Let's derive the equation sin2x + cos2x=1 from the pythagorean formula.
A right triangle with angle x will have a leg that is adjacent to angle x and it will have a leg that is opposite to angle x. There will also be a hypotenuse.
From trigonometry, a right triangle with a given angle x, can defined as follows:
adjacent/hypotenuse=cosx
opposite/hypotenuse=sinx
Following is the pythagorean identity for our right triangle:
adjacent2 + opposite2=hypotenuse2
If we divide the above equation by hypotenuse2, we have:
(adjacent2 + opposite2=hypotenuse2)/hypotenuse2 or
adjacent2/hypotenuse2 + opposite2/hypotenuse2=1
Substitute in sinx and cosx into the above equation:
cos2x+sin2x=1
Can you derive the other two Pythagorean trigonometric identities?
cot2x + 1=csc2x
tan2x + 1=sec2x
Hint:...
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I was working with a student today, and as we worked through the section in his book dealing with Trigonometric Identities and Pythagorean Identities, we stumbled across a problem that gave us a bit of trouble. The solution is not so complicated, but it sure had us stumped earlier.
The problem was presented as such:
Factor and simplify the following using Trigonometric and Pythagorean Identities:
sec3(x) - sec2(x) - sec(x) + 1
We tried a couple of different approaches, such as factoring sec(x) from each term:
sec(x) * [ sec2(x) - sec(x) - 1 + 1/sec(x) ]
and factoring sec2(x) from each term:
sec2(x) * [ sec(x) - 1 - 1/sec(x) + 1/sec2(x) ]
We followed these approaches through a few steps, but nothing we were attempting led to the solution. After doing some reading online, I found that the solution required a simple...
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