slightly to work with modern sines. Let O to be the center of a circle of radius 1, and take one of the lines, AC, to be a diameter of the circle. We’ll interpret each of the lines AC, BD, AB, CD, AD, and BC in terms of sines and cosines of angles. We already know AC = 2.
Let α be ∠BAC. Recall that the sine of an angle is half the chord of twice the angle. Euclid’s proposition III.20 says that the angle at the center of a circle twice the angle at the circumference, therefore ∠BOC equals 2α. Thus, the sine of α is half the chord of ∠BOC, so it equals BC/2, and so BC = 2 sin α.
Let β be ∠CAD. That’s half of ∠COD, so sin β equals CD/2, and CD = 2 sin β.
Then α + β is ∠BAD, so BD = 2 sin (α + β).
We still have to interpret AB and AD. The line segment AB is twice the sine of ∠ACB. Triangle ABC is a right triangle by Thale’s theorem (Euclid’s proposition III.31: an angle in a semicircle is right). Therefore sin ∠ACB cos α. Hence, AB = 2 cos α. Likewise, AD = 2 cos β.
Now we can write Ptolemy’s theorem
AC BD = AB CD + AD BC.
in terms of sines and cosines. It says
(2) (2 sin (α + β)) = (2 cos α) (2 sin β) + (2 cos β) (2 sin α).
After dividing by 4, we get the addition formula for sines.
sin (α + β) = cos α sin β + cos β sin α.
Other sum and difference formulas
You can directly show that the sum formula for cosines and the two difference formulas hold by taking one of the line segments in Ptolemy’s theorem to be the diameter of a circle, interpreting the other ones as chords of central angles, that is, twice the sines of angles on the circumference, and using Thale’s theorem to convert between sines and cosines.
For example, take AD to be a diameter, α to be ∠BAD, and β to be ∠CAD, then you can directly show the difference formula for sines.
Alternatively, you can show the other three formulas starting with the sum formula for sines that we’ve already proved. If you replace β by −β, you’ll get the difference formula. If you replace certain angles by their complements, then you can derive the sum and difference formulas for cosines.