# Trig Identities

• Jul 28th 2008, 02:35 PM
ahling
Trig Identities
Prove that
1-sin^2X/1+cotX - cos^2X/1+tanX = sinXcosX

• Jul 28th 2008, 05:49 PM
TKHunny
Quote:

Originally Posted by ahling
Prove that
1-sin^2X/1+cotX - cos^2X/1+tanX = sinXcosX

Notice:

1 + x / x - 1 = 1 + (x / x) - 1 = 1 + (1) - 1 = 2 - 1 = 1

However,

(1 + x) / (x - 1) is a very different animal.
• Jul 28th 2008, 06:16 PM
ahling
Yes, I understand, but .. in the original equation, there are no parantheses. I typed the question exactly as it is shown on the paper given to me.
• Jul 28th 2008, 07:23 PM
Soroban
Hello, ahling!

Quote:

Prove that: .$\displaystyle 1 - \frac{\sin^2\!x}{1 + \cot x}- \frac{\cos^2\!x}{1 + \tan x} \:=\:\sin x\cos x$
We have: .$\displaystyle 1 - \frac{\sin^2\!x}{1 + \dfrac{\cos x}{\sin x}} - \frac{\cos^2\!x}{1 + \dfrac{\sin x}{\cos x}}$

Multiply the first fraction by $\displaystyle \frac{\sin x}{\sin x}$ , the second by $\displaystyle \frac{\cos x}{\cos x}$

. . $\displaystyle 1 - \frac{\sin^3\!x}{\sin x + \cos x} - \frac{\cos^3\!x}{\cos x + \sin x} \;\;=\;\;1 - \frac{sin^3\!x + \cos^3\!x}{\sin x + \cos x}$

. . $\displaystyle = \;\;1 - \frac{(\sin x + \cos x)(\sin^2\!x - \sin x\cos x + \cos^2\!x)}{\sin x + \cos x}$

. . $\displaystyle = \;\;1 - (\underbrace{\sin^2\!x + \cos^2\!x}_{\text{This is 1}} - \sin x\cos x) \;\;=\;\;1 - (1 - \sin x\cos x)$

. . $\displaystyle = \;\sin x\cos x$ . . . ta-DAA!

And LateX is working !!

• Jul 29th 2008, 06:24 AM
TKHunny
Quote:

Originally Posted by ahling
in the original equation, there are no parantheses. I typed the question exactly as it is shown on the paper given to me.

I have a little trouble believing that, but let's go with it...

1-sin^2X/1+cotX - cos^2X/1+tanX = sinXcosX

The "/1" are meaningless in this form, so...

1-sin^2X+cotX - cos^2X+tanX = sinXcosX

1-sin^2X = cos^2X <== Identity to Substitute

cos^2X+cotX - cos^2X+tanX = sinXcosX

cotX+tanX = sinXcosX

Since cotX = (cosX)/(sinX) and tanX = (sinX)/(cosX)
And Since cos^2X + sin^2X = 1

1/(sinXcosX) = sinXcosX

I think you would have a hard time selling that as an identity.

Note: I would do this on an exam. It is technically correct, but may have points taken off for being a smart aleck. (Evilgrin) I would also rewrite it as it probably is intended and work it that way. This may avoid the smart aleck points. Don't put up with sloppy notation. Certainly don't write any yourself. Everyone needs reminders to be clear - particularly if they want to be understood generally.

LaTeX Check: $\displaystyle \cos^{2}(x) + \sin^{2}(x) = 1$