What simpler indentity is equal to sin(x) - cos(x)

1. Dec 28, 2003

Matt Jacques

What simpler indentity is equal to sin(x) - cos(x) ?

Trig Identities have come back to haunt me!!!

2. Dec 28, 2003

Bob3141592

There isn't one. Plot out the two curves and look at their differences and you'll see it's not simpler than the basic sine curve.

There are lots of websites to check out trig identities if you need references. A quick google search will show more than you need, but most contain the same information. Here's three is you want to check them out:

http://www.math2.org/math/trig/identities.htm

http://aleph0.clarku.edu/~djoyce/java/trig/identities.html

http://www.mathwizz.com/algebra/help/help32.htm

3. Dec 29, 2003

Matt Jacques

Then how do I solve for theta in a physics problem that contains that identity?

4. Dec 29, 2003

Muzza

Approximation? Square both sides of the equation (to get sin^2(x) + cos^2(x) - 2sin(x)cos(x) = 1 - sin(2x))? You're being much too vague ;)

5. Dec 29, 2003

mathman

cos(x+y)=cos(x)cos(y)-sin(x)sin(y). Lety=45o. Net result
sin(x)-cos(x)=-sqrt(2)cos(x+y).

Is that simple enough?

6. Dec 29, 2003

NateTG

Just nitpicking -- shouldn't that be

$$\sin(x)-\cos(x)=\sqrt{2}\cos(x+45)$$

7. Dec 29, 2003

Matt Jacques

Thanks everyone!

8. Dec 30, 2003

jcm15

need a help with log problem

I'm a bit confused with this problem can you help me to workout and explain it to me on the way. thanks.

log(e)x=a log(e0y=c express log(e){(100x^3y^-1/2)/(y^2)} in terms of a and c.

my interpretation is that you separate the function then workout by using loga(mn)=logam+logan law. thanks for your guys.

9. Dec 30, 2003

himanshu121

Is it
$$\log(\frac{100x^3y^{\frac{-1}{2}}}{y^2})$$

Last edited: Dec 30, 2003
10. Dec 30, 2003

Hurkyl

Staff Emeritus
jcm; you should start a new post when you want to ask an unrelated question.

11. Jan 1, 2004