Thread: Proving Trig Equations
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Old October 11th, 2007, 08:51 PM
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Quote:
Originally Posted by Blue Griffin View Post
s tan squared x.

2) {sin^3x + cos^3x}/{1 - 2cos^2x} = {secx - sinx}/{tanx - 1}

In words, sin cubed x minus cos cubed x all divided by one minus two cos squared x equals sec x minus sin x all divided by tanx minus one.
Hmmmm....

How about
\frac{sin^3(x) + cos^3(x)}{1 - 2cos^2(x)} = \frac{(sin(x) + cos(x))(sin^2(x) - sin(x)cos(x) + cos^2(x))}{1 - 2cos^2(x)}

\frac{sin^3(x) + cos^3(x)}{1 - 2cos^2(x)} = \frac{(sin(x) + cos(x))(1 - sin(x)cos(x))}{1 - 2cos^2(x)}

Divide both the numerator and denominator by cos^2(x). In the numerator divide each factor by cos(x):
= \frac{(tan(x) + 1)(sec(x) - sin(x))}{sec^2(x) - 2}

Now, sec^2(x) - 1 = tan^2(x), so
= \frac{(tan(x) + 1)(sec(x) - sin(x))}{tan^2(x) - 1}

= \frac{(tan(x) + 1)(sec(x) - sin(x))}{(tan(x) + 1)(tan(x) - 1)}

= \frac{sec(x) - sin(x)}{tan(x) - 1}

-Dan

(Thank you. I enjoyed doing that one! )
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