Question 428624: (1 + cos theta)(1 - cos theta) is that like a difference of squares? So how is it simplified?
Found 3 solutions by stanbon, Theo, solver91311:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! (1 + cos theta)(1 - cos theta) is that like a difference of squares? So how is it simplified?
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= 1-cos^2(theta)
= sin^2(theta)
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Cheers,
Stan H.
Answer by Theo(13342)  (Show Source):
You can put this solution on YOUR website! (1 + cos) * (1 - cos) should result in:
1^2 - cos + cos - cos^2 which should result in 1^2 - cos^2 which should result in 1 - cos^2.
from the pythagorean theorem, we get sin^2 + cos^2 = 1 (theta, or the angle in question, is implied).
if we subtract cos^2 from both sides of this equation, we get 1 - cos^2 = sin^2
so what resulted in 1 - cos^2 should be equivalent to sin^2 which yields to the conclusion that:
(1 + cos) * (1 - cos) = sin^2
let's see if this holds water.
assume our angle is 30 degrees.
(1 + cos(30)) * (1 - cos(30)) will be equal to .25
sin (30) = .5
sin^2 (30) = .5^2 = .25
looks like it works !!!!!
let's try another angle at random.
let's try an angle of 53 degrees
(1 - cos(53)) * (1 + cos(53) = .637818678
sin(53) = .79863551
sin^2(53) = .79863551^2 = .637818678
it works again !!!!!
the answer to your question i that, yes, algebraically, this kind of looks like the difference of square, i.e. (x-a) * (x+a) = x^2 - a^2.
the operation is similar.
the translation, however, does depend on knowledge of the pythagorean formula of sin^2 + cos^2 = 1 in order to get the final answer.
Answer by solver91311(24713)  (Show Source):
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