# Simplification of Trig FUnctions

• September 7th 2006, 10:52 AM
classicstrings
Simplification of Trig FUnctions

Simplify

1/(sin@ + cos@)
• September 7th 2006, 12:30 PM
rgep
There's a standard technique for dealing with combinations such as sin(X) + cos(X), or, more generally, A.sin(X) + B.cos(X) for constants A and B (you have A=B=1). The idea is to reverse engineer the simplification sin(X+Y) = sin(X).cos(Y) + cos(X).sin(Y). So your first shot would be that you want A = cos(Y) and B = sin(Y), but that doesn't always happen, since sin^2(Y)+cos^2(Y) = 1, but it might not happen that A^2+B^2 = 1. So take out a factor of C = sqrt(A^2+B^2), and then look for a Y such that C.sin(Y) = A and C.cos(Y) = B. Dividing, tan(Y) = A/B, so that Y is arctan(A/B).

In your case, A=B=1 so Y = arctan(1/1) = pi/4 and C = sqrt(1^2+1^2) = sqrt(2). Thus sin(X) + cos(X) = sqrt(2).sin(X+pi/4).
• September 7th 2006, 03:39 PM
JakeD
Quote:

Originally Posted by rgep
There's a standard technique for dealing with combinations such as sin(X) + cos(X), or, more generally, A.sin(X) + B.cos(X) for constants A and B (you have A=B=1). The idea is to reverse engineer the simplification sin(X+Y) = sin(X).cos(Y) + cos(X).sin(Y). So your first shot would be that you want A = cos(Y) and B = sin(Y), but that doesn't always happen, since sin^2(Y)+cos^2(Y) = 1, but it might not happen that A^2+B^2 = 1. So take out a factor of C = sqrt(A^2+B^2), and then look for a Y such that C.sin(Y) = A and C.cos(Y) = B. Dividing, tan(Y) = A/B, so that Y is arctan(A/B).

In your case, A=B=1 so Y = arctan(1/1) = pi/4 and C = sqrt(1^2+1^2) = sqrt(2). Thus sin(X) + cos(X) = sqrt(2).sin(X+pi/4).

rgep, nice explanation!

This kind of question is very commonly posed on math help sites. I'd say I see it once or twice a week. Everyone who answers it must derive the solution, either from the sum formula as you have done or other ways. I'm looking for a good link to refer people to that contains your derivation.

For example, this thread had this question in another guise. I found a link with the following explanation. But this explanation is not as complete as yours and the explanation is buried in the middle of a long page. Can you (or anyone else) give me more information on this technique (say names for it) to help me find a better link? Thank you.

http://www.mathhelpforum.com/math-he...ethod-gash.jpg

I found these formulas in Wikipedia, but there is no derivation.

Sine formula.

Cosine formula.

• September 7th 2006, 05:54 PM
ThePerfectHacker
Quote:

Originally Posted by JakeD

but there is no derivation.

I am not satisfied with CaptainBlank's link.
---
You have,
$A\sin x+B\cos x$
Write as, ( $A,B\not = 0$)
$\sqrt{A^2+B^2}\left( \frac{A}{\sqrt{A^2+B^2}}\sin x+\frac{B}{\sqrt{A^2+B^2}}\cos x\right)$
Let,
$\phi=\tan^{-1} B/A$
Then,
$\cos \phi=\frac{A}{\sqrt{A^2+B^2}}$
And,
$\sin \phi=\frac{B}{\sqrt{A^2+B^2}}$
Thus,
$\frac{1}{\sqrt{A^2+B^2}}(\cos \phi\sin x+\sin x \cos \phi)$
• September 7th 2006, 11:26 PM
CaptainBlack
Quote:

Originally Posted by ThePerfectHacker
I am not satisfied with CaptainBlank's link.
---

1. It's not my link it's JakeD's, I was only pointing out that it had information
needed to answer the original question in the thread it was posted in;
because every one was ignoring it and in consequence going off at half-cock.

2. I posted the image file because the needed information from the link was
a long way down the page and might easily be missed.

3. It was posted to answer the question: "what is the auxiliary angle method?".

RonL