SOLUTION: Hi. I have a question about Pythagorean identities. I do not understrand how to write the espression as an integer. The problem is: {{{Csc^2}}}{{{(3alpha)}}}{{{""-""}}}{{{Cot^2}

Algebra ->  Trigonometry-basics -> SOLUTION: Hi. I have a question about Pythagorean identities. I do not understrand how to write the espression as an integer. The problem is: {{{Csc^2}}}{{{(3alpha)}}}{{{""-""}}}{{{Cot^2}      Log On


   



Question 276355: Hi. I have a question about Pythagorean identities. I do not understrand how to write the espression as an integer. The problem is:
Csc%5E2%283alpha%29%22%22-%22%22Cot%5E2%283alpha%29
Thanks in advance.

Found 2 solutions by Alan3354, Edwin McCravy:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
I do not understrand how to write the espression as an integer. The problem is:
csc^2 3alpha - cot^2 3alpha
----------------------------
I'm not sure what you mean by "write...as an integer."
This can be simplified:
csc^2 3alpha - cot^2 3alpha
= 1/sin^2 - cos^2/sin^2
= (1-cos^2)/sin^2
= sin^2/sin^2
= 1

Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!

First let's learn where the Pythagorean identities came from,
namely the Pythagorean theorem:



By the Pythagorean theorem:

adjacent%5E2+%2B+opposite%5E2+=+hypotenuse%5E2

Divide through by hypotenuse%5E2





Cos%5E2theta+%2B+Sin%5E2theta+=+1

That's the first Pythagorean identity:

-------------------------------------

adjacent%5E2+%2B+opposite%5E2+=+hypotenuse%5E2

Divide through by adjacent%5E2



1+%2B+%28%28opposite%29%2F%28hypotenuse%29%29%5E2+=+%28%28hypotenuse%29%2Fadjacent%29%29%5E2

1+%2B+Tan%5E2theta+=+Sec%5E2theta

That's the second Pythagorean identity:

-------------------------------------

adjacent%5E2+%2B+opposite%5E2+=+hypotenuse%5E2

Divide through by opposite%5E2





Cot%5E2theta+%2B+1+=+Csc%5E2theta

That's the third Pythagorean identity:

-------------------------------------

Be sure to memorize all three of them.  Usually we write them
this way, so they're easier to learn:

Sin%5E2theta+%2B+Cos%5E2theta+=+1
1+%2B+Tan%5E2theta+=+Sec%5E2theta
1+%2B+Cot%5E2theta+=+Csc%5E2theta

--------------------------------------

Those identities are true for ALL angles theta

So for

Csc%5E2%283alpha%29%22%22-%22%22Cot%5E2%283alpha%29

Since the third Pythagorean identity

1+%2B+Cot%5E2theta+=+Csc%5E2theta 

is true for ALL angles theta, let theta+=+3alpha, then

1+%2B+Cot%5E2%283alpha%29%22%22=%22%22Csc%5E2%283alpha%29

So we can replace Csc%5E2%283alpha%29 by 1+%2B+Cot%5E2%283alpha%29

So your problem:

Csc%5E2%283alpha%29%22%22-%22%22Cot%5E2%283alpha%29

becomes:

1+%2B+Cot%5E2%283alpha%29%22%22-%22%22Cot%5E2%283alpha%29

then the "cotangent squared"s cancel:

1+%2B+cross%28Cot%5E2%29%28cross%283alpha%29%29%22%22-%22%22cross%28Cot%5E2%29%28cross%283alpha%29%29

and that just leaves 1, which is an integer.

Edwin