How would I go about this one.?

Simplify

1/(sin@ + cos@)

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- Sep 7th 2006, 09:52 AMclassicstringsSimplification of Trig FUnctions
How would I go about this one.?

Simplify

1/(sin@ + cos@) - Sep 7th 2006, 11:30 AMrgep
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). - Sep 7th 2006, 02:39 PMJakeDQuote:

Originally Posted by**rgep**

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.

http://upload.wikimedia.org/math/7/8...fcc5b26f77.png

http://upload.wikimedia.org/math/7/2...ccdb2845bc.png

Cosine formula.

http://upload.wikimedia.org/math/3/a...0453ec349c.png - Sep 7th 2006, 04:54 PMThePerfectHackerQuote:

Originally Posted by**JakeD**

**Blank**'s link.

---

You have,

$\displaystyle A\sin x+B\cos x$

Write as, ($\displaystyle A,B\not = 0$)

$\displaystyle \sqrt{A^2+B^2}\left( \frac{A}{\sqrt{A^2+B^2}}\sin x+\frac{B}{\sqrt{A^2+B^2}}\cos x\right)$

Let,

$\displaystyle \phi=\tan^{-1} B/A$

Then,

$\displaystyle \cos \phi=\frac{A}{\sqrt{A^2+B^2}}$

And,

$\displaystyle \sin \phi=\frac{B}{\sqrt{A^2+B^2}}$

Thus,

$\displaystyle \frac{1}{\sqrt{A^2+B^2}}(\cos \phi\sin x+\sin x \cos \phi)$ - Sep 7th 2006, 10:26 PMCaptainBlackQuote:

Originally Posted by**ThePerfectHacker**

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