Search

Blogs Blogs

Pythagorean Identity Blogs

Newest Most Active

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:... read more

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... read more

Pythagorean Identity Blogs RSS feed