[SOLVED] I'm confused on how to simplify trig expressions..

**Megaman** · February 17th 2009, 04:59 PM

Well, I know I must be missing something 'cause I just don't understand how to simplify these trig expressions.

a) sin^3 + cos^3 / sin + cos

The book says the answer is.. 1 - sin * cos or something of the sort, don't really remember. Another one that's confusing me is

b) 2 - tan / 2csc - sec

The book says the answer to this one is sin.

In general, I really don't quite understand how to simplify these. I'm just confused.. can someone explain to me how to get this done? Any input would be appreciated 'cause this is starting to give me a headache. ={

**skeeter** · February 17th 2009, 05:19 PM

for the first one, try the factoring pattern for the sum of two cubes ...

$a^3 + b^3 = (a + b)(a^2 - ab + b^2)$

$\frac{2 - \tan{x}}{2\csc{x} - \sec{x}}$

change every trig term to its basic form using sine and/or cosine ...

$\frac{2 - \frac{\sin{x}}{\cos{x}}}{\frac{2}{\sin{x}} - \frac{1}{\cos{x}}}$

common denominator in the numerator and denominator ...

$\frac{\frac{2\cos{x} -\sin{x}}{\cos{x}}}{\frac{2\cos{x} -\sin{x}}{\sin{x}\cos{x}}}$

$\frac{2\cos{x} -\sin{x}}{\cos{x}} \cdot \frac{\sin{x}\cos{x}}{2\cos{x} -\sin{x}} = \sin{x}$

**Megaman** · February 17th 2009, 05:22 PM

Aah, Skeeter thanks so much. I was doing the second one right just didn't know what to do with all the fractions. Gracias.

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