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bellatrix · Dec 11, 2010, 04:12 AM

Proving trigonometric identities

galactus · Dec 11, 2010, 12:57 PM

$\frac{cot(x)}{1-tan(x)}+\frac{tan(x)}{1-cot(x)}=csc(x)sec(x)+1$

Transform everything into sin and cos. This is a good idea when solving most any trig identity.

$\frac{\frac{sin(x)}{cos(x)}}{1-\frac{cos(x)}{sin(x)}}+\frac{\frac{cos(x)}{sin(x)} }{1-\frac{sin(x)}{cos(x)}}$

$\frac{\frac{sin(x)}{cos(x)}}{\frac{sin(x)-cos(x)}{sin(x)}}$ $+\frac{\frac{cos(x)}{sin(x)}}{\frac{cos(x)-sin(x)}{cos(x)}}$

$\frac{sin^{2}(x)}{cos(x)(sin(x)cos(x))}+\frac{cos^ {2}(x)}{sin(x)(cos(x)-sin(x))}$

$\frac{sin^{3}(x)-cos^{3}(x)}{sin(x)cos(x)(sin(x)-cos(x))}$

Note the difference of two cubes in the numerator. Factoring and cancelling leads to:

$\frac{sin^{2}(x)+sin(x)cos(x)+cos^{2}(x)}{sin(x)co s(x)}$

$\frac{1+sin(x)cos(x)}{sin(x)cos(x)}$

$\frac{1}{sin(x)cos(x)}+\frac{sin(x)cos(x)}{sin(x)c os(x)}$

$sec(x)csc(x)+1$

bellatrix · Dec 12, 2010, 05:59 PM

Thanks for the answer, but how come the minus sign is missing? Also, how do you solve for the least common denominator in these instances, other than just multiplying it? Thanks!
[IMG]C:\Documents and Settings\052\Desktop\trig.ident.jpeg[/IMG]

Unknown008 · Dec 13, 2010, 02:57 AM

If you mean the minus sign in the 4th line, this is surely a typing mistake, the line should indeed be

$\frac{\sin^2(x)}{\cos(x)(\sin(x) - \cos(x))} + \frac{\cos^2(x)}{\sin(x)(\sin(x) - \cos(x))}$

The LCM in this case will be $\cos(x)\sin(x)(\sin(x) - \cos(x))$ . You don't multiply again the (sin(x) - cos(x)) because it is present in both denominators, if you see what I mean.

bellatrix · Dec 13, 2010, 06:44 PM

but on the left side, the denominator is cos(x)(sin(x) - cos(x)) and on the right side its sin(x)(cos(x) - sin(x))

is (sin(x) - cos(x)) and (cos(x) - sin(x)) considered the same?

galactus · Dec 14, 2010, 03:10 AM

No, one is the negative of the other. Thus, the sign change.
-(sin(x)-cos(x))=cos(x)-sin(x)