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

[Matt Jacques]

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

Trig Identities have come back to haunt me!!!

[Bob3141592]

Originally posted by Matt Jacques
What simpler indentity is equal to sin(x) - cos(x) ?

Trig Identities have come back to haunt me!!!

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

[Matt Jacques]

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

[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 ;)

[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?

[NateTG]

Originally posted by 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?

Just nitpicking -- shouldn't that be

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

[Matt Jacques]

Thanks everyone!

[jcm15]

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.

[himanshu121]

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

[Hurkyl]

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

[fffbone]

If I read your equations correctly, it should be ln100+3a-(5/2)c